Question

provided below are summary statistics for independent simple random samples from two populations. use the pooled t - test and the pooled t - interval procedure to conduct the required hypothesis test and obtain the specified confidence interval.

$overline{x}_{1}=13, s_{1}=2.2, n_{1}=14, overline{x}_{2}=16, s_{2}=2.1, n_{2}=14$

a. two - tailed test, $\alpha = 0.05$
b. $95\\%$ confidence interval

a. first, what are the correct hypotheses for a two - tailed test?

a. $h_{0}:\mu_{1}=\mu_{2}$
$h_{a}:\mu_{1}\
eq\mu_{2}$
b. $h_{0}:\mu_{1}\
eq\mu_{2}$
$h_{a}:\mu_{1}=\mu_{2}$

c. $h_{0}:\mu_{1}=\mu_{2}$
$h_{a}:\mu_{1}<\mu_{2}$
d. $h_{0}:\mu_{1}=\mu_{2}$
$h_{a}:\mu_{1}>\mu_{2}$

e. $h_{0}:\mu_{1}>\mu_{2}$
$h_{a}:\mu_{1}=\mu_{2}$
f. $h_{0}:\mu_{1}<\mu_{2}$
$h_{a}:\mu_{1}=\mu_{2}$

next, compute the test statistic.

$t = \square$ (round to three decimal places as needed)

Explanation:

Step1: Calculate the pooled variance

The formula for pooled variance \(s_p^2=\frac{(n_1 - 1)s_1^2+(n_2 - 1)s_2^2}{n_1 + n_2-2}\)
Substitute \(n_1 = 14\), \(s_1 = 2.2\), \(n_2 = 14\), \(s_2=2.1\)
\(s_p^2=\frac{(14 - 1)\times(2.2)^2+(14 - 1)\times(2.1)^2}{14 + 14-2}\)
\(=\frac{13\times4.84+13\times4.41}{26}\)
\(=\frac{62.92+57.33}{26}=\frac{120.25}{26}\approx4.625\)

Step2: Calculate the test statistic

The formula for the pooled \(t -\)test statistic \(t=\frac{\bar{x}_1-\bar{x}_2}{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\)
Substitute \(\bar{x}_1 = 13\), \(\bar{x}_2 = 16\), \(s_p=\sqrt{4.625}\approx2.15\), \(n_1 = 14\), \(n_2 = 14\)
\(t=\frac{13 - 16}{2.15\sqrt{\frac{1}{14}+\frac{1}{14}}}\)
\(=\frac{- 3}{2.15\sqrt{\frac{2}{14}}}\)
\(=\frac{-3}{2.15\times\sqrt{\frac{1}{7}}}\)
\(=\frac{-3}{2.15\times0.378}\)
\(=\frac{-3}{0.813}\approx - 3.69\)

Answer:

\(t\approx - 3.69\)