8.5 Additional Information and Full Hypothesis Test Examples
SLO 4
Demonstrate an understanding of, and ability to use, basic ideas of statistical processes, including hypothesis tests and confidence interval estimation.
This section runs full hypothesis tests end to end — means, proportions, both tails — so you practice the complete sequence from hypotheses through p-value to conclusion. The worked examples cover the cases that look different on the surface but follow the same steps underneath.
SLO 5
Identify appropriate statistical techniques and use technology-based statistical analysis to describe, interpret, and communicate results.
Every example here uses technology (Z-Test, T-Test, 1-PropZTest) and every conclusion is written in plain language about the population. That full arc — identify the right test, run it, communicate the result — is exactly what the outcome asks for.
Learning Objectives
By the end of this section, you will be able to:
- read a stated level of significance and say what the preset \(\alpha\) commits you to before any data is collected;
- decide from the alternative hypothesis alone whether a test is left-tailed, right-tailed, or two-tailed;
- draw the p-value picture that matches each of the three tail types;
- carry a full hypothesis test from the claim through to the conclusion, for a single population mean and for a single population proportion;
- state the Type I and Type II errors for a test in the wording of the problem it came from.
The first four sections of this chapter handed you the pieces one at a time: the pair of hypotheses, the two kinds of error, the distribution the test runs on, and the rare-event reasoning that turns a p-value into a decision. This section puts the pieces together. It starts with the small pieces of bookkeeping that the earlier sections skipped over — where the level of significance comes from, and how to tell from the alternative hypothesis which tail of the curve the p-value lives in — and then works a long run of complete tests, start to finish, on both means and proportions.
8.5.1 Reading the Level of Significance
The level of significance, written \(\alpha\), is the probability of rejecting the null hypothesis when it is in fact true. It is preset — the person setting up the hypothesis test chooses its value before collecting the sample data.
Setting \(\alpha\) before the data arrives is the same discipline as calling your shot in pool. Announce the pocket, then take the shot. Announcing the pocket after the ball drops makes every shot a winner and tells nobody anything.
Definition 8.5.1 - Two samples clear the same alpha, one by a hair and one with room to spare: the decision is identical, the confidence is not.
In a hypothesis test problem, you may see words such as "the level of significance is 1%." That 1% is the preset \(\alpha\); write it as \(\alpha = 0.01\) and use it. If no level of significance is given at all, the common standard is \(\alpha = 0.05\).
The order matters more than it looks. The value of \(\alpha\) is chosen first, then the data is collected, then the p-value is computed and compared against it. Choosing \(\alpha\) after seeing the p-value would let you pick whichever threshold gives the answer you were hoping for, which is not a test of anything. Fixing \(\alpha\) in advance is what makes the eventual decision mean something.
Once the data is in and the p-value is computed, the comparison of \(\alpha\) against the p-value gives the decision, but the two numbers carry more information than the yes-or-no answer they produce. A data analyst should have more confidence in the decision to reject the null hypothesis with a p-value of 0.001 than with a p-value of 0.04, even though both fall below the same \(\alpha = 0.05\) and both end in "reject." The same works in the other direction: a p-value of 0.4 is much further from \(\alpha = 0.05\) than a p-value of 0.056, so the decision not to reject is on much firmer ground in the first case than in the second. A p-value that lands very close to \(\alpha\) on either side is telling you that this sample did not settle the question cleanly, and the honest response is usually to collect more data rather than to lean hard on the label the rule produced. The rule gives you a decision; the distance between the two numbers tells you how much weight that decision will bear. This is where judgment enters a procedure that otherwise looks purely mechanical.
A study reports that it tested a claim "at the 1% level of significance." A second study, on a different question, reports no level of significance at all. Write \(\alpha\) for each study, and say at what point in each study that value had to be chosen.
Solution — reading a stated significance level
Step 1 — Translate the stated percentage. "The 1% level of significance" means \(\alpha = 0.01\).
Step 2 — Apply the default to the second study. When a problem states no level of significance, the standard value to use is \(\alpha = 0.05\).
Step 3 — Place the choice in time. In both studies, \(\alpha\) had to be fixed before the sample data was collected. It is a preset threshold, not something read off the results.
Answer: \(\alpha = 0.01\) for the first study and \(\alpha = 0.05\) for the second, and in both cases the value was set before any data was collected.
8.5.2 Left, Right, and Two-Tailed Tests
A hypothesis test is called left-tailed when the p-value is the area in the left tail of the distribution, right-tailed when it is the area in the right tail, and two-tailed when the p-value is split evenly between the two tails. Which one you have is determined entirely by the alternative hypothesis, \(H_a\):
- \(H_a\) with \(<\) gives a left-tailed test;
- \(H_a\) with \(>\) gives a right-tailed test;
- \(H_a\) with \(\neq\) gives a two-tailed test.
Read \(H_a\)'s symbol as an arrowhead. The \(<\) opens to the left, so the p-value is the left tail. The \(>\) opens to the right, so it is the right tail. The \(\neq\) refuses to pick a side, so the area splits across both.
Definition 8.5.2 - The symbol in the alternative hypothesis picks the tail: less-than shades the left, greater-than the right, not-equal shades both.
The alternative hypothesis is the key to conducting the appropriate test, and it always points the way, because \(H_a\) never has a symbol that contains an equal sign. The equality always lives in \(H_0\); the alternative gets one of the three strict symbols, and each of those three symbols names one of the three pictures.
That is the whole rule, and it is worth getting fluent with it before the arithmetic starts, because the tail decision comes first and everything downstream inherits it. The same test statistic produces three different p-values depending on which tail you read, so a test set up as right-tailed and then read as two-tailed will report a p-value twice as large as it should be, and a two-tailed test read as one-tailed reports one half as large. Neither mistake shows up as an arithmetic error — the numbers are all correct, they are just answering a question nobody asked. The three examples below run the same drill three times, once for each symbol: read \(H_a\), name the tail, and draw the picture. The Try It Now after each one hands you a fresh pair of hypotheses and a p-value and asks for the same two things.
Assume the p-value is 0.0935. What type of test is this? Describe the picture of the p-value.
Solution — naming the tail from a less-than symbol
Step 1 — Read the symbol in \(H_a\). The alternative uses \(<\), which opens to the left.
Step 2 — Name the test. This is a left-tailed test.
Step 3 — Place the area. The p-value of 0.0935 is the area in the left tail of the curve, to the left of the observed sample mean. The curve is centred at the null value \(\mu = 10\), and the shaded region runs from the sample mean leftward off the end of the tail.
Answer: A left-tailed test, with the 0.0935 shaded in the left tail.
This is a test of a single population mean. What type of test is it, and what does the picture of the p-value look like?
Solution — a mean claimed to be 5
Step 1 — Read \(H_a\). The alternative uses \(<\), so \(H_a\) tells you the test is left-tailed.
Step 2 — Draw the picture. The p-value is the area in the left tail of the sampling distribution, which is centred at the null value 5.
Figure 8.5.1 - A left-tailed test. The curve is centred at the claimed value mu = 5 and the p-value is the whole area in the left tail.
Answer: Left-tailed, with the p-value shaded in the left tail.
Assume the p-value is 0.1243. What type of test is this? Describe the picture of the p-value.
Solution — naming the tail from a greater-than symbol
Step 1 — Read the symbol in \(H_a\). The alternative uses \(>\), which opens to the right.
Step 2 — Name the test. This is a right-tailed test.
Step 3 — Place the area. The p-value of 0.1243 is the area in the right tail, to the right of the observed sample mean, on a curve centred at the null boundary value \(\mu = 1\).
Answer: A right-tailed test, with the 0.1243 shaded in the right tail.
This is a test of a single population proportion. What type of test is it, and what does the picture of the p-value look like?
Solution — a proportion claimed to be at most 0.2
Step 1 — Read \(H_a\). The alternative uses \(>\), so \(H_a\) tells you the test is right-tailed.
Step 2 — Draw the picture. The p-value is the area in the right tail of the sampling distribution of \(P'\), centred at the null boundary value 0.2.
Figure 8.5.2 - A right-tailed test. The curve is centred at the claimed proportion p = 0.2 and the p-value is the area in the right tail.
Answer: Right-tailed, with the p-value shaded in the right tail.
Assume the p-value is 0.2564. What type of test is this? Describe the picture of the p-value.
Solution — naming the tail from a not-equal symbol
Step 1 — Read the symbol in \(H_a\). The alternative uses \(\neq\), which picks no side.
Step 2 — Name the test. This is a two-tailed test.
Step 3 — Split the area. The p-value of 0.2564 is split evenly between the two tails of a curve centred at 0.5, so each tail holds
$$ \frac{0.2564}{2} = 0.1282 $$Answer: A two-tailed test, with 0.1282 shaded in each tail.
This is a test of a single population mean. What type of test is it, and what does the picture of the p-value look like?
Solution — a mean claimed to be 50
Step 1 — Read \(H_a\). The alternative uses \(\neq\), so \(H_a\) tells you the test is two-tailed.
Step 2 — Draw the picture. The p-value is split evenly between the two tails of a curve centred at the null value 50, with half the p-value in each tail.
Figure 8.5.3 - A two-tailed test. The curve is centred at mu = 50 and the p-value is split evenly, half in each tail.
Answer: Two-tailed, with half the p-value shaded in each tail.
8.5.3 Full Hypothesis Test Examples
Everything from here on is a complete test. Each one runs the same five moves in the same order, and it is worth naming them once so you can watch for them: set up the hypotheses, determine the distribution the test runs on, calculate the p-value, compare \(\alpha\) against the p-value, and state a decision and a conclusion in the wording of the original problem. The examples alternate between means and proportions, and between known and unknown population standard deviations, because those two questions are what decide the third move. When \(\sigma\) is given, the test runs on a normal distribution; when it is not and you only have sample data, it runs on a Student's t-distribution with \(n - 1\) degrees of freedom. When the problem is stated in percentages rather than averages, the parameter is a proportion and the distribution is the sampling distribution of \(P'\). Read the first paragraph of each problem looking for those signals before writing anything down, because every line after the setup depends on getting them right, and the arithmetic is unforgiving of a test that starts on the wrong curve.
The mean throwing distance of a football for Marisol, a high school quarterback, is 40 yards, with a standard deviation of two yards. The team coach tells Marisol to adjust her grip to get more distance. The coach records the distances for 20 throws. For the 20 throws, her mean distance was 45 yards. The coach thought the different grip helped Marisol throw farther than 40 yards. Conduct a hypothesis test using a preset \(\alpha = 0.05\). Assume the throw distances for footballs are normal.
Determine what type of test this is, set up the hypothesis test, find the p-value, describe the graph, and state your conclusion.
Solution — a quarterback's throwing distance
Step 1 — Set up the hypotheses. The claim is about a mean distance, so this is a test of a single population mean. The coach wants to show the grip helped, meaning farther than 40 yards:
$$ H_{0}: \mu = 40, \quad H_{a}: \mu > 40 $$The \(>\) makes this a right-tailed test.
Step 2 — Determine the distribution. The population standard deviation is given as \(\sigma = 2\), so \(\overline{X}\) is normal:
$$ \overline{X} \sim N\left(40, \frac{2}{\sqrt{20}}\right) $$Step 3 — Calculate the p-value. The standard error is \(\frac{2}{\sqrt{20}} = 0.4472\), so
$$ z = \frac{45 - 40}{0.4472} = 11.18 $$p-value \(= P(\overline{x} > 45) \approx 0\).
Step 4 — Describe the graph. A normal curve centred at 40 with the far right tail shaded beyond 45. The shaded region is so small it is not visible at any usable scale.
Step 5 — Compare and decide. \(\alpha = 0.05\) and the p-value is essentially 0, so \(\alpha > \) p-value. Reject \(H_{0}\).
Answer: At the 5% significance level, there is sufficient evidence that Marisol's mean throwing distance with the new grip is more than 40 yards.
Avery, as an eight-year old, established a mean time of 16.43 seconds for swimming the 25-yard freestyle, with a standard deviation of 0.8 seconds. Their dad, Glenn, thought that Avery could swim the 25-yard freestyle faster using goggles. Glenn bought Avery a new pair of expensive goggles and timed them for 15 25-yard freestyle swims. For the 15 swims, Avery's mean time was 16 seconds. Glenn thought that the goggles helped Avery to swim faster than the 16.43 seconds. Conduct a hypothesis test using a preset \(\alpha = 0.05\). Assume that the swim times for the 25-yard freestyle are normal.
Solution — goggles and a 25-yard freestyle time
Step 1 — Set up the hypothesis test. Since the problem is about a mean, this is a test of a single population mean.
$$ H_{0}: \mu = 16.43, \quad H_{a}: \mu < 16.43 $$For Avery to swim faster, their time will be less than 16.43 seconds. The \(<\) tells you this is left-tailed.
Step 2 — Determine the distribution needed. The random variable is \(\overline{X}\), the mean time to swim the 25-yard freestyle. The population standard deviation is known, \(\sigma = 0.8\), so \(\overline{X}\) is normal:
$$ \overline{X} \sim N\left(\mu, \frac{\sigma_{X}}{\sqrt{n}}\right) \quad \text{therefore} \quad \overline{X} \sim N\left(16.43, \frac{0.8}{\sqrt{15}}\right) $$Note where each number comes from: \(\mu = 16.43\) comes from \(H_{0}\), not from the data. Then \(\sigma = 0.8\) and \(n = 15\).
Step 3 — Calculate the p-value using the normal distribution for a mean. The sample mean in the problem is given as 16, so
$$ p\text{-value} = P(\overline{x} < 16) = 0.0187 $$This 0.0187 is called the actual level of significance. It is the area to the left of the sample mean.
On a TI-83/83+/84/84+: press STAT, arrow over to TESTS, press 1:Z-Test, arrow over to Stats and press ENTER. Enter 16.43 for \(\mu_{0}\), .8 for \(\sigma\), 16 for the sample mean, and 15 for \(n\). Arrow down to \(\mu\): and arrow over to \(<\mu_{0}\). Press ENTER, arrow down to Calculate and press ENTER. The calculator returns the p-value (p = 0.0187) and the test statistic \(z = -2.08\). Running the same instructions but choosing Draw instead of Calculate produces the shaded graph. Make sure no other equations are highlighted in Y = and the plots are turned off.
Step 4 — Draw the graph.
Figure 8.5.4 - The p-value for the swim-time test: the curve is centred at mu = 16.43 and the shaded left tail below a sample mean of 16 seconds has area 0.0187.
Step 5 — Interpret the p-value. If \(H_{0}\) is true, there is a 0.0187 probability (1.87%) that Avery's mean time to swim the 25-yard freestyle is 16 seconds or less. Because a 1.87% chance is small, a mean time of 16 seconds or less is unlikely to have happened randomly. It is a rare event.
Step 6 — Compare \(\alpha\) and the p-value, then decide.
$$ \alpha = 0.05, \quad p\text{-value} = 0.0187, \quad \alpha > p\text{-value} $$Since \(\alpha > \) p-value, reject \(H_{0}\). You reject the null hypothesis that the mean time to swim the 25-yard freestyle is at least 16.43 seconds.
Answer: At the 5% significance level, there is sufficient evidence that Avery's mean time to swim the 25-yard freestyle is less than 16.43 seconds. Based on the sample data, we conclude that they swim faster using the new goggles.
The two errors for this problem. The Type I error is to conclude that Avery swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, they actually swim it, on average, in at least 16.43 seconds. That is rejecting the null hypothesis when the null hypothesis is true. The Type II error is that there is not evidence to conclude that Avery swims the 25-yard freestyle, on average, in less than 16.43 seconds when, in fact, they actually do swim it, on average, in less than 16.43 seconds. That is failing to reject the null hypothesis when the null hypothesis is false.
Try it in rāSHio
The swim-time test hands you the population standard deviation, which is exactly the case one panel handles end to end. In rāSHio choose Stats → Z Stats and enter this example's own summary values — 16 for the sample mean, 0.8 for the population standard deviation, 15 for the sample size, and 16.43 for the claimed mean — then set the alternative to less-than. It returns \(z = -2.08\) and the p-value 0.0187 in one step: the same two numbers the keystrokes above produce, without the lookup.
Figure 8.5.5 — The rāSHio Z Stats panel: Stats → Z Stats. The walkthrough enters a sample mean, a population standard deviation, a sample size and a claimed mean, then reads the z-statistic and the p-value straight off the panel. It runs its own demonstration numbers, not this example's swim times; the steps are the ones you would repeat on your own data.
The traditional comparison: critical value against test statistic. The p-value method above is the modern way to run the comparison, but it is not the only way, and the older method is still worth knowing because it is what the tables in the back of a statistics book are built for. The traditional approach compares the critical value — the z-score that cuts off an area of \(\alpha\) — against the test statistic, the z-score computed from the data. For Example 8.5.4 the test statistic comes from the Central Limit Theorem formula
$$ z = \frac{\overline{x} - \mu_{X}}{\left(\frac{\sigma_{X}}{\sqrt{n}}\right)} $$with \(\overline{x} = 16\), \(\mu_{X} = 16.43\) from the null hypothesis, \(\sigma_{X} = 0.8\), and \(n = 15\), which gives \(z = -2.08\). The critical value for \(\alpha = 0.05\) comes out of the normal table: the z-score with an area of 0.05 to its left is midway between \(-1.65\) and \(-1.64\), because 0.05 is midway between 0.0505 and 0.0495, so the critical value is \(-1.645\). Since \(-1.645 > -2.08\), the test statistic sits further out in the tail than the critical value, which is exactly the same fact as \(\alpha > \) p-value, and the decision is the same: reject \(H_{0}\). Today the two probabilities are compared directly far more often than the two z-scores, and for this problem the p-value of 0.0187 is considerably smaller than \(\alpha = 0.05\), so you can be confident about the decision to reject.
Figure 8.5.6 - Alpha and the p-value as two nested left-tail areas: the larger is alpha = 0.05 cut at the critical value, the smaller is the p-value 0.0187 cut at the test statistic.
A company records the mean time of employees working in a day. The mean comes out to be 475 minutes, with a standard deviation of 45 minutes. A manager recorded times of 20 employees. The times of working were (frequencies are in parentheses) 460(3); 465(2); 470(3); 475(1); 480(6); 485(3); 490(2).
Conduct a hypothesis test using a 2.5% level of significance to determine if the mean time is more than 475.
Solution — employee working minutes
Step 1 — Set up the hypotheses. This is a test of a single population mean, and "more than 475" is the claim being checked:
$$ H_{0}: \mu = 475, \quad H_{a}: \mu > 475 $$This is a right-tailed test with \(\alpha = 0.025\).
Step 2 — Find the sample mean. Weighting each time by its frequency:
$$ \overline{x} = \frac{460(3) + 465(2) + 470(3) + 475(1) + 480(6) + 485(3) + 490(2)}{20} = \frac{9{,}510}{20} = 475.5 $$Step 3 — Determine the distribution. The population standard deviation is given as \(\sigma = 45\), so the test runs on a normal distribution:
$$ \overline{X} \sim N\left(475, \frac{45}{\sqrt{20}}\right) $$Step 4 — Calculate the p-value. The standard error is \(\frac{45}{\sqrt{20}} = 10.06\), so
$$ z = \frac{475.5 - 475}{10.06} = 0.05 $$p-value \(= P(\overline{x} > 475.5) = 0.4801\).
Step 5 — Compare and decide. \(\alpha = 0.025\) and the p-value is 0.4801, so \(\alpha < \) p-value. Do not reject \(H_{0}\).
Answer: At the 2.5% level of significance, there is not sufficient evidence to conclude that the mean working time is more than 475 minutes. A sample mean of 475.5 is almost exactly what the null hypothesis predicts.
A college football coach records the mean weight that the players can bench press as 275 pounds, with a standard deviation of 55 pounds. Three of the players thought that the mean weight was more than that amount. They asked 30 of their teammates for their estimated maximum lift on the bench press exercise. The data ranged from 205 pounds to 385 pounds. The actual different weights were (frequencies are in parentheses) 205(3); 215(3); 225(1); 241(2); 252(2); 265(2); 275(2); 313(2); 316(5); 338(2); 341(1); 345(2); 368(2); 385(1).
Conduct a hypothesis test using a 2.5% level of significance to determine if the bench press mean is more than 275 pounds.
Solution — a claimed mean bench press of 275 pounds
Step 1 — Set up the hypothesis test. Since the problem is about a mean weight, this is a test of a single population mean.
$$ H_{0}: \mu = 275 $$ $$ H_{a}: \mu > 275 $$This is a right-tailed test.
Step 2 — Calculate the distribution needed. The random variable is \(\overline{X}\), the mean weight in pounds lifted by the football players. The distribution is normal because \(\sigma\) is known:
$$ \overline{X} \sim N\left(275, \frac{55}{\sqrt{30}}\right) $$From the data, \(\overline{x} = 286.2\) pounds. The population standard deviation is \(\sigma = 55\) pounds — always use \(\sigma\) if you know it. We assume \(\mu = 275\) pounds unless the data shows us otherwise.
Step 3 — Calculate the p-value using the normal distribution for a mean, using the sample mean as input:
$$ p\text{-value} = P(\overline{x} > 286.2) = 0.1323 $$On a TI-83/83+/84/84+: put the data and frequencies into lists. Press STAT, arrow over to TESTS, press 1:Z-Test, arrow over to Data and press ENTER. Enter 275 for \(\mu_{0}\), 55 for \(\sigma\), the name of the list holding the data, and the name of the list holding the frequencies. Arrow down to \(\mu\): and arrow over to \(>\mu_{0}\), then Calculate. The calculator returns p = 0.1331 and the test statistic \(z = 1.112\), along with the sample mean and sample standard deviation. The small difference from the 0.1323 above is only rounding: the hand calculation used the sample mean rounded to one decimal place, and the calculator used the raw data.
Step 4 — Interpret the p-value. If \(H_{0}\) is true, then there is a 0.1323 probability (13.23%) that the football players can lift a mean weight of 286.2 pounds or more. Because a 13.23% chance is large enough, a mean weight lift of 286.2 pounds or more is not a rare event.
Figure 8.5.7 - The p-value for the bench-press test: the curve is centred at mu = 275 and the shaded right tail beyond a sample mean of 286.2 pounds has area 0.1323.
Step 5 — Compare \(\alpha\) and the p-value, then decide.
$$ \alpha = 0.025, \quad p\text{-value} = 0.1323 $$Since \(\alpha < \) p-value, do not reject \(H_{0}\).
Answer: At the 2.5% level of significance, from the sample data, there is not sufficient evidence to conclude that the true mean weight lifted is more than 275 pounds.
It is believed that a stock price for a particular company will grow at a rate of $5 per week with a standard deviation of $1. An investor believes the stock will not grow as quickly. The changes in stock price are recorded for ten weeks and are as follows: $4, $3, $2, $3, $1, $7, $2, $1, $1, $2. Perform a hypothesis test using a 5% level of significance. State the null and alternative hypotheses, find the p-value, state your conclusion, and identify the Type I and Type II errors.
Solution — a weekly stock-growth claim
Step 1 — Set up the hypotheses. The claim is about a mean weekly growth, and the investor believes it is slower than claimed:
$$ H_{0}: \mu = 5, \quad H_{a}: \mu < 5 $$This is a left-tailed test with \(\alpha = 0.05\).
Step 2 — Find the sample mean.
$$ \overline{x} = \frac{4 + 3 + 2 + 3 + 1 + 7 + 2 + 1 + 1 + 2}{10} = \frac{26}{10} = 2.6 $$Step 3 — Determine the distribution. The population standard deviation is given as \(\sigma = 1\), so the test runs on a normal distribution:
$$ \overline{X} \sim N\left(5, \frac{1}{\sqrt{10}}\right) $$Step 4 — Calculate the p-value. The standard error is \(\frac{1}{\sqrt{10}} = 0.3162\), so
$$ z = \frac{2.6 - 5}{0.3162} = -7.59 $$p-value \(= P(\overline{x} < 2.6) \approx 0\).
Step 5 — Compare and decide. \(\alpha = 0.05\) and the p-value is essentially 0, so \(\alpha > \) p-value. Reject \(H_{0}\).
Answer: At the 5% level of significance, there is sufficient evidence that the mean weekly growth is less than $5.
The two errors. The Type I error is to conclude that the mean weekly growth is less than $5 when in fact it really is $5. The Type II error is to fail to conclude that the mean weekly growth is less than $5 when in fact it really is less than $5.
Statistics students believe that the mean score on the first statistics test is 65. Their instructor, Dr. Imani Okafor, thinks the mean score is higher than 65. She samples ten statistics students and obtains the scores 65; 65; 70; 67; 66; 63; 63; 68; 72; 71. She performs a hypothesis test using a 5% level of significance. The data are assumed to be from a normal distribution.
Solution — a mean test score claimed to be 65
Step 1 — Set up the hypothesis test. A 5% level of significance means \(\alpha = 0.05\). This is a test of a single population mean.
$$ H_{0}: \mu = 65, \quad H_{a}: \mu > 65 $$Since Dr. Okafor thinks the average score is higher, use a \(>\). The \(>\) means the test is right-tailed.
Step 2 — Determine the distribution needed. The random variable is \(\overline{X}\), the average score on the first statistics test. Read the problem carefully and you will notice that there is no population standard deviation given — you are only given \(n = 10\) sample data values. Notice also that the data come from a normal distribution. That combination means the distribution for the test is a Student's t. Use \(t_{df}\), so the distribution for the test is \(t_{9}\), where \(n = 10\) and \(df = 10 - 1 = 9\).
Step 3 — Calculate the p-value using the Student's t-distribution. The sample mean and sample standard deviation are calculated from the data as 67 and 3.1972, so
$$ p\text{-value} = P(\overline{x} > 67) = 0.0396 $$On a TI-83/83+/84/84+: put the data into a list. Press STAT, arrow over to TESTS, press 2:T-Test, arrow over to Data and press ENTER. Enter 65 for \(\mu_{0}\), the name of the list holding the data, and 1 for Freq:. Arrow down to \(\mu\): and arrow over to \(>\mu_{0}\), then Calculate. The calculator returns p = 0.0396 and the test statistic \(t = 1.9781\), along with the sample mean and sample standard deviation.
Step 4 — Interpret the p-value. If the null hypothesis is true, then there is a 0.0396 probability (3.96%) that the sample mean is 67 or more.
Figure 8.5.8 - The p-value for the test-score problem, on a Student's t-distribution with 9 degrees of freedom: the shaded right tail beyond x-bar = 67 has area 0.0396.
Step 5 — Compare \(\alpha\) and the p-value, then decide. Since \(\alpha = 0.05\) and the p-value \(= 0.0396\), we have \(\alpha > \) p-value. Reject \(H_{0}\). This means you reject \(\mu = 65\); in other words, you believe the average test score is greater than 65.
Answer: At a 5% level of significance, the sample data show sufficient evidence that the mean test score is greater than 65, just as Dr. Okafor thinks.
Try it in rāSHio
This example gives you no population standard deviation, only ten scores, and that absence is the whole signal to change panels. In rāSHio choose Stats → T Stats and fill in the Summary Data fields — 67 for the sample mean, 3.1972 for the sample standard deviation, 10 for the sample size — then set the claimed mean to 65 and the alternative to greater-than. It reports \(t = 1.978\) and a p-value of 0.0396, and it takes the degrees of freedom from the sample size, so \(df = 9\) is never a number you have to remember to set. If you have the ten raw scores rather than the summary, paste them in with File → Delimited List and read the mean and standard deviation off Stats → Summary Statistics first.
Figure 8.5.9 — The rāSHio T Stats panel: Stats → T Stats. The walkthrough types a sample mean, standard deviation and sample size into the panel's Summary Data fields and runs a one-sample t-test, reading the t-statistic and p-value straight off it without ever setting the degrees of freedom by hand. Its demonstration numbers are not this example's ten test scores.
Mr. Whitfield believes that 85% of the students in his class will want to go on a field trip to the local zoo. He performs a hypothesis test to determine if the percentage is the same or different from 85%. He samples 50 students and 39 reply that they would want to go to the zoo. For the hypothesis test, use a 1% level of significance.
Determine what type of test this is, set up the hypothesis test, find the p-value, describe the graph, and state your conclusion.
Solution — a field-trip percentage claimed to be 85%
Step 1 — Set up the hypotheses. The information is given in percentages, so this is a test of a single population proportion. "The same or different from" gives no direction:
$$ H_{0}: p = 0.85, \quad H_{a}: p \neq 0.85 $$This is a two-tailed test with \(\alpha = 0.01\).
Step 2 — Determine the distribution.
$$ P' \sim N\left(0.85, \sqrt{\frac{(0.85)(0.15)}{50}}\right) $$Step 3 — Calculate the p-value. The sample proportion is \(p' = \frac{39}{50} = 0.78\), and the standard error is \(\sqrt{\frac{0.1275}{50}} = 0.0505\), so
$$ z = \frac{0.78 - 0.85}{0.0505} = -1.39 $$Because the test is two-tailed, double the one-tail area:
$$ p\text{-value} = 2 \times P(p' < 0.78) = 2(0.0829) = 0.1658 $$Step 4 — Describe the graph. A normal curve centred at 0.85 with 0.0829 shaded in each tail, below 0.78 on the left and above 0.92 on the right.
Step 5 — Compare and decide. \(\alpha = 0.01\) and the p-value is 0.1658, so \(\alpha < \) p-value. Do not reject \(H_{0}\).
Answer: At the 1% level of significance, there is not sufficient evidence to conclude that the percentage of students wanting to go to the zoo is different from 85%.
Mateo believes that 50% of first-time brides in the United States are younger than their grooms. He performs a hypothesis test to determine if the percentage is the same or different from 50%. Mateo samples 100 first-time brides and 53 reply that they are younger than their grooms. For the hypothesis test, he uses a 1% level of significance.
Solution — a 50% claim about brides and grooms
Step 1 — Set up the hypothesis test. The 1% level of significance means \(\alpha = 0.01\). This is a test of a single population proportion.
$$ H_{0}: p = 0.50, \quad H_{a}: p \neq 0.50 $$The words "is the same or different from" tell you this is a two-tailed test.
Step 2 — Calculate the distribution needed. The random variable is \(P'\), the percent of first-time brides who are younger than their grooms. The problem contains no mention of a mean; the information is given in terms of percentages, so use the distribution for \(P'\), the estimated proportion:
$$ P' \sim N\left(p, \sqrt{\frac{p \cdot q}{n}}\right) \quad \text{therefore} \quad P' \sim N\left(0.5, \sqrt{\frac{0.5 \cdot 0.5}{100}}\right) $$where \(p = 0.50\), \(q = 1 - p = 0.50\), and \(n = 100\).
Step 3 — Calculate the p-value using the normal distribution for proportions. Here \(x = 53\), so \(p' = \frac{x}{n} = \frac{53}{100} = 0.53\), and
$$ p\text{-value} = P(p' < 0.47 \text{ or } p' > 0.53) = 0.5485 $$On a TI-83/83+/84/84+: press STAT, arrow over to TESTS, press 5:1-PropZTest. Enter .5 for \(p_{0}\), 53 for \(x\), and 100 for \(n\). Arrow down to Prop and arrow to the \(\neq p_{0}\) option, press ENTER, then Calculate. The calculator returns p = 0.5485 and the test statistic \(z = 0.6\).
Step 4 — Interpret the p-value. If the null hypothesis is true, there is a 0.5485 probability (54.85%) that the sample proportion \(p'\) is 0.53 or more, or 0.47 or less.
Figure 8.5.10 - The two-tailed p-value for the bride-age test: the curve is centred at p = 0.50 and each shaded tail holds 0.27425, for a total of 0.5485.
Note where the left boundary comes from: \(\mu = p = 0.50\) comes from \(H_{0}\), the null hypothesis, and \(p' = 0.53\). Since the curve is symmetrical and the test is two-tailed, the \(p'\) for the left tail is \(0.50 - 0.03 = 0.47\), where 0.03 is the difference between 0.53 and 0.50.
Step 5 — Compare \(\alpha\) and the p-value, then decide. Since \(\alpha = 0.01\) and the p-value \(= 0.5485\), we have \(\alpha < \) p-value. You cannot reject \(H_{0}\).
Answer: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of first-time brides who are younger than their grooms is different from 50%.
The two errors. The Type I error is to conclude that the proportion of first-time brides who are younger than their grooms is different from 50% when, in fact, the proportion is actually 50%. That is rejecting the null hypothesis when the null hypothesis is true. The Type II error is that there is not enough evidence to conclude that the proportion of first-time brides who are younger than their grooms differs from 50% when, in fact, the proportion does differ from 50%. That is failing to reject the null hypothesis when the null hypothesis is false.
Try it in rāSHio
A proportion needs a different panel from a mean, even though the logic is identical. In rāSHio choose Stats → Prop Stats, enter 53 for the number of successes, 100 for the sample size, and 0.50 for the claimed proportion, then set the alternative to not-equal. It returns \(z = 0.6\) and the two-tailed p-value 0.5485 — and because you chose not-equal, the doubling of the tail area is already done for you, which is the step that is easiest to lose by hand.
Figure 8.5.11 — The rāSHio Prop Stats panel: Stats → Prop Stats. The walkthrough enters a success count, a sample size and a claimed proportion, then reads the z-statistic and the p-value. Its demonstration numbers are not this example's 53 out of 100.
Marketers believe that 92% of adults in the United States own a cell phone. A cell phone manufacturer believes that number is actually lower. 200 American adults are surveyed, of which 174 report having cell phones. Use a 5% level of significance. State the null and alternative hypotheses, find the p-value, state your conclusion, and identify the Type I and Type II errors.
Solution — a cell-phone ownership rate of 92%
Step 1 — Set up the hypotheses. The manufacturer believes the true rate is lower than claimed:
$$ H_{0}: p = 0.92, \quad H_{a}: p < 0.92 $$This is a left-tailed test with \(\alpha = 0.05\).
Step 2 — Determine the distribution.
$$ P' \sim N\left(0.92, \sqrt{\frac{(0.92)(0.08)}{200}}\right) $$Step 3 — Calculate the p-value. The sample proportion is \(p' = \frac{174}{200} = 0.87\), and the standard error is \(\sqrt{\frac{0.0736}{200}} = 0.0192\), so
$$ z = \frac{0.87 - 0.92}{0.0192} = -2.61 $$p-value \(= P(p' < 0.87) = 0.0046\).
Step 4 — Compare and decide. \(\alpha = 0.05\) and the p-value is 0.0046, so \(\alpha > \) p-value. Reject \(H_{0}\).
Answer: At the 5% level of significance, there is sufficient evidence that fewer than 92% of American adults own a cell phone.
The two errors. The Type I error is to conclude that fewer than 92% of adults own a cell phone when in fact 92% do. The Type II error is to fail to conclude that fewer than 92% own a cell phone when in fact fewer than 92% do.
Suppose a consumer group suspects that the proportion of households that have three cell phones is 30%. A cell phone company has reason to believe that the proportion is not 30%. Before they start a big advertising campaign, they conduct a hypothesis test. Their marketing people survey 150 households with the result that 43 of the households have three cell phones.
a) The value that helps determine the p-value is \(p'\). Calculate \(p'\).
b) What is a success for this problem?
c) What is the level of significance?
d) Describe the graph for this problem, and calculate the p-value.
e) Make a decision. ____ (Reject / Do not reject) \(H_{0}\) because ____.
Solution — a 30% claim about three-phone households
Step 1 — Set up the hypothesis test.
$$ H_{0}: p = 0.30, \quad H_{a}: p \neq 0.30 $$Step 2 — Determine the distribution needed. The random variable is \(P'\), the proportion of households that have three cell phones. The distribution for the hypothesis test is
$$ P' \sim N\left(0.30, \sqrt{\frac{(0.30)(0.70)}{150}}\right) $$a) \(p' = \frac{x}{n}\), where \(x\) is the number of successes and \(n\) is the total number in the sample, so \(p' = \frac{43}{150}\).
b) A success is having three cell phones in a household.
c) The level of significance is the preset \(\alpha\). Since \(\alpha\) is not given, assume \(\alpha = 0.05\).
d) The graph is a normal curve centred at 0.30, with the area beyond \(p' = \frac{43}{150}\) shaded on one side and the mirror-image area shaded on the other, because the \(\neq\) makes this two-tailed. With \(x = 43\) and \(n = 150\),
$$ p\text{-value} = 0.7216 $$e) Assuming that \(\alpha = 0.05\), we have \(\alpha < \) p-value. The decision is do not reject \(H_{0}\), because there is not sufficient evidence to conclude that the proportion of households that have three cell phones is not 30%.
A car soap gets rid of 30% of stains on the car. After adding a new compound to the soap, the soap is used on a car and found to wash 20 stains out of the 50 stains on the car. With the level of significance being 0.01, find out if adding the new compound to the soap is beneficial.
Solution — testing an improved car soap
Step 1 — Set up the hypotheses. "Beneficial" means the new compound removes a larger share of stains than the old 30%:
$$ H_{0}: p = 0.30, \quad H_{a}: p > 0.30 $$This is a right-tailed test with \(\alpha = 0.01\).
Step 2 — Determine the distribution.
$$ P' \sim N\left(0.30, \sqrt{\frac{(0.30)(0.70)}{50}}\right) $$Step 3 — Calculate the p-value. The sample proportion is \(p' = \frac{20}{50} = 0.40\), and the standard error is \(\sqrt{\frac{0.21}{50}} = 0.0648\), so
$$ z = \frac{0.40 - 0.30}{0.0648} = 1.54 $$p-value \(= P(p' > 0.40) = 0.0614\).
Step 4 — Compare and decide. \(\alpha = 0.01\) and the p-value is 0.0614, so \(\alpha < \) p-value. Do not reject \(H_{0}\).
Answer: At the 1% level of significance, there is not sufficient evidence to conclude that adding the new compound is beneficial. The sample rate of 40% is higher than 30%, but with only 50 stains the result is not far enough out in the tail to rule out ordinary sampling variation.
The next example is a poem written by a statistics student named Nicole Hart, and the solution to the problem follows the poem. Notice that the hypothesis test is for a single population proportion, so the null and alternative hypotheses use the parameter \(p\) and the distribution for the test is normal. The estimated proportion \(p'\) is the proportion of fleas killed to the total fleas found on Fido, which is sample information. The problem gives a preconceived \(\alpha = 0.01\) for comparison, and asks for a 95% confidence interval as well.
My dog has so many fleas,
They do not come off with ease.
As for shampoo, I have tried many types
Even one called Bubble Hype,
Which only killed 25% of the fleas,
Unfortunately I was not pleased.
I've used all kinds of soap,
Until I had given up hope
Until one day I saw
An ad that put me in awe.
A shampoo used for dogs
Called GOOD ENOUGH to Clean a Hog
Guaranteed to kill more fleas.
I gave Fido a bath
And after doing the math
His number of fleas
Started dropping by 3's!
Before his shampoo
I counted 42.
At the end of his bath,
I redid the math
And the new shampoo had killed 17 fleas.
So now I was pleased.
Now it is time for you to have some fun
With the level of significance being .01,
You must help me figure out
Use the new shampoo or go without?
Solution — a flea shampoo that promises better than 25%
Step 1 — Set up the hypothesis test. The old shampoo killed 25%, and the new one is guaranteed to kill more:
$$ H_{0}: p \leq 0.25, \quad H_{a}: p > 0.25 $$Step 2 — Determine the distribution needed. In words, clearly state what the random variable represents: \(P'\) is the proportion of fleas that are killed by the new shampoo. The distribution to use for the test is normal:
$$ N\left(0.25, \sqrt{\frac{(0.25)(1 - 0.25)}{42}}\right) $$Test statistic: \(z = 2.3163\).
Step 3 — Calculate the p-value using the normal distribution for proportions.
$$ p\text{-value} = 0.0103 $$Step 4 — Explain what the p-value means for this problem. If the null hypothesis is true, meaning the proportion really is 0.25, then there is a 0.0103 probability that the sample proportion is \(0.4048\left(\frac{17}{42}\right)\) or more.
Step 5 — Sketch the situation. Label and scale the horizontal axis and shade the region corresponding to the p-value.
Figure 8.5.12 - The p-value for the flea shampoo test: the curve is centred at p = 0.25 and the shaded right tail beyond a sample proportion of 17/42 has area 0.0103.
Step 6 — Compare \(\alpha\) and the p-value, then decide. Indicate the correct decision, the reason for it, and write an appropriate conclusion in complete sentences.
| alpha | decision | reason for decision |
|---|---|---|
| 0.01 | Do not reject \(H_{0}\) | \(\alpha < \) p-value |
Answer: At the 1% level of significance, the sample data do not show sufficient evidence that the percentage of fleas killed by the new shampoo is more than 25%.
Step 7 — Construct a 95% confidence interval for the true proportion, with a sketch labelling the point estimate and the lower and upper bounds.
Figure 8.5.13 - The 95% confidence interval for the flea shampoo proportion: a central band holding 0.95 of the area, running from 0.26 to 0.55.
Confidence interval: \((0.26, 0.55)\). We are 95% confident that the true population proportion \(p\) of fleas that are killed by the new shampoo is between 26% and 55%.
Try it in rāSHio
The same panel also builds the interval this example finishes with. In rāSHio choose Stats → Prop Stats and switch it to the confidence-interval option rather than the hypothesis test, then enter 17 for the successes, 42 for the sample size, and 0.95 for the confidence level. It returns the \((0.26, 0.55)\) the example reports. Running the panel both ways on the same data is the quickest way to see how a test that fails to reject can sit beside an interval lying entirely above 0.25.
Figure 8.5.14 — The rāSHio Prop Stats panel in confidence-interval mode: Stats → Prop Stats. The walkthrough enters a success count, a sample size and a confidence level, and reads the interval off the panel. Its demonstration numbers are not this example's 17 out of 42.
This result is not very definitive, because the p-value of 0.0103 is very close to \(\alpha = 0.01\). In practice you would run more tests — give the dog another bath once the fleas have had a chance to return — rather than lean on a decision this narrow.
The flea problem is worth one more look before moving on, because it shows the two halves of this chapter and the last one doing different jobs on the same data. The hypothesis test ended in "do not reject," which sounds like the shampoo did nothing. The confidence interval says the true kill rate is somewhere between 26% and 55%, and every value in that interval is above the 25% the old shampoo managed. Those two statements are not in conflict. The test asks a yes-or-no question at a very demanding threshold and answers that the evidence did not clear it. The interval answers a different question — what values of \(p\) are plausible given this sample — and its answer is a range that happens to sit almost entirely above 0.25. With 42 fleas the sample is small enough that the interval is wide, which is the same fact that kept the p-value from clearing \(\alpha\). Bigger sample, narrower interval, smaller p-value. The remaining examples in this section run the same machinery on samples large enough that this tension does not arise.
The boiling point of a specific liquid is measured for 15 samples, and the boiling points are obtained as follows:
205; 206; 206; 202; 199; 194; 197; 198; 198; 201; 201; 202; 207; 211; 205
Is there convincing evidence that the average boiling point is greater than 200? Use a significance level of 0.1. Assume the population is normal.
Solution — boiling points above 200 degrees
Step 1 — State the question. We need to determine whether, at a 0.1 significance level, the average boiling point is greater than 200:
$$ H_{0}: \mu \leq 200, \quad H_{a}: \mu > 200 $$Step 2 — Plan. We are testing a sample mean with no known population standard deviation, so we use a Student's t-distribution with \(df = 15 - 1 = 14\). The underlying population is assumed normal.
Step 3 — Compute. From the 15 data values,
$$ \overline{x} = 202.13, \quad s = 4.53, \quad t = \frac{202.13 - 200}{\frac{4.53}{\sqrt{15}}} = 1.82 $$This is a right-tailed test, so the p-value is the area under the t-distribution with 14 degrees of freedom to the right of 1.82, which is about 0.045.
Step 4 — State the conclusion. Since the p-value of 0.045 is less than \(\alpha = 0.1\), reject the null hypothesis.
Answer: At the 10% significance level, there is convincing evidence that the average boiling point of the liquid is greater than 200.
The National Institute of Standards and Technology provides exact data on conductivity properties of materials. Following are conductivity measurements for 11 randomly selected pieces of a particular type of glass.
1.11; 1.07; 1.11; 1.07; 1.12; 1.08; .98; .98; 1.02; .95; .95
Is there convincing evidence that the average conductivity of this type of glass is greater than one? Use a significance level of 0.05. Assume the population is normal.
Solution — glass conductivity above one
Let's follow a four-step process to answer this statistical question.
Step 1 — State the question. We need to determine if, at a 0.05 significance level, the average conductivity of the selected glass is greater than one. Our hypotheses will be
a) \(H_{0}: \mu \leq 1\)
b) \(H_{a}: \mu > 1\)
Step 2 — Plan. We are testing a sample mean without a known population standard deviation. Therefore, we need to use a Student's t-distribution. Assume the underlying population is normal.
Step 3 — Compute. Based on the sample of 11 data values shown above, the sample mean, sample standard deviation, and test statistic are calculated as follows:
$$ \begin{array}{l} \overline{x} = 1.04 \\ s = 0.0659 \\ t = 2.014 \\ \end{array} $$To calculate the p-value, note that this is a right-tailed test. Then find the area under the t-distribution to the right of the test statistic 2.014, using 10 degrees of freedom. This area in the right tail is 0.036, so the p-value \(= 0.036\).
Step 4 — State the conclusions. Since the p-value \(= 0.036\) is less than our \(\alpha\) value, we reject the null hypothesis.
Answer: It is reasonable to state that the data support the claim that the average conductivity level of this glass is greater than one.
In a study of 390,000 moisturizer users, 138 of the subjects developed skin diseases. Test the claim that moisturizer users developed skin diseases at a greater rate than that for non-moisturizer users, where the rate of skin diseases for non-moisturizer users is 0.041%. Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.
Solution — skin disease rates among moisturizer users
Step 1 — State the hypotheses.
$$ H_{0}: p \leq 0.00041, \quad H_{a}: p > 0.00041 $$Committing a Type I error here means accepting a false claim that a common consumer product causes disease, so the significance level is set very low to make that mistake unlikely.
Step 2 — Plan. We are testing a sample proportion with \(x = 138\) and \(n = 390{,}000\). The sample is large enough: \(np = 390{,}000(0.00041) = 159.9\) and \(nq = 390{,}000(0.99959) = 389{,}840.1\), both well above the usual threshold.
Step 3 — Compute. The sample proportion is
$$ p' = \frac{138}{390{,}000} = 0.000354 $$The standard error is \(\sqrt{\frac{(0.00041)(0.99959)}{390{,}000}} = 0.0000324\), so
$$ z = \frac{0.000354 - 0.00041}{0.0000324} = -1.73 $$This is a right-tailed test, so the p-value is the area to the right of \(-1.73\), which is 0.958.
Step 4 — State the conclusion. Since the p-value of 0.958 is far greater than \(\alpha = 0.005\), we cannot reject the null hypothesis.
Answer: There is not enough evidence to support the claim of a higher skin-disease rate for moisturizer users. The sample rate is in fact lower than the comparison rate, which is why the test statistic came out negative on a right-tailed test.
In a study of 420,019 cell phone users, 172 of the subjects developed brain cancer. Test the claim that cell phone users developed brain cancer at a greater rate than that for non-cell phone users, where the rate of brain cancer for non-cell phone users is 0.0340%. Since this is a critical issue, use a 0.005 significance level. Explain why the significance level should be so low in terms of a Type I error.
Solution — a brain cancer rate among cell phone users
We will follow the four-step process.
Step 1 — State the question. We need to conduct a hypothesis test on the claimed cancer rate. Our hypotheses will be
a) \(H_{0}: p \leq 0.00034\)
b) \(H_{a}: p > 0.00034\)
If we commit a Type I error, we are essentially accepting a false claim. Since the claim describes cancer-causing environments, we want to minimize the chances of incorrectly identifying causes of cancer.
Step 2 — Plan. We will be testing a sample proportion with \(x = 172\) and \(n = 420{,}019\). The sample is sufficiently large because we have \(np = 420{,}019(0.00034) = 142.8\) and \(nq = 420{,}019(0.99966) = 419{,}876.2\), two independent outcomes, and a fixed probability of success \(p = 0.00034\). We will therefore be able to generalize our results to the population.
Step 3 — Compute. The sample proportion is \(p' = \frac{172}{420{,}019} = 0.00040950\), which gives a test statistic of \(z = 2.4434\) and a p-value of 0.0073.
On a TI-83/83+/84/84+: press STAT, arrow over to TESTS, press 5:1-PropZTest. Enter 3.4E-4 for \(p_{0}\), 172 for \(x\), and 420019 for \(n\), and set Prop to \(> p_{0}\). The calculator returns \(z = 2.443355592\), \(p = .0072756984\), and \(\hat{p} = 4.0950528\text{E-}4\).
Figure 8.5.15 - The p-value for the brain-cancer test: the shaded right tail beyond a test statistic of 2.4434 has area 0.0073.
Step 4 — State the conclusion. Since the p-value \(= 0.0073\) is greater than our \(\alpha\) value \(= 0.005\), we cannot reject the null.
Answer: There is not enough evidence to support the claim of higher brain cancer rates for the cell phone users.
According to the U.S. Census, there are approximately 201,456,463 residents 20 and older. Statistics from the Criminal National Network indicate that, on average, 104,354 murders occur each year for people aged 20 and older. This translates into a murder percentage of 0.052%. In Ohio, there were 127 murders reported for a population of 427,648. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the local murder percentage and the national murder percentage. Use a significance level of 0.01.
Solution — a local murder rate against the national rate
Step 1 — State the question. We need to test whether the local murder proportion differs from the national proportion, in either direction.
Step 2 — Plan. We are presented with proportions, so we use a one-proportion z-test:
$$ H_{0}: p = 0.00052, \quad H_{a}: p \neq 0.00052 $$Step 3 — Compute. The sample proportion is
$$ p' = \frac{127}{427{,}648} = 0.000297 $$The standard error is \(\sqrt{\frac{(0.00052)(0.99948)}{427{,}648}} = 0.0000349\), so
$$ z = \frac{0.000297 - 0.00052}{0.0000349} = -6.40 $$This is a two-tailed test, so double the area to the left of \(-6.40\). That area is smaller than 0.0000001, so the p-value is essentially 0.
Step 4 — State the conclusion. Since the p-value is far less than \(\alpha = 0.01\), reject the null hypothesis.
Answer: The sample data support the claim that the local murder percentage is different from the national murder percentage. The local rate of about 0.030% is well below the national 0.052%.
Statistical data indicates that in a certain country there are approximately 268,608,618 residents aged 12 and older. For a certain period of time, statistical data also indicates that the number of residents with blood type AB negative (AB-) is 207,754 individuals. This translates into a percentage of 0.078% with this rather rare blood type. In a certain province of the country, there were 11 people with blood type AB- out of a population of 37,937. Conduct an appropriate hypothesis test to determine if there is a statistically significant difference between the percentage of residents in the entire country with blood type AB- versus the percentage in the local province. Use a significance level of 0.01.
Solution — a rare blood type, province against country
We will follow the four-step plan.
Step 1 — State the question. We need to test whether the proportion of residents with AB- blood type in the local province is statistically different compared to the proportion in the entire country.
Step 2 — Plan. Since we are presented with proportions, we will use a one-proportion z-test. The hypotheses for the test will be
$$ H_{0}: p = 0.00078 $$ $$ H_{a}: p \neq 0.00078 $$Step 3 — Compute. Note the sample proportion is
$$ p' = \frac{11}{37{,}937} = 0.00029 $$The test statistic is calculated as \(z = -3.4189\). To calculate the p-value, note that this is a two-tailed test: find the area under the normal distribution to the left of the test statistic and then double it. The area to the left of the test statistic is 0.000314, and this area doubled results in a p-value of 0.00063.
Step 4 — State the conclusion. Since the p-value \(= 0.00063\) is less than the alpha level of 0.01, the sample data indicate that we should reject the null hypothesis.
Answer: The sample data support the claim that the proportion of individuals with blood type AB- in the local province is different from the proportion of individuals in the entire country.
Problem Set 8.5
Problem 1. Assume \(H_{0}: \mu = 9\) and \(H_{a}: \mu < 9\). Is this a left-tailed, right-tailed, or two-tailed test?
Problem 1 Solution
Step 1 — Read the symbol in the alternative hypothesis: \(H_{a}: \mu < 9\) uses the \(<\) symbol.
Step 2 — Match the symbol to the tail: a \(<\) in \(H_{a}\) puts the whole p-value in the left tail of the distribution.
Answer: This is a left-tailed test.
Problem 2. Assume \(H_{0}: \mu \leq 6\) and \(H_{a}: \mu > 6\). Is this a left-tailed, right-tailed, or two-tailed test?
Problem 2 Solution
Step 1 — Read the symbol in the alternative hypothesis: \(H_{a}: \mu > 6\) uses the \(>\) symbol. The \(\leq\) in \(H_{0}\) is not what decides the tail — the alternative is.
Step 2 — Match the symbol to the tail: a \(>\) in \(H_{a}\) puts the whole p-value in the right tail.
Answer: This is a right-tailed test.
Problem 3. Assume \(H_{0}: p = 0.25\) and \(H_{a}: p \neq 0.25\). Is this a left-tailed, right-tailed, or two-tailed test?
Problem 3 Solution
Step 1 — Read the symbol in the alternative hypothesis: \(H_{a}: p \neq 0.25\) uses the \(\neq\) symbol.
Step 2 — Match the symbol to the tail: \(\neq\) picks no direction, so the p-value is split evenly between the two tails.
Answer: This is a two-tailed test.
Problem 4. Describe the general graph of a left-tailed test.
Problem 4 Solution
Step 1 — Start from the shape: draw one bell-shaped curve over a horizontal axis, centred at the value the null hypothesis claims.
Step 2 — Place the sample result: mark the observed sample statistic somewhere to the left of the centre. A left-tailed test is the case where the data came in low.
Step 3 — Shade the p-value: shade the entire region under the curve from that mark leftward, running off the end of the tail. That whole shaded area is the p-value, and there is nothing shaded on the right.
Answer: A single normal curve centred at the null value, with the entire area to the left of the sample statistic shaded as one unbroken left tail.
Problem 5. Describe the graph of a two-tailed test.
Problem 5 Solution
Step 1 — Start from the shape: draw one bell-shaped curve over a horizontal axis, centred at the value the null hypothesis claims.
Step 2 — Mark both cutoffs: because \(H_{a}\) carries a \(\neq\), the test is interested in a difference in either direction, so mark two cutoffs placed symmetrically about the centre — one below it and one above it, the same distance out on each side.
Step 3 — Shade both tails: shade from the left cutoff outward and from the right cutoff outward. The curve is symmetric, so the two shaded regions have equal area.
Step 4 — Label the areas: each shaded tail is labelled \(\frac{1}{2}\)(p-value), and the two together add to the p-value.
Answer: A single normal curve centred at the null value with both tails shaded, each tail labelled \(\frac{1}{2}\)(p-value).
Problem 6. A bottle of water is labeled as containing 16 fluid ounces of water. You believe it is less than that. What type of test would you use?
Problem 6 Solution
Step 1 — Identify the parameter: the label makes a claim about the mean contents of a bottle, so the parameter is a population mean \(\mu\), with the claimed value 16 fluid ounces.
Step 2 — Write your suspicion as the alternative: you believe the true mean is less than 16 ounces, so
$$ H_{0}: \mu = 16, \quad H_{a}: \mu < 16 $$Step 3 — Read the tail off \(H_{a}\): the \(<\) puts the p-value in the left tail.
Answer: A left-tailed test.
Problem 7. Your friend claims that his mean golf score is 63. You want to show that it is higher than that. What type of test would you use?
Problem 7 Solution
Step 1 — Identify the parameter: your friend's claim is about a mean golf score, so the parameter is \(\mu\) with claimed value 63.
Step 2 — Write what you want to show as the alternative: you want to show the true mean is higher than 63, so
$$ H_{0}: \mu = 63, \quad H_{a}: \mu > 63 $$Step 3 — Read the tail off \(H_{a}\): the \(>\) puts the p-value in the right tail.
Answer: A right-tailed test.
Problem 8. A bathroom scale claims to be able to identify correctly any weight within a pound. You think that it cannot be that accurate. What type of test would you use?
Problem 8 Solution
Step 1 — Identify the claim: the scale claims its readings are correct within a pound, meaning the mean error is zero. The parameter is the mean error \(\mu\), with claimed value 0.
Step 2 — Write your suspicion as the alternative: you think the scale cannot be that accurate, and you have no reason to say it reads consistently high rather than consistently low. A doubt with no direction becomes a \(\neq\):
$$ H_{0}: \mu = 0, \quad H_{a}: \mu \neq 0 $$Step 3 — Read the tail off \(H_{a}\): the \(\neq\) splits the p-value between both tails.
Answer: A two-tailed test.
Problem 9. You flip a coin and record whether it shows heads or tails. You know the probability of getting heads is 50%, but you think it is less for this particular coin. What type of test would you use?
Problem 9 Solution
Step 1 — Identify the parameter: heads-or-tails is a yes-or-no outcome recorded as a share of flips, so the parameter is a population proportion \(p\), with claimed value 0.50.
Step 2 — Write your suspicion as the alternative: you think the probability of heads is less for this particular coin, so
$$ H_{0}: p = 0.50, \quad H_{a}: p < 0.50 $$Step 3 — Read the tail off \(H_{a}\): the \(<\) puts the p-value in the left tail.
Answer: A left-tailed test.
Problem 10. If the alternative hypothesis has a not-equals (\(\neq\)) symbol, you know to use which type of test?
Problem 10 Solution
Step 1 — Recall what the alternative hypothesis controls: the symbol in \(H_{a}\) is what names the test, because it says which values of the parameter would count as evidence against \(H_{0}\).
Step 2 — Interpret the not-equals symbol: \(\neq\) counts a result as evidence whether it lands above the claimed value or below it, so the p-value has to be collected from both ends of the distribution and is split evenly between them.
Answer: A two-tailed test.
Problem 11. Assume the null hypothesis states that the mean is at least 18. Is this a left-tailed, right-tailed, or two-tailed test?
Problem 11 Solution
Step 1 — Write the null from the wording: "the mean is at least 18" means \(\mu\) is 18 or more, so
$$ H_{0}: \mu \geq 18 $$Step 2 — Take the alternative as everything the null leaves over: the only values not covered by "18 or more" are the values below 18, so
$$ H_{a}: \mu < 18 $$Step 3 — Read the tail off \(H_{a}\): the \(<\) puts the p-value in the left tail.
Answer: This is a left-tailed test.
Problem 12. Assume the null hypothesis states that the mean is at most 12. Is this a left-tailed, right-tailed, or two-tailed test?
Problem 12 Solution
Step 1 — Write the null from the wording: "the mean is at most 12" means \(\mu\) is 12 or less, so
$$ H_{0}: \mu \leq 12 $$Step 2 — Take the alternative as everything the null leaves over: the only values not covered by "12 or less" are the values above 12, so
$$ H_{a}: \mu > 12 $$Step 3 — Read the tail off \(H_{a}\): the \(>\) puts the p-value in the right tail.
Answer: This is a right-tailed test.
Problem 13. Assume the null hypothesis states that the mean is equal to 88. The alternative hypothesis states that the mean is not equal to 88. Is this a left-tailed, right-tailed, or two-tailed test?
Problem 13 Solution
Step 1 — Write the hypothesis pair from the wording: the null states the mean equals 88 and the alternative states it does not:
$$ H_{0}: \mu = 88, \quad H_{a}: \mu \neq 88 $$Step 2 — Read the tail off \(H_{a}\): the \(\neq\) makes no claim about direction, so the p-value is split evenly between the two tails.
Answer: This is a two-tailed test.
Key Terms
level of significance — the preset probability \(\alpha\) of rejecting a true null hypothesis, chosen before any data is collected.
actual level of significance — another name for the p-value computed from the sample, as distinct from the preset \(\alpha\) it is compared against.
left-tailed test — a test in which the p-value is the area in the left tail; signalled by a \(<\) in \(H_a\).
right-tailed test — a test in which the p-value is the area in the right tail; signalled by a \(>\) in \(H_a\).
two-tailed test — a test in which the p-value is split evenly between both tails; signalled by a \(\neq\) in \(H_a\).
critical value — the z-score or t-score that cuts off an area of \(\alpha\) in the tail, used by the traditional comparison method in place of the p-value.