10.7 Lab 1: Chi-Square Goodness-of-Fit

Step 1 — Find the expected count for five intervals. Equal-probability fifths each hold one fifth of the sample:

$$ E = \frac{30}{5} = 6 $$

Step 2 — Check that against the condition. Six is at least five, so five intervals is legal — but only barely.

Step 3 — Find the expected count for six intervals.

$$ E = \frac{30}{6} = 5 $$

Step 4 — Check that against the condition. Exactly five, which satisfies "at least five" with nothing to spare.

Step 5 — Find the limit. Seven intervals would give \(\frac{30}{7} \approx 4.3\), below the floor. So 30 observations support at most six equal-probability intervals.

Answer: Five intervals give \(E = 6\) and six give \(E = 5\). Six is the maximum this sample size supports, which is why the lab asks for five to six.

Step 1 — Consider what a cashier's last ten amounts represent. They are ten consecutive transactions at one register, so they share a time of day and a register type. Consecutive transactions at one till are not independent draws from all shoppers.

Step 2 — Consider what 30 departing customers represent. They come from every open register across the time you are standing there, so the sample spreads over register types and shopper types.

Step 3 — Explain the express lane instruction. Express lanes cap the number of items, which caps the receipt total. Omitting them removes most of the small receipts and truncates the low end of the distribution.

Step 4 — Connect this to the test. Both distributions being tested depend on the whole spread of values. Cutting off the small totals would shift the interval boundaries and the expected counts, so the test would be answering a question about express-lane-free shopping rather than about shopping.

Answer: Consecutive transactions at one register are not independent and share a time and register type, while the express lane holds the small receipts. Excluding either one biases the shape of the distribution the test is about.

Step 1 — Compute the sample variance. The variance is the square of the standard deviation:

$$ s^{2} = (31.85)^{2} \approx 1014.42 $$

Step 2 — Say what \(\overline{x}\) is for. The sample mean supplies the decay parameter for the exponential test, \(m = \frac{1}{\overline{x}} = \frac{1}{42.60} \approx 0.0235\). Without it that half of the lab cannot start.

Step 3 — Say what \(s\) and \(s^{2}\) are for. They describe the spread of the receipts and let you compare the two fitted distributions against the data's actual variability. Neither enters the chi-square statistic itself.

Step 4 — Identify what the uniform test uses. The uniform test needs only the lowest and highest values, which set \(X \sim U(a, b)\), and then counts within fifths. It never uses \(\overline{x}\), \(s\) or \(s^{2}\).

Answer: \(s^{2} \approx 1014.42\). The mean supplies the exponential decay parameter; \(s\) and \(s^{2}\) describe spread but enter neither test statistic. The uniform test uses none of the three — only the minimum and maximum.

Step 1 — Write the hypotheses.

\(H_{0}\): The grocery receipt amounts fit the uniform distribution.

\(H_{a}\): The grocery receipt amounts do not fit the uniform distribution.

Step 2 — Name the distribution for the test. A goodness-of-fit test uses the chi-square distribution, and it is right-tailed.

Step 3 — Find the degrees of freedom. There are five categories, one per fifth:

$$ df = \text{number of categories} - 1 = 5 - 1 = 4 $$

so the distribution for the test is \(\chi_{4}^{2}\).

Step 4 — Explain the identical expected counts. A fifth of a distribution holds 20% of its probability by construction, whatever the shape of the interval widths. With 30 receipts, each fifth is therefore expected to hold \((0.20)(30) = 6\).

Answer: \(H_{0}\): the receipts fit the uniform distribution; \(H_{a}\): they do not; use \(\chi_{4}^{2}\) with \(df = 4\). Every expected count is 6 because each fifth carries 20% of the probability by definition.

Step 1 — Find the width of the interval. The range runs from 4 to 144, so its width is \(144 - 4 = 140\).

Step 2 — Use the uniform percentile rule. For \(X \sim U(a,b)\), the \(k\)th percentile sits a fraction \(k\) of the way along the interval:

$$ P_{k} = a + k(b - a) $$

Step 3 — Compute each cut point.

$$ P_{20} = 4 + 0.20(140) = 32 \qquad P_{40} = 4 + 0.40(140) = 60 $$ $$ P_{60} = 4 + 0.60(140) = 88 \qquad P_{80} = 4 + 0.80(140) = 116 $$

Step 4 — State the five intervals. They are $4–$32, $32–$60, $60–$88, $88–$116 and $116–$144, each 28 dollars wide.

Answer: The cut points are 32, 60, 88 and 116. Under a uniform distribution the fifths are equal in WIDTH as well as in probability, which is what makes this the easy case.

Step 1 — Find the expected counts. Under the uniform hypothesis each fifth is expected to hold \(\frac{30}{5} = 6\) receipts.

Step 2 — Check the condition. The condition constrains the EXPECTED counts, and every one of them is 6, which is at least five. The condition is satisfied.

Step 3 — Address the small observed counts. Observed counts of 3 and 1 are not a violation. They are the evidence — they are exactly the mismatch the test exists to measure.

Step 4 — State the consequence. Nothing needs to be merged, and the degrees of freedom stays at \(df = 5 - 1 = 4\).

Step 5 — Anticipate the result. With counts sliding from 12 down to 1 against a flat expectation of 6, the terms \(\frac{(O-E)^{2}}{E}\) will be large and the test will very likely reject the uniform hypothesis.

Answer: The condition IS satisfied, because every expected count is 6. No merging is needed and \(df\) remains 4. The small observed counts are the signal, not a problem with the setup.

Step 1 — Compute the decay parameter. The lab specifies \(m = \frac{1}{\overline{x}}\):

$$ m = \frac{1}{42.60} \approx 0.0235 $$

Step 2 — Write the distribution. \(X \sim Exp(0.0235)\).

Step 3 — Compare with the uniform setup. The uniform test also took its endpoints from the data, using the lowest and highest observed values as \(a\) and \(b\). Both tests therefore borrow from the sample.

Step 4 — Note what that costs. A distribution whose parameters were estimated from the same data being tested will fit that data better than one specified in advance, which makes the test more reluctant to reject. This lab keeps the degrees of freedom at cells minus one for simplicity, but it is the reason a strict treatment subtracts one further degree of freedom per estimated parameter.

Answer: \(m \approx 0.0235\), so \(X \sim Exp(0.0235)\). Both tests estimate their parameters from the sample, which biases each toward fitting.

Step 1 — Find the expected count. Six equal-probability cells over 30 receipts give

$$ E = \frac{30}{6} = 5 $$

which meets the at-least-five condition exactly.

Step 2 — Find the degrees of freedom. \(df = 6 - 1 = 5\), so the distribution for the test is \(\chi_{5}^{2}\).

Step 3 — Consider what equal-width intervals would do. An exponential distribution is heavily right-skewed. Cutting the range into six equal widths would pile most of the data into the first interval and leave the last ones nearly empty, producing expected counts far below five at the top end.

Step 4 — Explain the percentile cuts. Cutting at percentiles makes every cell carry the same probability under the claimed distribution, so every expected count is equal and the condition is satisfied everywhere at once.

Answer: \(E = 5\) per cell with \(df = 5\). Percentile cut points give equal expected counts, while equal-width intervals on a skewed distribution would starve the upper cells.

Step 1 — Decide the uniform test. \(\alpha = 0.05\) and the p-value is 0.31, so \(\alpha <\) p-value. Do not reject \(H_{0}\).

Step 2 — Decide the exponential test. \(\alpha = 0.05\) and the p-value is 0.42, so \(\alpha <\) p-value. Do not reject \(H_{0}\).

Step 3 — State what that does establish. Neither test found sufficient evidence that the receipts fail to fit its distribution.

Step 4 — State what it does not establish. Failing to reject is not proof that a distribution fits. With only 30 observations the test has little power, so it will fail to reject many distributions — including ones the data genuinely do not follow.

Step 5 — Answer the discussion question directly. Two non-rejections do not mean the data follow both distributions. They mean the sample is too small to rule either out.

Answer: Do not reject in both cases. The correct conclusion is that this sample provides insufficient evidence against either distribution, not that the receipts follow both.