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use the given statistics to complete parts (a) and (b). assume that the…

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 } \\)
0.52 (round to two decimal places as needed)
determine the p - value for this hypothesis test.
0.305 (round to three decimal places as needed)
state the conclusion for this hypothesis test.
a. reject \\( h _ { 0 } \\). there is not sufficient evidence at the \\( \alpha = 0.01 \\) level of significance to conclude that \\( \mu _ { 1 } > \mu _ { 2 } \\)
b. reject \\( h _ { 0 } \\). there is sufficient evidence at the \\( \alpha = 0.01 \\) level of significance to conclude that \\( \mu _ { 1 } > \mu _ { 2 } \\)
c. do not reject \\( h _ { 0 } \\). there is sufficient evidence at the \\( \alpha = 0.01 \\) level of significance to conclude that \\( \mu _ { 1 } > \mu _ { 2 } \\)
d. do not reject \\( h _ { 0 } \\). there is not sufficient evidence at the \\( \alpha = 0.01 \\) level of significance to conclude that \\( \mu _ { 1 } > \mu _ { 2 } \\)
(b) the \\( 99 \\% \\) confidence interval about \\( \mu _ { 1 } - \mu _ { 2 } \\) is the range from a lower bound of \\( \square \\) to an upper bound of \\( \square \\)
(round to three decimal places as needed.)

Explanation:

Step1: Calculate the degrees of freedom

For two - sample \(t\) - test, the degrees of freedom \(df=\min(n_1 - 1,n_2 - 1)\). Here \(n_1 = 26\), \(n_2=18\). So \(df=\min(26 - 1,18 - 1)=17\)

Step2: Determine the critical value for confidence interval

For a \(99\%\) confidence interval and \(df = 17\), the critical value \(t_{\alpha/2}\) (where \(\alpha=1 - 0.99 = 0.01\), \(\alpha/2=0.005\)) from the \(t\) - distribution table is \(t_{0.005,17}=2.898\)

Step3: Calculate the standard error

The formula for the standard error \(SE=\sqrt{\frac{s_1^{2}}{n_1}+\frac{s_2^{2}}{n_2}}\), where \(s_1 = 4.0\), \(n_1 = 26\), \(s_2=9.8\), \(n_2 = 18\)

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Step4: Calculate the lower and upper bounds of the confidence interval

The point estimate is \(\bar{x}_1-\bar{x}_2=45.7 - 44.4 = 1.3\)
The lower bound \(LB=(\bar{x}_1-\bar{x}_2)-t_{\alpha/2}\times SE\)

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The upper bound \(UB=(\bar{x}_1-\bar{x}_2)+t_{\alpha/2}\times SE\)

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Answer:

The \(99\%\) confidence interval about \(\mu_1-\mu_2\) is the range from a lower bound of \(-5.771\) to an upper bound of \(8.371\)