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. (a) options: a. h₀: μ₁ ≥ μ₂, hₐ: μ₁ < μ₂; b. h₀: μ₁ = μ₂, hₐ: μ₁ ≠ μ₂; c. h₀: μ₁ ≤ μ₂, hₐ: μ₁ > μ₂; d. h₀: μ₁ ≠ μ₂, hₐ: μ₁ = μ₂; e. h₀: μ₁ > μ₂, hₐ: μ₁ ≤ μ₂; f. h₀: μ₁ < μ₂, hₐ: μ₁ ≥ μ₂ (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

Explanation:

Step1: Determine Test Type

This is a two - tailed z - test (since \(H_a:\mu_1
eq\mu_2\)) with \(\alpha = 0.01\). For a two - tailed test, we split \(\alpha\) into two equal parts: \(\alpha/2=0.005\) in each tail.

Step2: Find Critical Values

We look up the z - value that corresponds to a cumulative probability of \(1 - 0.005=0.995\) (for the right - tail) and \(0.005\) (for the left - tail) in the standard normal distribution table. The z - value for a cumulative probability of \(0.995\) is \(z = 2.58\) and for \(0.005\) is \(z=- 2.58\).

Step3: Identify Rejection Regions

In a two - tailed z - test, we reject the null hypothesis when the test statistic \(z\) is less than the left - tailed critical value (\(z < - 2.58\)) or greater than the right - tailed critical value (\(z>2.58\)).

Answer:

D. \(z < - 2.58,z>2.58\)