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?
\\( \checkmark \\) a. \\( \

$$\begin{array} { l } { h _ { 0 } : \\mu _ { 1 } = \\mu _ { 2 } } \\\\ { h _ { a } : \\mu _ { 1 } \ eq \\mu _ { 2 } } \\end{array}$$

\\)
\\( \bigcirc \\) c. \\( \

$$\begin{array} { l } { h _ { 0 } : \\mu _ { 1 } = \\mu _ { 2 } } \\\\ { h _ { a } : \\mu _ { 1 } < \\mu _ { 2 } } \\end{array}$$

\\)
\\( \bigcirc \\) e. \\( \

$$\begin{array} { l } { h _ { 0 } : \\mu _ { 1 } > \\mu _ { 2 } } \\\\ { h _ { a } : \\mu _ { 1 } = \\mu _ { 2 } } \\end{array}$$

\\)
\\( \bigcirc \\) b. \\( \

$$\begin{array} { l } { h _ { 0 } : \\mu _ { 1 } \ eq \\mu _ { 2 } } \\\\ { h _ { a } : \\mu _ { 1 } = \\mu _ { 2 } } \\end{array}$$

\\)
\\( \bigcirc \\) d. \\( \

$$\begin{array} { l } { h _ { 0 } : \\mu _ { 1 } = \\mu _ { 2 } } \\\\ { h _ { a } : \\mu _ { 1 } > \\mu _ { 2 } } \\end{array}$$

\\)
\\( \bigcirc \\) e. \\( \

$$\begin{array} { l } { h _ { 0 } : \\mu _ { 1 } < \\mu _ { 2 } } \\\\ { h _ { a } : \\mu _ { 1 } = \\mu _ { 2 } } \\end{array}$$

\\)
next, compute the test statistic.
\\( t = - 3.69 \\) (round to three decimal places as needed.)
now determine the critical values.
\\( \pm t _ { \alpha / 2 } = \pm 2.056 \\) (round to three decimal places as needed.)
what is the conclusion of the hypothesis test?
since the test statistic is in the rejection region, reject \\( h _ { 0 } \\).
b. the \\( 95 \\% \\) confidence interval is from \\( \square \\) to \\( \square \\).
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the pooled variance \(s_p^2\)

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

Step2: Calculate the margin of error \(E\)

The formula for the margin of error for a confidence interval is \(E = t_{\alpha/2}s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}\)
We know \(t_{\alpha/2}=2.056\) (from part a, for a 95% confidence interval, \(\alpha=0.05\) and \(\alpha/2 = 0.025\)), \(s_p=\sqrt{4.625}\approx2.151\), \(n_1=n_2 = 14\)
\(\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}=\sqrt{\frac{1}{14}+\frac{1}{14}}=\sqrt{\frac{2}{14}}\approx0.378\)
\(E=2.056\times2.151\times0.378\approx2.056\times0.813\approx1.672\)

Step3: Calculate the confidence interval

The formula for the confidence interval for \(\mu_1-\mu_2\) is \((\bar{x}_1-\bar{x}_2)-E<\mu_1 - \mu_2<(\bar{x}_1-\bar{x}_2)+E\)
We know \(\bar{x}_1 = 13\), \(\bar{x}_2 = 16\), so \(\bar{x}_1-\bar{x}_2=13 - 16=- 3\)
The lower limit is \(-3-1.672=-4.672\)
The upper limit is \(-3 + 1.672=-1.328\)

Answer:

The 95% confidence interval is from \(-4.672\) to \(-1.328\)