Question

the mean exam score for 45 male high school students is 24.1 and the population standard deviation is 4.5. the mean exam score for 52 female high school students is 21.1 and the population standard deviation is 4.2. at \\( \alpha = 0.01 \\), can you reject the claim that male and female high school students have equal exam scores? complete parts (a) through (e). click here to view page 1 of the standard normal distribution table. click here to view page 2 of the standard normal distribution table. a. male high school students have lower exam scores than female students. b. male and female high school students have equal exam scores. c. male high school students have greater exam scores than female students. d. male and female high school students have different exam scores. what are \\( h_0 \\) and \\( h_a \\)? a. \\( \begin{array} { l } { h _ { 0 } : mu _ { 1 } = mu _ { 2 } } \\\\ { h _ { a } : mu _ { 1 }
eq mu _ { 2 } } end{array} \\) b. \\( \begin{array} { l } { h _ { 0 } : mu _ { 1 } geq mu _ { 2 } } \\\\ { h _ { a } : mu _ { 1 } < mu _ { 2 } } end{array} \\) c. \\( \begin{array} { l } { h _ { 0 } : mu _ { 1 } leq mu _ { 2 } } \\\\ { h _ { a } : mu _ { 1 } > mu _ { 2 } } end{array} \\) d. \\( \begin{array} { l } { h _ { 0 } : mu _ { 1 } > mu _ { 2 } } \\\\ { h _ { a } : mu _ { 1 } leq mu _ { 2 } } end{array} \\) e. \\( \begin{array} { l } { h _ { 0 } : mu _ { 1 } < mu _ { 2 } } \\\\ { h _ { a } : mu _ { 1 } geq mu _ { 2 } } end{array} \\) f. \\( \begin{array} { l } { h _ { 0 } : mu _ { 1 }
eq mu _ { 2 } } \\\\ { h _ { a } : mu _ { 1 } = mu _ { 2 } } end{array} \\) (b) find the critical value(s) and identify the rejection region(s). the critical value(s) is/are (round to two decimal places as needed. use a comma to separate answers as needed.)

Explanation:

Step1: Determine the type of test

Since the claim is that \(\mu_1=\mu_2\) (male and female have equal scores) and the alternative hypothesis is \(\mu_1
eq\mu_2\), this is a two - tailed test.

Step2: Find the critical values

For a two - tailed test with \(\alpha = 0.01\), the significance level is split between the two tails. So \(\frac{\alpha}{2}=0.005\) in each tail.
Looking up in the standard normal distribution table (z - table), the z - score corresponding to an area of \(1 - 0.005=0.995\) is \(z = 2.58\) and the z - score corresponding to an area of \(0.005\) is \(z=- 2.58\)

Answer:

\(-2.58,2.58\)