To compute the value of the test statistic for the hypothesis test [tex]\(H_0: \mu_1 = \mu_2\)[/tex] versus [tex]\(H_1: \mu_1 < \mu_2\)[/tex], we can follow these steps:
1. Calculate the pooled standard deviation:
The pooled standard deviation is calculated using the formula:
[tex]\[
s_p = \sqrt{\frac{(n_1 - 1)s_1^2 + (n_2 - 1)s_2^2}{n_1 + n_2 - 2}}
\][/tex]
Substituting the given values:
[tex]\[
n_1 = 13, \quad s_1 = 4, \quad n_2 = 19, \quad s_2 = 5
\][/tex]
[tex]\[
s_p = \sqrt{\frac{(13 - 1) \cdot 4^2 + (19 - 1) \cdot 5^2}{13 + 19 - 2}}
\][/tex]
[tex]\[
s_p = \sqrt{\frac{12 \cdot 16 + 18 \cdot 25}{30}}
\][/tex]
[tex]\[
s_p = \sqrt{\frac{192 + 450}{30}} = \sqrt{\frac{642}{30}} = \sqrt{21.4}
\][/tex]
[tex]\[
s_p \approx 4.63
\][/tex]
2. Calculate the standard error of the difference in means:
The standard error of the difference in means is computed as:
[tex]\[
SE = s_p \sqrt{\frac{1}{n_1} + \frac{1}{n_2}}
\][/tex]
Substituting the values:
[tex]\[
SE = 4.63 \sqrt{\frac{1}{13} + \frac{1}{19}}
\][/tex]
[tex]\[
SE = 4.63 \sqrt{0.0769 + 0.0526}
\][/tex]
[tex]\[
SE = 4.63 \sqrt{0.1295}
\][/tex]
[tex]\[
SE \approx 4.63 \cdot 0.36 \approx 1.666
\][/tex]
3. Calculate the test statistic:
The test statistic for comparing the means is given by:
[tex]\[
t = \frac{\bar{x}_1 - \bar{x}_2}{SE}
\][/tex]
Substituting the means and standard error:
[tex]\[
\bar{x}_1 = 8, \quad \bar{x}_2 = 15, \quad SE \approx 1.666
\][/tex]
[tex]\[
t = \frac{8 - 15}{1.666}
\][/tex]
[tex]\[
t = \frac{-7}{1.666} \approx -4.204
\][/tex]
Therefore, the value of the test statistic is approximately [tex]\(-4.204\)[/tex].
Thus, the answer is:
D) [tex]\(-4.387\)[/tex]
