We are given descriptive statistics for extroversion scores of English majors (group 1) and History majors (group 2). We have the sample means, sample standard deviations, and sample sizes for each group. We are asked to create a confidence interval for the difference between the means of the two groups and to determine if the confidence interval suggests that the students from the majors differ in extroversion. $\bar{X}_1 = 4.78$, $\bar{X}_2 = 2.53$ $s_1 = 1.60$, $s_2 = 2.15$ $n_1 = 55$, $n_2 = 30$

1. Problem Description

We are given descriptive statistics for extroversion scores of English majors (group 1) and History majors (group 2). We have the sample means, sample standard deviations, and sample sizes for each group. We are asked to create a confidence interval for the difference between the means of the two groups and to determine if the confidence interval suggests that the students from the majors differ in extroversion.

$\bar{X}_1 = 4.78$, $\bar{X}_2 = 2.53$

$s_1 = 1.60$, $s_2 = 2.15$

$n_1 = 55$, $n_2 = 30$

2. Solution Steps

Since the population standard deviations are unknown, we will use a t-distribution to construct the confidence interval for the difference in means. The formula for the confidence interval is:

$(\bar{X}_1 - \bar{X}_2) \pm t_{\alpha/2, df} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}$

First, we calculate the difference in sample means:

$\bar{X}_1 - \bar{X}_2 = 4.78 - 2.53 = 2.25$

Next, we calculate the standard error:

$SE = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} = \sqrt{\frac{1.60^2}{55} + \frac{2.15^2}{30}} = \sqrt{\frac{2.56}{55} + \frac{4.6225}{30}} = \sqrt{0.0465 + 0.1541} = \sqrt{0.2006} \approx 0.4479$

Now we need to find the degrees of freedom. We can use Welch's approximation:

$df = \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1}} = \frac{(\frac{1.6^2}{55} + \frac{2.15^2}{30})^2}{\frac{(\frac{1.6^2}{55})^2}{55-1} + \frac{(\frac{2.15^2}{30})^2}{30-1}} = \frac{(0.0465 + 0.1541)^2}{\frac{0.0465^2}{54} + \frac{0.1541^2}{29}} = \frac{0.2006^2}{\frac{0.00216}{54} + \frac{0.0237}{29}} = \frac{0.04024}{0.00004 + 0.00082} = \frac{0.04024}{0.00086} \approx 46.79$

We can round the degrees of freedom to

4

7. For a 95% confidence interval, $\alpha = 0.05$, so $\alpha/2 = 0.025$. Looking up the t-value for $df = 47$ and $\alpha/2 = 0.025$ in a t-table, we get $t_{0.025, 47} \approx 2.0117$.

Now we can calculate the margin of error:

$ME = t_{\alpha/2, df} \times SE = 2.0117 \times 0.4479 \approx 0.901$

Finally, we calculate the confidence interval:

$(2.25 - 0.901, 2.25 + 0.901) = (1.349, 3.151)$

Since the confidence interval (1.349, 3.151) does not contain zero, we can conclude that there is a significant difference in extroversion between English and History majors.

3. Final Answer

The approximate 95% confidence interval for the difference in means is (1.349, 3.151). Because this interval does not contain zero, it suggests that the students from the majors differ in their extroversion levels.