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To solve this problem, we are tasked with conducting a two-sample hypothesis test to compare the population means of two supermarket chains, assuming normality in the shopping expenses. Here's how we approach it step by step:

### Given Data:
- **Chain 1 (Miller's)**:
  - Sample size: $n_1 = 10$
  - Sample mean: $\bar{x}_1 = 121.92$
  - Sample standard deviation: $s_1 = 1.40$

- **Chain 2 (Albert's)**:
  - Sample size: $n_2 = 10$
  - Sample mean: $\bar{x}_2 = 114.81$
  - Sample standard deviation: $s_2 = 1.84$

- Significance level: $\alpha = 0.05$

We are testing whether the population standard deviations differ by conducting a two-sample $t$-test. Now, let’s break down the steps:

### (a) Calculate the test statistic

The test statistic for a two-sample $t$-test assuming unequal variances (Welch's $t$-test) is given by:

\[
t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}
\]

Plugging in the values:

\[
t = \frac{121.92 - 114.81}{\sqrt{\frac{1.40^2}{10} + \frac{1.84^2}{10}}}
\]
\[
t = \frac{7.11}{\sqrt{\frac{1.96}{10} + \frac{3.3856}{10}}}
\]
\[
t = \frac{7.11}{\sqrt{0.196 + 0.33856}} = \frac{7.11}{\sqrt{0.53456}} = \frac{7.11}{0.7312} \approx 9.72
\]

So, the test statistic $t \approx 9.72$.

### (b) Calculate the critical value

For a two-tailed test at a significance level of $\alpha = 0.05$ with degrees of freedom (calculated using the Welch-Satterthwaite equation), we need to estimate the critical value.

The degrees of freedom $df$ for Welch's test is approximated by:

\[
df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1-1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2-1}}
\]

Substituting the values:

\[
df = \frac{\left(0.196 + 0.33856\right)^2}{\frac{(0.196)^2}{9} + \frac{(0.33856)^2}{9}} = \frac{0.53456^2}{\frac{0.038416}{9} + \frac{0.114633}{9}}
\]
\[
df = \frac{0.28576}{0.004268 + 0.012737} \approx \frac{0.28576}{0.017005} \approx 16.8
\]

Using a $t$-distribution table or calculator for a two-tailed test at $\alpha = 0.05$ with approximately 17 degrees of freedom, the critical value is around $t_{0.025} \approx 2.11$.

### (c) Conclusion

Since the calculated $t$-statistic (9.72) is much larger than the critical value (2.11), we **reject the null hypothesis** $H_0$.

### Summary:
- **(a)** Test statistic: $ t \approx 9.72 $
- **(b)** Critical value: $ t_{0.025} \approx 2.11 $
- **(c)** Conclusion: **Reject** $H_0$

Would you like further clarification or detailed calculations? Here are 5 questions to further explore this topic:

1. How is the test statistic formula derived for unequal variances?
2. What is the difference between a pooled $t$-test and Welch's $t$-test?
3. Why is it important to check the assumption of normality in hypothesis testing?
4. What are the implications of rejecting the null hypothesis in this context?
5. How can we interpret degrees of freedom in a Welch's $t$-test?

**Tip:** Always check whether the assumptions for the statistical test are met, such as normality and equal variances.

A marketing research firm wishes to compare the prices charged by two supermarket chains—Miller’s and Albert’s. Using a standardized one-week shopping plan, the firm makes identical purchases at 10 stores of each chain. It is found that the mean and the standard deviation of the shopping expenses at the 10 Miller’s stores are x1 = $121.92 and s1 = 1.40, while at the 10 Albert’s stores, the mean and standard deviation are x2 = $114.81 and s2 = 1.84. Test if the corresponding population standard deviations differ by setting alpha equal to 0.05. Calculate the test statistic, critical value, and conclusion.

Explore a two-sample hypothesis test comparing prices between Miller's and Albert's supermarkets. This problem involves calculating the test statistic, critical value, and determining the conclusion based on the significance level of 0.05. Suitable for undergraduate statistics students.

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