To determine the $p$-value for this test, let's proceed with a hypothesis test comparing two means from independent samples, specifically focusing on whether high school girls miss more days of school than boys. We will use a one-tailed t-test for independent means.

Given dataTo determine the $p$-value for this test, let's proceed with a hypothesis test comparing two means from independent samples, specifically focusing on whether high school girls miss more days of school than boys. We will use a one-tailed t-test for independent means.

Given

• Sample size for girls ($n_1$) = 16

• Mean for girls ($\bar{x}_1$) = 3.9

• Standard deviation for girls ($s_1$) = 0.6

• Sample size for boys ($n_2$) = 22

• Mean for boys ($\bar{x}_2$) = 3.6

• Standard deviation for boys ($s_2$) = 0.8

• Significance level ($\alpha$) = 0.01

Step 1: Set up the hypotheses

• Null Hypothesis ($H_0$): $\mu_1 = \mu_2$ (There is no difference in the average days missed between girls and boys).
• Alternative Hypothesis ($H_a$): $\mu_1 > \mu_2$ (Girls miss more days than boys).

Step 2: Calculate the test statistic

The test statistic for comparing two means with independent samples is given by: $t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}}$

Substituting the values:

1. Difference in sample means: $\bar{x}_1 - \bar{x}_2 = 3.9 - 3.6 = 0.3$

2. Standard error: $\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{0.6^2}{16} + \frac{0.8^2}{22}} = \sqrt{\frac{0.36}{16} + \frac{0.64}{22}}$ Calculating this gives: $\sqrt{0.0225 + 0.0291} = \sqrt{0.0516} \approx 0.2272$

3. Test statistic $t$: $t = \frac{0.3}{0.2272} \approx 1.32$

Step 3: Determine degrees of freedom

The degrees of freedom $df$ for this test can be approximated using the formula: $df = \frac{\left(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}\right)^2}{\frac{\left(\frac{s_1^2}{n_1}\right)^2}{n_1 - 1} + \frac{\left(\frac{s_2^2}{n_2}\right)^2}{n_2 - 1}}$ Using this formula, we calculate $df \approx 32$ (approximately).

Step 4: Calculate the $p$-value

Using a t-distribution table or calculator for a one-tailed test with $t = 1.32$ and $df \approx 32$, we find that:

$p \approx 0.097$

Conclusion

With $p = 0.097$ and $\alpha = 0.01$, we fail to reject the null hypothesis because $p > \alpha$. There is insufficient evidence to support the claim that high school girls miss more days than high school boys at the 1% significance level.

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Would you like more details on any specific part of this calculation? Here are some related questions you might consider:

1. What are the implications of failing to reject the null hypothesis?
2. How would the conclusion change with a higher significance level, like 0.05?
3. How is the degrees of freedom calculated, and why does it use the approximate method?
4. What assumptions are required for a two-sample t-test to be valid?
5. How could the sample size affect the reliability of this test?

Tip: When performing hypothesis tests, always verify if assumptions (like normality and independence) are met to ensure the results are valid.