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 hi 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.
what are \\( h _ { 0 } \\) and \\( h _ { a } \\) ?
a. \\( h _ { 0 } : \mu _ { 1 } = \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } \
eq \mu _ { 2 } \\)
b. \\( h _ { 0 } : \mu _ { 1 } \geq \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } < \mu _ { 2 } \\)
c. \\( h _ { 0 } : \mu _ { 1 } \leq \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } > \mu _ { 2 } \\)
d. \\( h _ { 0 } : \mu _ { 1 } > \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } \leq \mu _ { 2 } \\)
e. \\( h _ { 0 } : \mu _ { 1 } < \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } \geq \mu _ { 2 } \\)
f. \\( h _ { 0 } : \mu _ { 1 } \
eq \mu _ { 2 } \\)
\\( h _ { a } : \mu _ { 1 } = \mu _ { 2 } \\)
(b) find the critical value(s) and identify the rejection region(s).
the critical value(s) is/are \\( \pm 2.58 \\).
(round to two decimal places as needed. use a comma to separate answers as needed.)
what is/are the rejection region(s)?
a. \\( z < - 1.64 , z > 1.64 \\)
b. \\( z < - 2.33 , z > - 2.33 \\)
c. \\( z > 2.58 \\)
d. \\( z < 1.64 \\)
e. \\( z < - 2.33 \\)
f. \\( z > - 3.08 \\)
g. \\( z < - 2.58 , z > 2.58 \\)
h. \\( z < - 3.08 , z > - 3.08 \\)
(c) find the standardized test statistic \\( z \\) for \\( \mu _ { 1 } - \mu _ { 2 } \\).
\\( z = \square \\) (round to two decimal places as needed.)

Explanation:

Step1: Calculate the difference in means

The difference in sample means is \(\bar{x}_1-\bar{x}_2 = 24.1 - 21.1=3\)

Step2: Calculate the standard error

The formula for the standard error \(SE\) when population standard deviations \(\sigma_1\) and \(\sigma_2\) are known is \(SE=\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}\)
Substitute \(\sigma_1 = 4.5\), \(n_1=45\), \(\sigma_2 = 4.2\), \(n_2 = 52\)

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Step3: Calculate the z - statistic

The formula for the z - statistic is \(z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1-\mu_2)}{SE}\)
Under \(H_0:\mu_1=\mu_2\), so \((\mu_1 - \mu_2)=0\)
\(z=\frac{3 - 0}{0.89}\approx3.37\)

Answer:

\(z = 3.37\)