8.4 Rare Events, the Sample, Decision and Conclusion
SLO 4
Demonstrate an understanding of, and ability to use, basic ideas of statistical processes, including hypothesis tests and confidence interval estimation.
This is where a hypothesis test actually runs: you compute the test statistic and the p-value, compare the p-value to alpha, and write the conclusion in the words of the problem. The whole process — rare-event rule, decision, interpretation — is the outcome in action.
SLO 5
Identify appropriate statistical techniques and use technology-based statistical analysis to describe, interpret, and communicate results.
You use the calculator to find the p-value, but the real work is interpreting it: saying what a p-value of 0.03 means to a non-statistician, and writing the conclusion so it answers the original claim rather than reciting statistical vocabulary.
Learning Objectives
By the end of this section, you will be able to:
- explain why a sample result that would be very unlikely under \(H_{0}\) counts as evidence against \(H_{0}\);
- state what a p-value measures and shade the region of the sampling distribution it stands for;
- compare a p-value against a preset significance level \(\alpha\) and make the reject / do not reject decision;
- write the conclusion of a hypothesis test in plain sentences about the original claim.
Knowing the shape of the distribution, the sample size, and whether the standard deviation is known gets you as far as setting up a hypothesis test. It does not tell you how the test actually decides anything. This section supplies the missing piece: a rule for turning one sample into a verdict about the population. The rule rests on a single idea — if the data you collected would be strange under the assumption you started with, the assumption is what should give way.
8.4.1 Rare Events
A rare event is a sample outcome whose probability of occurring is very small if the null hypothesis is true. Because a rare event almost never happens by chance alone, observing one gives us reason to doubt the assumption that made it rare.
Nothing in a hypothesis test measures whether a claim is true. It measures how surprised you should be by your data if the claim were true. Enough surprise, and you stop believing the claim.
Suppose you make an assumption about a property of the population. That assumption is the null hypothesis. Then you gather sample data randomly. If the sample has properties that would be very unlikely to occur when the assumption is true, you would conclude that your assumption about the population is probably incorrect. Remember what each side of that sentence is: your assumption is only an assumption, not a fact, and it may or may not be true. Your sample data are real, and they are showing you a fact that seems to contradict the assumption. When a fact and an assumption disagree, the assumption is the one that has to move.
Here is what that looks like away from any formula. Mai and Tou are at the birthday party of a very wealthy friend. They hurry to be first in line to grab a prize from a tall basket they cannot see inside, because they will be blindfolded. There are 200 plastic bubbles in the basket, and Mai and Tou have been told that only one of them holds a $100 bill. Mai reaches in first and pulls out a bubble. Her bubble contains a $100 bill. The probability of that happening is
$$ \frac{1}{200} = 0.005 $$Five chances in a thousand. Because that is so unlikely, Tou is hoping that what the two of them were told is wrong and that there are more $100 bills in the basket. A rare event has occurred — Mai getting the $100 bill on the first try — so he doubts the assumption that only one $100 bill was in the basket.
Notice that Tou cannot prove anything. It is entirely possible that there was exactly one $100 bill and Mai simply got lucky. Rare things do happen; that is what makes them rare rather than impossible. What Tou has is a reason to doubt, and the strength of that reason is exactly the smallness of 0.005. That is the whole logic of a hypothesis test, and everything that follows in this section is a way of putting a number on how strong the doubt is.
Definition 8.4.1 - One prize among 200 identical bubbles: the draw that finds it is a rare event.
A jar is said to hold 500 marbles, exactly 2 of which are red. You draw one marble at random without looking and it is red. State the assumption being tested, compute the probability of your draw under that assumption, and say what the draw suggests about the assumption.
Solution — a red marble on the first draw
Step 1 — Name the assumption. The claim on trial is that the jar holds 500 marbles of which exactly 2 are red. That is the null hypothesis.
Step 2 — Find the probability of what you saw, assuming the claim is true. Two of the 500 marbles are red, so
$$ P(\text{red on one random draw}) = \frac{2}{500} = 0.004 $$Step 3 — Judge it. Four chances in a thousand is a rare event. Getting a red marble on a single blind draw would almost never happen if only 2 of the 500 were red.
Answer: the draw is a rare event under the stated claim, so it gives us reason to doubt that only 2 of the 500 marbles are red — the jar probably holds more red marbles than we were told. It does not prove that, because a genuinely lucky draw is still possible.
In the party scenario above, suppose Mai had drawn a bubble with no bill in it, and then Tou drew a bubble that did hold a $100 bill. Is his draw still a rare event under the same assumption?
Solution — the second blindfolded draw
Step 1 — Restate the assumption. Exactly one of the 200 bubbles holds a $100 bill.
Step 2 — Work out the probability for the second draw. Mai removed one empty bubble, so 199 bubbles remain and one of them holds the bill:
$$ P(\text{Tou draws the bill}) = \frac{1}{199} \approx 0.005 $$Step 3 — Compare. That is essentially the same tiny probability she faced.
Answer: yes. Under the assumption that only one bubble holds a bill, any single blindfolded draw producing the bill is a rare event, whether it is the first draw or the second. The doubt it raises about the assumption is the same either way.
8.4.2 Using the Sample to Test the Null Hypothesis
The p-value is the probability that, if the null hypothesis is true, the results from another randomly selected sample will be as extreme or more extreme than the results obtained from the given sample.
It answers one question: if the null hypothesis really were true, how often would a random sample look at least this unusual? A small answer means your sample is hard to explain away as luck.
Read that definition slowly, because two things in it are easy to slide past. First, the p-value is computed while assuming \(H_{0}\) is true — it is a conditional probability, not the probability that \(H_{0}\) is true. Second, it covers results "as extreme or more extreme", not just the exact result you got, which is why it always corresponds to an area in the tail of a curve rather than to a single point on it.
A large p-value calculated from the data indicates that we should not reject the null hypothesis. The smaller the p-value, the more unlikely the outcome, and the stronger the evidence is against the null hypothesis. We would reject the null hypothesis if the evidence is strongly against it. The p-value is not a switch, it is a dial: a value of 0.4 says your sample is unremarkable, a value of 0.04 says it is unusual, and a value of 0.0001 says it is nearly impossible to produce by chance under \(H_{0}\).
Draw a graph that shows the p-value. Every hypothesis test is easier to perform if you use a graph, because you see the problem more clearly. Sketch the sampling distribution the null hypothesis predicts, mark the value your sample actually produced, and shade everything at least that far out. The shaded area is the p-value, and once it is on paper you can usually tell before computing anything whether the area is going to be large or vanishingly small.
Definition 8.4.2 - The p-value is the shaded tail beyond what you observed, and it shrinks as the observation moves out.
A normal distribution has a standard deviation of 1. We want to verify a claim that the mean is greater than 12. A sample of 36 is taken with a sample mean of 12.5.
$$ H_{0}: \mu \leq 12 $$ $$ H_{a}: \mu > 12 $$The p-value is 0.0013. Draw a graph that shows the p-value, and say what it tells you.
Solution — shading a right tail at 12.5
Step 1 — Find the distribution to draw. By the Central Limit Theorem, with \(\sigma = 1\) and \(n = 36\), the sample means follow
$$ \overline{X} \sim N\!\left(12, \frac{1}{\sqrt{36}}\right) = N(12, 0.1667) $$Step 2 — Draw the curve. Sketch a normal curve centered at 12, the value the null hypothesis puts on the boundary. Mark 12.5 on the horizontal axis, well to the right of center — it sits three standard errors out, since \(\frac{12.5 - 12}{0.1667} = 3\).
Step 3 — Shade the right tail. The alternative hypothesis points right, so shade everything from 12.5 outward to the right. That sliver of area is the p-value, 0.0013.
Answer: the shaded right tail has area 0.0013. Only about 13 samples in 10,000 would produce a mean of 12.5 or higher if the true mean were really 12, so the sample is strong evidence that the mean is greater than 12.
Try it in rāSHio
You do not need a calculator to get that 0.0013. In rāSHio choose Distributions → Normal, set the mean to 12 and the standard deviation to 0.1667 — that second number is the standard error \(1/\sqrt{36}\), not the population standard deviation, and typing the standard error is the whole of what the Central Limit Theorem asks of you here. Enter 12.5 as the cutoff and the panel shades the region and reports its area, which is the p-value you were handed.
Figure 8.4.1 — Reading a p-value off the curve in rāSHio: Distributions → Normal. The clip runs a single left-tail area read on its own demo numbers; a right-tail question like this one is the same panel, read from the other side.
Gordon Whitfield, who runs a neighborhood bakery with his husband, claims that his bread height is more than 15 cm, on average. Several of his customers do not believe him. To persuade them that he is right, Gordon decides to do a hypothesis test. He bakes 10 loaves of bread. The mean height of the sample loaves is 17 cm. He knows from baking hundreds of loaves of bread that the standard deviation for the height is 0.5 cm, and that the distribution of heights is normal.
State the hypotheses, find the p-value, draw the graph that shows it, and write the conclusion.
Solution — testing a claimed loaf height of 15 cm
Step 1 — Write the hypotheses. The words "is more than" translate to a \(>\), so \(\mu > 15\) goes into the alternate hypothesis. The null hypothesis must contradict the alternate hypothesis:
$$ H_{0}: \mu \leq 15 \qquad H_{a}: \mu > 15 $$Step 2 — Find the distribution of the sample mean. By the Central Limit Theorem, the distribution of sample means for samples of 10 loaves is normal with mean \(\mu = 15\) and standard error
$$ \frac{\sigma}{\sqrt{n}} = \frac{0.5}{\sqrt{10}} = 0.16 $$Step 3 — Ask the test's question. Suppose the null hypothesis is true, so the mean height of the loaves is no more than 15 cm. Is the 17 cm mean computed from the sample unexpectedly large? The hypothesis test works by asking exactly how unlikely that sample mean would be if the null hypothesis were true. A graph shows how far out the sample mean sits on the normal curve, and the p-value is the probability that any other sample mean would fall at least as far out as 17 cm.
Step 4 — Compute and shade the p-value. The p-value is the probability that a sample mean is 17 cm or greater when the population mean is in fact 15 cm. We can calculate this probability using the normal distribution for means:
$$ p\text{-value} = P(\overline{x} > 17) \approx 0 $$
Figure 8.4.2 - The sampling distribution of the mean loaf height under H0, centred at 15 cm, with the region beyond the observed 17 cm shaded.
Step 5 — Interpret it. A p-value of approximately zero tells us it is highly unlikely that a loaf of bread rises no more than 15 cm on average. Almost 0% of all loaves of bread would be at least as high as 17 cm purely by chance had the population mean height really been 15 cm.
Answer: because the outcome of 17 cm is so unlikely — meaning it is happening not by chance alone — we conclude that the evidence is strongly against the null hypothesis that the mean height is at most 15 cm. There is sufficient evidence that the true mean height for the population of Gordon's loaves of bread is greater than 15 cm.
Try it in rāSHio
The baker knows \(\sigma\), which is exactly the case one panel handles end to end. In rāSHio choose Stats → Z Stats and enter this example's summary values — 17 for the sample mean, 0.5 for the population standard deviation, 10 for the sample size, and 15 for the claimed mean — and it returns the z-statistic and the p-value in one step, instead of your standardising by hand and then looking up a tail. The verdict is still yours to write.
Figure 8.4.3 — The Z Stats panel in rāSHio: Stats → Z Stats. One panel covers both jobs a known \(\sigma\) allows, a confidence interval and a hypothesis test; the walkthrough runs the hypothesis-test side, which is the side this example needs.
8.4.3 Decision and Conclusion
The significance level, written \(\alpha\), is a probability chosen before the data are examined. It is the probability of a Type I error — rejecting the null hypothesis when the null hypothesis is true. It may or may not be given to you at the beginning of a problem.
Choosing the threshold before you see the data is what makes the decision a test rather than an opinion. Move the line after the fact and you can always land on the answer you wanted.
A systematic way to decide whether to reject the null hypothesis is to compare the p-value against that preset \(\alpha\). Setting \(\alpha\) in advance is what keeps the procedure honest. If you were allowed to look at the p-value first and then pick a threshold, you could justify any verdict you liked, which is why the significance level is chosen before the sample is examined and left alone afterward.
When you make a decision to reject or not reject \(H_{0}\), do as follows.
- If \(\alpha > p\)-value, reject \(H_{0}\). The results of the sample data are significant. There is sufficient evidence to conclude that \(H_{0}\) is an incorrect belief and that the alternative hypothesis, \(H_{a}\), may be correct.
- If \(\alpha \leq p\)-value, do not reject \(H_{0}\). The results of the sample data are not significant. There is not sufficient evidence to conclude that the alternative hypothesis, \(H_{a}\), may be correct.
- When you do not reject \(H_{0}\), it does not mean that you should believe \(H_{0}\) is true. It simply means that the sample data have failed to provide sufficient evidence to cast serious doubt about the truthfulness of \(H_{0}\).
After you make your decision, write a thoughtful conclusion about the hypotheses in terms of the given problem. That last step is not a formality. The decision line — reject or do not reject — is written in the language of \(H_{0}\) and \(H_{a}\), and nobody outside a statistics class talks that way. The conclusion translates the verdict back into the words of the original question, and it has to carry the same caution the decision does: enough evidence, or not enough evidence, never proof.
Definition 8.4.3 - Alpha is drawn on the curve before any data arrive; which side the sample lands on is the decision.
It's a Boy Genetics Labs claims their procedures improve the chances of a boy being born. Alex Nava, who is weighing the service with their partner, wants to know whether the evidence behind that claim holds up. The results for a test of a single population proportion are
$$ H_{0}: p = 0.50, \qquad H_{a}: p > 0.50 $$ $$ \alpha = 0.01 $$ $$ p\text{-value} = 0.025 $$Interpret the results and state a conclusion in simple, non-technical terms.
Solution — a genetics lab's claim about boys
Step 1 — Compare \(\alpha\) with the p-value. Here \(\alpha = 0.01\) and the p-value is 0.025, so
$$ \alpha = 0.01 \leq 0.025 = p\text{-value} $$Step 2 — Apply the decision rule. Since \(\alpha \leq p\)-value, we do not reject \(H_{0}\).
Step 3 — Translate the decision. A p-value of 0.025 does say the sample leaned toward more boys than a 50-50 split, but the lab set a strict threshold of 0.01, and 0.025 does not clear it.
Answer: at the 1% significance level, there is not enough evidence to conclude that the lab's procedures improve the chances of a boy being born. In plain terms, the study did not produce results unusual enough — under the strict standard the test was held to — to support the lab's claim, so Alex has no statistical grounds to take it at its word.
When using the p-value to evaluate a hypothesis test, it is sometimes useful to use the following memory device.
If the p-value is low, the null must go.
If the p-value is high, the null must fly.
This memory aid relates a p-value less than the established alpha (the p is low) to rejecting the null hypothesis, and likewise relates a p-value higher than the established alpha (the p is high) to not rejecting the null hypothesis.
Fill in the blanks.
Reject the null hypothesis when ____. The results of the sample data ____.
Do not reject the null hypothesis when ____. The results of the sample data ____.
Solution — filling in the decision sentences
Step 1 — Handle the low-p case. A p-value below \(\alpha\) means the sample would be a rare event under \(H_{0}\), so the null goes.
Reject the null hypothesis when the p-value is less than the established alpha value. The results of the sample data support the alternative hypothesis.
Step 2 — Handle the high-p case. A p-value above \(\alpha\) means the sample is unremarkable under \(H_{0}\), so the null flies.
Do not reject the null hypothesis when the p-value is greater than the established alpha value. The results of the sample data do not support the alternative hypothesis.
Answer: reject when the p-value is less than alpha and the data support \(H_{a}\); do not reject when the p-value is greater than alpha and the data do not support \(H_{a}\).
Problem Set 8.4
Problem 1. When do you reject the null hypothesis?
Problem 1 Solution
Step 1 — Recall what the decision compares: the test is decided by holding the p-value up against the significance level \(\alpha\) that was chosen before the data were examined.
Step 2 — Apply the rule: if \(\alpha > p\)-value, the sample would be a rare event under \(H_{0}\), so the null goes.
Answer: you reject the null hypothesis when the p-value is less than the preset significance level \(\alpha\). The results of the sample data are then significant, and there is sufficient evidence to conclude that \(H_{a}\) may be correct.
Problem 2. The probability of winning the grand prize at a particular carnival game is 0.005. Is the outcome of winning very likely or very unlikely?
Problem 2 Solution
Step 1 — Read the probability: the chance of winning the grand prize on one play is 0.005.
Step 2 — Put it on a familiar scale: 0.005 is 5 chances in 1,000, or 1 chance in 200. Out of 200 people who play, you would expect about one winner.
Answer: winning is very unlikely.
Problem 3. The probability of winning the grand prize at a particular carnival game is 0.005. Michele wins the grand prize. Is her win considered a rare or common event? Why?
Problem 3 Solution
Step 1 — Compare the probability with what "rare" means: an outcome is a rare event when its probability under the stated assumption is very small. Here that probability is 0.005.
Step 2 — Judge Michele's win: an event that happens about once in every 200 plays almost never happens to any one player on any one try.
Answer: her win is a rare event, because the probability of it happening is only 0.005 — about one chance in 200. Rare does not mean impossible, so a single win does not prove the game is rigged in the player's favor; it just means something very unlikely happened.
Problem 4. It is believed that the mean height of high school students who play basketball on the school team is 73 inches with a standard deviation of 1.8 inches. A random sample of 40 players is chosen. The sample mean was 71 inches, and the sample standard deviation was 1.5 inches. Do the data support the claim that the mean height is less than 73 inches? The p-value is almost zero. State the null and alternative hypotheses and interpret the p-value.
Problem 4 Solution
Step 1 — Identify the claim being tested: the claim is that the mean height is less than 73 inches, so "less than" points the alternative hypothesis to the left. The null must contradict it and carry the equality.
$$ H_{0}: \mu \geq 73 \qquad H_{a}: \mu < 73 $$Step 2 — Read the p-value: the p-value is given as almost zero. That is the probability of drawing a sample of 40 players whose mean height is 71 inches or lower if the population mean really were 73 inches.
Step 3 — Interpret it: almost none of the possible samples of 40 players would produce a mean of 71 inches or below when the true mean is 73 inches. The sample is a rare event under \(H_{0}\), so the evidence against \(H_{0}\) is strong. At the 5% level (indeed at any usual level) we reject \(H_{0}\).
Answer: \(H_{0}: \mu \geq 73\) and \(H_{a}: \mu < 73\). The p-value is almost zero, which means there is sufficient evidence to conclude that the mean height of high school students who play basketball on the school team is less than 73 inches. The data do support the claim.
Problem 5. The mean age of graduate students at a University is at most 31 years with a standard deviation of two years. A random sample of 15 graduate students is taken. The sample mean is 32 years and the sample standard deviation is three years. Are the data significant at the 1% level? The p-value is 0.0264. State the null and alternative hypotheses and interpret the p-value.
Problem 5 Solution
Step 1 — Write the hypotheses: "at most 31 years" is the claim on trial, so it goes in the null with its equality, and the alternative takes what is left over.
$$ H_{0}: \mu \leq 31 \qquad H_{a}: \mu > 31 $$Step 2 — Compare the p-value with the stated level: the question asks about the 1% level, so \(\alpha = 0.01\), and the p-value is 0.0264.
$$ \alpha = 0.01 \leq 0.0264 = p\text{-value} $$Step 3 — Apply the decision rule: since \(\alpha \leq p\)-value, we do not reject \(H_{0}\).
Answer: \(H_{0}: \mu \leq 31\) and \(H_{a}: \mu > 31\). The p-value of 0.0264 says that if the mean age really were 31 years, about 2.64% of samples of 15 graduate students would have a mean age of 32 years or higher. That is unusual, but it does not clear the strict 1% bar, so the data are not significant at the 1% level and we do not reject \(H_{0}\). (Note that at the 5% level the same data would be significant — the threshold you set in advance decides the verdict.)
Problem 6. Does the shaded region represent a low or a high p-value compared to a level of significance of 1%?
Exercise Figure 8.4.1 - The shaded region Problem 8.4.6 asks you to judge against a 1% level of significance.
Problem 6 Solution
Step 1 — Read the shaded area: the shaded sliver sits far out in the right tail, and the label on the figure reports the area as approximately zero.
Step 2 — Compare with the level of significance: the level of significance is 1%, or 0.01. An area of approximately 0 is smaller than 0.01.
Answer: the shaded region represents a low p-value. Since it is smaller than \(\alpha = 0.01\), the decision would be to reject \(H_{0}\).
Problem 7. What should you do when \(\alpha > p\)-value?
Problem 7 Solution
Step 1 — Translate the inequality: \(\alpha > p\)-value means the p-value fell below the threshold you set in advance, so the sample result is rarer than the standard you agreed to be surprised by.
Answer: reject \(H_{0}\). The results of the sample data are significant, and there is sufficient evidence to conclude that \(H_{a}\) may be correct.
Problem 8. What should you do if \(\alpha = p\)-value?
Problem 8 Solution
Step 1 — Find which side of the rule this lands on: the decision rule rejects only when \(\alpha > p\)-value. Equality is not "greater than", so \(\alpha = p\)-value falls under the \(\alpha \leq p\)-value branch.
Answer: do not reject \(H_{0}\). The results of the sample data are not significant, so there is not sufficient evidence to conclude that \(H_{a}\) may be correct.
Problem 9. If you do not reject the null hypothesis, then it must be true. Is this statement correct? State why or why not in complete sentences.
Problem 9 Solution
Step 1 — Check what "do not reject" actually claims: the test only measures how surprising the sample would be if \(H_{0}\) were true. A large p-value says the sample was unsurprising, not that \(H_{0}\) is correct.
Step 2 — Name the missing possibility: an unsurprising sample can also come from a small sample size, a large amount of natural variation, or a real effect too small for this test to detect. Any of those produce a large p-value while \(H_{0}\) is false.
Answer: no, the statement is not correct. Failing to reject the null hypothesis means only that the sample data did not provide enough evidence to cast serious doubt on \(H_{0}\); it never shows that \(H_{0}\) is true. That is why we say "do not reject \(H_{0}\)" rather than "accept \(H_{0}\)".
Use the following information to answer the next seven exercises: Suppose that a recent article stated that the mean time spent in jail by a first-time convicted burglar is 2.5 years. A study was then done to see if the mean time has increased in the new century. A random sample of 26 first-time convicted burglars in a recent year was picked. The mean length of time in jail from the survey was three years with a standard deviation of 1.8 years. Suppose that it is somehow known that the population standard deviation is 1.5. Conduct a hypothesis test to determine if the mean length of jail time has increased. Assume the distribution of the jail times is approximately normal.
Problem 10. Is this a test of means or proportions?
Problem 10 Solution
Step 1 — Look at what is being measured: the study records the length of time each burglar spent in jail, which is a quantitative measurement averaged across the sample, not a yes-or-no category counted as a share.
Answer: this is a test of means.
Problem 11. What symbol represents the random variable for this test?
Problem 11 Solution
Step 1 — Match the symbol to the quantity that varies from sample to sample: the test is about a mean, and the quantity that changes each time a new sample of 26 burglars is drawn is the sample mean.
Answer: \(\overline{X}\).
Problem 12. In words, define the random variable for this test.
Problem 12 Solution
Step 1 — Say what \(\overline{X}\) stands for in this context: name the units, the group, and the sample size.
Answer: \(\overline{X}\) is the mean time spent in jail, in years, for a sample of 26 first-time convicted burglars.
Problem 13. Is \(\sigma\) known and, if so, what is it?
Problem 13 Solution
Step 1 — Separate the two standard deviations in the setup: the sample's standard deviation is 1.8 years, but the problem also says it is "somehow known" that the population standard deviation is 1.5.
Answer: yes, \(\sigma\) is known, and \(\sigma = 1.5\) years.
Problem 14. Calculate the following:
a) \(\overline{x} = \) \(\underline{\hspace{6em}}\)
b) \(\sigma = \) \(\underline{\hspace{6em}}\)
c) \(s_{x} = \) \(\underline{\hspace{6em}}\)
d) \(n = \) \(\underline{\hspace{6em}}\)
Problem 14 Solution
Step 1 — Pull each value out of the problem statement: the survey mean is the sample mean, the "somehow known" value is the population standard deviation, the survey's own spread is the sample standard deviation, and the count of burglars is the sample size.
Answer:
a) \(\overline{x} = 3\) years
b) \(\sigma = 1.5\) years
c) \(s_{x} = 1.8\) years
d) \(n = 26\)
Problem 15. Since both \(\sigma\) and \(s_{x}\) are given, which should be used? In one to two complete sentences, explain why.
Problem 15 Solution
Step 1 — Rank the two by what they describe: \(s_{x}\) is an estimate of the spread built from just 26 burglars, while \(\sigma\) is the actual spread of the whole population.
Step 2 — Pick the better one: when the true population value is available there is nothing to estimate, so the estimate is not needed.
Answer: use \(\sigma = 1.5\). The population standard deviation is known, so we do not have to approximate it with the sample standard deviation, and the test statistic is built on the normal distribution rather than the Student's t.
Problem 16. State the distribution to use for the hypothesis test.
Problem 16 Solution
Step 1 — Center the distribution at the null value: the null hypothesis puts the mean jail time at 2.5 years.
Step 2 — Find the standard error: with \(\sigma = 1.5\) and \(n = 26\),
$$ \frac{\sigma}{\sqrt{n}} = \frac{1.5}{\sqrt{26}} \approx 0.2942 $$Answer:
$$ \overline{X} \sim N\!\left(2.5, \frac{1.5}{\sqrt{26}}\right) $$Problem 17. A random survey of 75 student loan recipients revealed that the mean length of time in repayment is 17.4 years with a standard deviation of 6.3 years. Conduct a hypothesis test to determine if the population mean time in repayment could likely be 15 years.
a) Is this a test of one mean or proportion?
b) State the null and alternative hypotheses. \(H_{0}\): \(\underline{\hspace{6em}}\) \(H_{a}\): \(\underline{\hspace{6em}}\)
c) Is this a right-tailed, left-tailed, or two-tailed test?
d) What symbol represents the random variable for this test?
e) In words, define the random variable for this test.
f) Is the population standard deviation known and, if so, what is it?
g) Calculate the following: i. \(\overline{x} = \) \(\underline{\hspace{6em}}\) ii. \(s = \) \(\underline{\hspace{6em}}\) iii. \(n = \) \(\underline{\hspace{6em}}\)
h) Which test should be used?
i) State the distribution to use for the hypothesis test.
j) Find the p-value.
k) At a pre-conceived \(\alpha = 0.05\), what is your i. decision, ii. reason for the decision, and iii. conclusion (write out in a complete sentence)?
Problem 17 Solution
Step 1 — Part a. Decide mean or proportion. Time in repayment is measured in years and averaged, so this is a test of one mean.
Step 2 — Part b. Write the hypotheses. The question asks whether the population mean "could likely be 15 years", which gives no direction, so the test is about equality:
$$ H_{0}: \mu = 15 \qquad H_{a}: \mu \neq 15 $$Step 3 — Part c. Find the tails. A \(\neq\) alternative puts evidence in both tails, so this is a two-tailed test.
Step 4 — Part d. Name the symbol. The random variable is \(\overline{X}\).
Step 5 — Part e. Define it in words. \(\overline{X}\) is the mean length of time in repayment, in years, for a sample of 75 student loan recipients.
Step 6 — Part f. Check the population standard deviation. Only the sample standard deviation of 6.3 years is given, so \(\sigma\) is not known.
Step 7 — Part g. List the summary values. \(\overline{x} = 17.4\), \(s = 6.3\), \(n = 75\).
Step 8 — Part h. Choose the test. Since \(\sigma\) is unknown and we are testing one mean, use a Student's t-test.
Step 9 — Part i. State the distribution. With \(n - 1 = 74\) degrees of freedom, \(\overline{X}\) follows a Student's t distribution centered at 15, with standard error
$$ \frac{6.3}{\sqrt{75}} \approx 0.7275 $$Step 10 — Part j. Find the p-value. The test statistic is
$$ t = \frac{\overline{x} - \mu_{0}}{s/\sqrt{n}} = \frac{17.4 - 15}{6.3/\sqrt{75}} = \frac{2.4}{0.7275} \approx 3.30 $$Because the test is two-tailed, double the area beyond \(t = 3.30\) on 74 degrees of freedom:
$$ p\text{-value} \approx 0.0015 $$Step 11 — Part k. Decide and conclude. Compare with \(\alpha = 0.05\): \(0.05 > 0.0015\), so \(\alpha > p\)-value.
Answer: a. one mean; b. \(H_{0}: \mu = 15\), \(H_{a}: \mu \neq 15\); c. two-tailed; d. \(\overline{X}\); e. the mean repayment time for a sample of 75 recipients; f. no, \(\sigma\) is not known; g. \(\overline{x} = 17.4\), \(s = 6.3\), \(n = 75\); h. a Student's t-test; i. \(t_{74}\) centered at 15 with standard error \(6.3/\sqrt{75} \approx 0.7275\); j. \(p\text{-value} \approx 0.0015\); k. i. reject \(H_{0}\), ii. because the p-value of 0.0015 is smaller than \(\alpha = 0.05\), iii. there is sufficient evidence at the 5% level to conclude that the mean length of time student loan recipients spend in repayment is not 15 years.
Problem 18. The National Institute of Mental Health published an article stating that in any one-year period, approximately 9.5% of American adults suffer from depression or a depressive illness. Suppose that in a survey of 100 people in a certain town, seven of them suffered from depression or a depressive illness. Conduct a hypothesis test to determine if the true proportion of people in that town suffering from depression or a depressive illness is lower than the percent in the general adult American population.
a) Is this a test of one mean or proportion?
b) State the null and alternative hypotheses. \(H_{0}\): \(\underline{\hspace{6em}}\) \(H_{a}\): \(\underline{\hspace{6em}}\)
c) Is this a right-tailed, left-tailed, or two-tailed test?
d) What symbol represents the random variable for this test?
e) In words, define the random variable for this test.
f) Calculate the following: i. \(x = \) \(\underline{\hspace{6em}}\) ii. \(n = \) \(\underline{\hspace{6em}}\) iii. \(p' = \) \(\underline{\hspace{6em}}\)
g) Calculate \(\sigma_{X} = \) \(\underline{\hspace{6em}}\). Show the formula set-up.
h) State the distribution to use for the hypothesis test.
i) Find the p-value.
j) At a pre-conceived \(\alpha = 0.05\), what is your i. decision, ii. reason for the decision, and iii. conclusion (write out in a complete sentence)?
Problem 18 Solution
Step 1 — Part a. Decide mean or proportion. Each person either does or does not suffer from depression, and the study reports a share of the group, so this is a test of one proportion.
Step 2 — Part b. Write the hypotheses. The claim being investigated is that the town's proportion is lower than the national 9.5%:
$$ H_{0}: p = 0.095 \qquad H_{a}: p < 0.095 $$Step 3 — Part c. Find the tail. The \(<\) in \(H_{a}\) puts all the evidence in the left tail, so this is a left-tailed test.
Step 4 — Part d. Name the symbol. The random variable is \(P'\), the sample proportion.
Step 5 — Part e. Define it in words. \(P'\) is the proportion of people in the surveyed town who suffer from depression or a depressive illness.
Step 6 — Part f. List the counts. \(x = 7\), \(n = 100\), and
$$ p' = \frac{x}{n} = \frac{7}{100} = 0.07 $$Step 7 — Part g. Compute the standard deviation. Under \(H_{0}\) the proportion is \(p = 0.095\), so \(q = 1 - p = 0.905\):
$$ \sigma_{X} = \sqrt{\frac{pq}{n}} = \sqrt{\frac{(0.095)(0.905)}{100}} = \sqrt{0.00085975} \approx 0.0293 $$Step 8 — Part h. State the distribution.
$$ P' \sim N\!\left(0.095, \sqrt{\frac{(0.095)(0.905)}{100}}\right) $$Step 9 — Part i. Find the p-value. The p-value is the left-tail area below the observed \(p' = 0.07\):
$$ z = \frac{0.07 - 0.095}{0.0293} \approx -0.85 \qquad p\text{-value} = P(P' < 0.07) \approx 0.1969 $$Step 10 — Part j. Decide and conclude. Compare with \(\alpha = 0.05\): \(0.05 \leq 0.1969\), so \(\alpha \leq p\)-value.
Answer: a. one proportion; b. \(H_{0}: p = 0.095\), \(H_{a}: p < 0.095\); c. left-tailed; d. \(P'\); e. the proportion of people in the town who suffer from depression or a depressive illness; f. \(x = 7\), \(n = 100\), \(p' = 0.07\); g. \(\sigma_{X} = \sqrt{(0.095)(0.905)/100} \approx 0.0293\); h. \(P' \sim N(0.095,\, 0.0293)\); i. \(p\text{-value} \approx 0.1969\); j. i. do not reject \(H_{0}\), ii. because the p-value of 0.1969 is larger than \(\alpha = 0.05\), iii. there is not sufficient evidence at the 5% level to conclude that the true proportion of people in that town suffering from depression or a depressive illness is lower than 9.5%.
Key Terms
rare event — a sample outcome that would almost never occur if the null hypothesis were true; observing one is what gives us grounds to doubt the null.
p-value — the probability, computed assuming \(H_{0}\) is true, of getting a sample result as extreme as or more extreme than the one observed.
significance level (\(\alpha\)) — the threshold chosen before the data are examined; it is the probability of rejecting a true null hypothesis, and the p-value is compared against it to make the decision.