Question

two samples are randomly selected from each population. the sample statistics to test the claim that $mu_1 > mu_2$. use $alpha = 0.05$.
$n_1 = 100, n_2 = 125, \bar{x}_1 = 615, \bar{x}_2 = 600, sigma_1 = 40, sigma_2 = 24$

a. 0.0005
b. 0.5105
c. 0.1015
d. 0.0505

Explanation:

Step1: Identify the test type

This is a two - sample z - test for the difference between two population means since the population standard deviations ($\sigma_1$ and $\sigma_2$) are known. The formula for the z - statistic in a two - sample z - test is $z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}$. The claim is $\mu_1>\mu_2$, so the null hypothesis $H_0:\mu_1=\mu_2$ (i.e., $\mu_1-\mu_2 = 0$) and the alternative hypothesis $H_a:\mu_1>\mu_2$.

Step2: Calculate the z - statistic

First, substitute the given values into the formula. We have $\bar{x}_1 = 615$, $\bar{x}_2=600$, $\sigma_1 = 40$, $\sigma_2=24$, $n_1 = 100$, $n_2 = 125$, and $(\mu_1-\mu_2)=0$.

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Step3: Find the p - value

Since the alternative hypothesis is one - tailed ( $\mu_1>\mu_2$), we need to find $P(Z > 3.30)$ where $Z$ follows the standard normal distribution. Looking up in the standard normal table or using a calculator, $P(Z\leq3.30)\approx0.9995$, so $P(Z > 3.30)=1 - 0.9995 = 0.0005$.

Answer:

A. 0.0005