Question

a college student is interested in investigating the claim that students who graduate with a masters degree earn higher salaries, on average, than those who finish with a bachelors degree. she surveys, at random, 34 recent graduates who completed their masters degrees, and finds that their mean salary is $39,800 per year. the standard deviation of annual salaries for the population of recent graduates who have masters degrees is known to be $900. she also surveys, at random, 31 recent graduates who completed their bachelors degrees, and finds that their mean salary is $39,000 per year. the standard deviation of annual salaries for the population of recent graduates with only bachelors degrees is known to be $1600. test the claim at the 0.05 level of significance. let recent graduates with a masters degree be population 1 and let recent graduates with a bachelors degree be population 2. step 2 of 3: compute the value of the test statistic. round your answer to two decimal places.

Explanation:

Step1: Identify the formula for two - sample z - test statistic

The formula for the two - sample z - test statistic when population standard deviations $\sigma_1$ and $\sigma_2$ are known is $z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}$. Under the null hypothesis $H_0:\mu_1-\mu_2 = 0$.

Step2: Define the given values

We have $\bar{x}_1 = 39800$, $\bar{x}_2=39000$, $\sigma_1 = 900$, $\sigma_2 = 1600$, $n_1 = 34$, $n_2=31$, and $\mu_1-\mu_2 = 0$ (under $H_0$).

Step3: Substitute the values into the formula

\[

$$\begin{align*} z&=\frac{(39800 - 39000)-0}{\sqrt{\frac{900^{2}}{34}+\frac{1600^{2}}{31}}}\\ &=\frac{800}{\sqrt{\frac{810000}{34}+\frac{2560000}{31}}}\\ &=\frac{800}{\sqrt{23823.53+82580.65}}\\ &=\frac{800}{\sqrt{106404.18}}\\ &=\frac{800}{326.196}\\ &\approx2.45 \end{align*}$$

\]

Answer:

$2.45$