Question

use the given statistics to complete parts (a) and (b). assume that the populations are normally distributed.
(a) test whether \\( \mu _ { 1 } > \mu _ { 2 } \\) at the \\( \alpha = 0.01 \\) level of significance for the given sample data.
(b) construct a \\( 99 \\% \\) confidence interval about \\( \mu _ { 1 } - \mu _ { 2 } \\).
(a) identify the null and alternative hypotheses for this test.
a. \\( h _ { 0 } : \mu _ { 1 } > \mu _ { 2 } \\)
\\( h _ { 1 } : \mu _ { 1 } = \mu _ { 2 } \\)
b. \\( h _ { 0 } : \mu _ { 1 } = \mu _ { 2 } \\)
\\( h _ { 1 } : \mu _ { 1 } > \mu _ { 2 } \\)
c. \\( h _ { 0 } : \mu _ { 1 } \
eq \mu _ { 2 } \\)
\\( h _ { 1 } : \mu _ { 1 } = \mu _ { 2 } \\)
d. \\( h _ { 0 } : \mu _ { 1 } = \mu _ { 2 } \\)
\\( h _ { 1 } : \mu _ { 1 } < \mu _ { 2 } \\)
e. \\( h _ { 0 } : \mu _ { 1 } < \mu _ { 2 } \\)
\\( h _ { 1 } : \mu _ { 1 } = \mu _ { 2 } \\)
f. \\( h _ { 0 } : \mu _ { 1 } = \mu _ { 2 } \\)
\\( h _ { 1 } : \mu _ { 1 } \
eq \mu _ { 2 } \\)
find the test statistic for this hypothesis test.
(round to two decimal places as needed.)

Explanation:

Step1: Calculate the test statistic formula

The formula for the test statistic \(t\) for two - sample \(t\) - test (assuming unequal variances) is \(t=\frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}}\)

Given \(\bar{x}_{1} = 45.7\), \(\bar{x}_{2}=44.4\), \(s_{1}=4.9\), \(s_{2}=9.8\), \(n_{1}=26\), \(n_{2}=18\)

Step2: Substitute the values into the formula

First, calculate the numerator: \(\bar{x}_{1}-\bar{x}_{2}=45.7 - 44.4=1.3\)

Then, calculate the denominator:

\(\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}=\frac{4.9^{2}}{26}+\frac{9.8^{2}}{18}=\frac{24.01}{26}+\frac{96.04}{18}\)

\(\frac{24.01}{26}\approx0.9235\), \(\frac{96.04}{18}\approx5.3356\)

\(\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}\approx0.9235 + 5.3356=6.2591\)

\(\sqrt{\frac{s_{1}^{2}}{n_{1}}+\frac{s_{2}^{2}}{n_{2}}}=\sqrt{6.2591}\approx2.50\)

Finally, \(t=\frac{1.3}{2.50}=0.52\)

Answer:

\(0.52\)