Statistics · Hypothesis test

Chi-Square Goodness of Fit: Are Coffee Sales Even Across Weekdays?

Testing whether observed daily sales differ from a uniform expectation

Difficulty 5/10~25 min to solvePublished Jun 2026Donated by students

Chi-square testGoodness-of-fitHypothesis testingp-value interpretation

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The problem

A coffee shop manager wants to know whether daily sales are evenly distributed across the five weekdays (Monday through Friday) or whether some days are significantly busier than others. The shop tracked sales for one randomly selected week and recorded the following number of cups sold per day: Monday 110, Tuesday 95, Wednesday 105, Thursday 120, Friday 170. Test, at the 5% significance level, the claim that daily sales are uniformly distributed across the five weekdays.

Step 1 — State the hypotheses

The claim being tested is that sales are uniformly distributed across the five weekdays. That gives us:

$H_0$ : Daily sales are uniformly distributed across the five weekdays.
$H_a$ : At least one weekday has a different sales rate.

Significance level: $\alpha = 0.05$ .

Step 2 — Compute the expected counts

Total cups sold over the week:

n = 110 + 95 + 105 + 120 + 170 = 600

Under $H_0$ (uniform distribution across $k = 5$ weekdays), the expected count for each day is the total divided by the number of categories:

E_i = \frac{n}{k} = \frac{600}{5} = 120 \text{ cups per day}

Step 3 — Compute the chi-square statistic

The test statistic is:

\chi^2 = \sum_{i=1}^{k} \frac{(O_i - E_i)^2}{E_i}

Plugging in each weekday:

Day	$O_i$	$E_i$	$O_i - E_i$	$(O_i - E_i)^2$	$(O_i - E_i)^2 / E_i$
Monday	110	120	−10	100	0.833
Tuesday	95	120	−25	625	5.208
Wednesday	105	120	−15	225	1.875
Thursday	120	120	0	0	0.000
Friday	170	120	50	2500	20.833
Sum					28.749

\chi^2 = 28.749

Step 4 — Determine the critical value

Degrees of freedom for a goodness-of-fit test:

df = k - 1 = 5 - 1 = 4

From a chi-square table at $\alpha = 0.05$ and $df = 4$ :

\chi^2_{\text{critical}} = 9.488

Step 5 — Decision and interpretation

Compare:

\chi^2 = 28.749 > 9.488 = \chi^2_{\text{critical}}

Since the test statistic exceeds the critical value, reject $H_0$ at the 5% significance level.

Conclusion

The data provide strong evidence that daily coffee sales are not uniformly distributed across the five weekdays. The largest deviation comes from Friday (170 observed vs. 120 expected), which alone contributes 20.8 out of the 28.7 total chi-square statistic. The manager should investigate what makes Friday different — pricing, foot traffic, promotions — rather than assume weekdays are interchangeable for staffing or inventory planning.

Read with the expert

Difficulty 5/102 techniques4 mistakes to watch4 expert's notes

Difficulty

5/10

Standard intro chi-square problem. Catch is recognizing "uniformly distributed" means equal expected counts across categories.

Techniques used

01
Chi-square goodness-of-fit test
Use this when comparing observed categorical counts to a theoretical distribution. Here, observed vs. uniform expected.
02
p-value interpretation
After computing the statistic, compare to the critical value at df = k - 1 and α = 0.05 to decide whether to reject H0.

Watch for

Computing expected counts as 0.20 (the probability) instead of total/k (the count). The chi-square statistic uses counts, not proportions.
Using degrees of freedom = k instead of k - 1. For goodness-of-fit, df = number of categories minus one.
Stopping at the test statistic without comparing it to the critical value or computing a p-value. The test conclusion requires both.

Expert's notes

Difficulty

5/10

Standard intro chi-square problem. Catch is recognizing "uniformly distributed" means equal expected counts across categories.

Techniques used

01
Chi-square goodness-of-fit test
Use this when comparing observed categorical counts to a theoretical distribution. Here, observed vs. uniform expected.
02
p-value interpretation
After computing the statistic, compare to the critical value at df = k - 1 and α = 0.05 to decide whether to reject H0.

Watch for

Computing expected counts as 0.20 (the probability) instead of total/k (the count). The chi-square statistic uses counts, not proportions.
Using degrees of freedom = k instead of k - 1. For goodness-of-fit, df = number of categories minus one.
Stopping at the test statistic without comparing it to the critical value or computing a p-value. The test conclusion requires both.

Expert's notes

Editor's analysis

What this solution demonstrates, and what to watch out for.

Chi-square goodness-of-fit test. Use this when comparing observed categorical counts to a theoretical distribution. Here, observed vs. uniform expected.

p-value interpretation. After computing the statistic, compare to the critical value at df = k - 1 and α = 0.05 to decide whether to reject H0.

Computing expected counts as 0.20 (the probability) instead of total/k (the count). The chi-square statistic uses counts, not proportions.

Using degrees of freedom = k instead of k - 1. For goodness-of-fit, df = number of categories minus one.

Stopping at the test statistic without comparing it to the critical value or computing a p-value. The test conclusion requires both.

Forgetting to state the conclusion in context. 'Reject H0' is necessary but not sufficient — say what it means for the coffee shop.

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