Question

the mean exam score for 43 male high school students is 21.5 and the population standard deviation is 4.9. the mean exam score for 52 female high school students is 20.1 and the population standard deviation is 4.1. at α = 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.

○ d. $h_0: \mu_1 \
eq \mu_2$ $h_a: \mu_1 = \mu_2$ ○ e. $h_0: \mu_1 > \mu_2$ $h_a: \mu_1 \leq \mu_2$ ○ f. $h_0: \mu_1 < \mu_2$ $h_a: \mu_1 \geq \mu_2$

(b) find the critical value(s) and identify the rejection region(s).
the critical value(s) is/are -2.58, 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. $z > -3.08$ ○ b. $z < -1.64, z > 1.64$ ○ c. $z > 2.58$ ○ d. $z < -2.58, z > 2.58$ ○ e. $z < -2.33$ ○ f. $z < -3.08, z > -3.08$ ○ g. $z < 1.64$ ○ h. $z < -2.33, z > -2.33$

(c) find the standardized test statistic $z$ for $\mu_1 - \mu_2$.
$z = \square$ (round to two decimal places as needed.)

Explanation:

Step1: Recall z - test formula for two means

The formula for the standardized test statistic \( z \) for the difference between two population means (\( \mu_1-\mu_2 \)) when the population standard deviations (\( \sigma_1,\sigma_2 \)) are known is:

$$ z=\frac{(\bar{x}_1 - \bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}} $$

Here, the claim is that \( \mu_1=\mu_2 \), so \( \mu_1 - \mu_2 = 0 \). Let \( \bar{x}_1 = 21.5 \) (mean for males), \( n_1 = 43 \), \( \sigma_1=4.9 \); \( \bar{x}_2 = 20.1 \) (mean for females), \( n_2 = 52 \), \( \sigma_2 = 4.1 \).

Step2: Substitute values into the formula

First, calculate the numerator: \( (\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)=(21.5 - 20.1)-0 = 1.4 \)
Then, calculate the denominator:

$$ \sqrt{\frac{\sigma_1^2}{n_1}+\frac{\sigma_2^2}{n_2}}=\sqrt{\frac{4.9^2}{43}+\frac{4.1^2}{52}} $$

Calculate \( \frac{4.9^2}{43}=\frac{24.01}{43}\approx0.5584 \)
Calculate \( \frac{4.1^2}{52}=\frac{16.81}{52}\approx0.3233 \)
Sum inside the square root: \( 0.5584 + 0.3233=0.8817 \)
Take the square root: \( \sqrt{0.8817}\approx0.939 \)

Step3: Calculate the z - statistic

Now, \( z=\frac{1.4}{0.939}\approx1.49 \) (rounded to two decimal places)

Answer:

\( 1.49 \)