Let's analyze Megan's hypothesis test step by step:

Part (a): Null and Alternative Hypotheses

We aim to test if there is a noticeable difference in average commute times between the two high schools. The hypotheses are:

• Null Hypothesis ($H_0$): The average commute times are equal.
  $H_0: \mu_1 = \mu_2$

• Alternative Hypothesis ($H_a$): The average commute times are not equal (two-tailed test).
  $H_a: \mu_1 \neq \mu_2$

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Part (b): Test Statistic

The test statistic for comparing two sample means assuming equal variances is computed using the formula:

$t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{S_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$

Where:

• $\bar{x}_1 = 21.0$, $\bar{x}_2 = 17.1$ (sample means)
• $s_1 = 3.6$, $s_2 = 4.4$ (sample standard deviations)
• $n_1 = n_2 = 51$ (sample sizes)
• $S_p^2$ is the pooled variance, calculated as: $S_p^2 = \frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}$

Step 1: Calculate the pooled variance ($S_p^2$)

$S_p^2 = \frac{(51 - 1)(3.6^2) + (51 - 1)(4.4^2)}{51 + 51 - 2}$ $S_p^2 = \frac{50(12.96) + 50(19.36)}{100} = \frac{648 + 968}{100} = 16.16$

Step 2: Calculate the standard error ($SE$)

$SE = \sqrt{S_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)} = \sqrt{16.16 \left(\frac{1}{51} + \frac{1}{51}\right)}$ $SE = \sqrt{16.16 \times \frac{2}{51}} = \sqrt{0.6337} \approx 0.796$

Step 3: Compute the $t$-statistic

$t = \frac{\bar{x}_1 - \bar{x}_2}{SE} = \frac{21.0 - 17.1}{0.796} = \frac{3.9}{0.796} \approx 4.899$

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Part (c): $p$-Value Range

From a $t$-distribution table with $df = n_1 + n_2 - 2 = 100$, and a $t$-statistic of $4.899$, the $p$-value is extremely small: $0.01 < p\text{-value} < 0.02$

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Part (d): Decision on $H_0$

Since the $p$-value is smaller than the significance level ($\alpha = 0.1$), we reject the null hypothesis.

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Part (e): Conclusion

The data provides sufficient evidence to conclude that the average commute time for students at High School 1 is noticeably different from that of High School 2.

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Summary of Results:

1. (a) $H_0: \mu_1 = \mu_2$; $H_a: \mu_1 \neq \mu_2$
2. (b) $t = 4.899$
3. (c) $0.01 < p\text{-value} < 0.02$
4. (d) Reject $H_0$
5. (e) The data supports the conclusion that commute times are different.

Do you need further explanation or have any questions?

Related Questions:

1. How would the result change if the significance level was $0.01$?
2. What assumptions must hold for this test to be valid?
3. What would the $t$-statistic be if the sample sizes were unequal?
4. How is the pooled variance calculated in hypothesis testing?
5. What is the effect of larger sample sizes on the test's power?

Tip:

When performing hypothesis tests, always confirm whether the assumptions of the test (e.g., equal variances, normality) are reasonable for the given data.