Question

two friends, karen and jodi, work different shifts for the same ambulance service. they wonder if the different shifts average different numbers of calls. looking at past records, karen determines from a random sample of 34 shifts that she had a mean of 5.3 calls per shift. she knows that the population standard deviation for her shift is 1.4 calls. jodi calculates from a random sample of 41 shifts that her mean was 4.7 calls per shift. she knows that the population standard deviation for her shift is 1.1 calls. test the claim that there is a difference between the mean numbers of calls for the two shifts at the 0.02 level of significance. let karens shifts be population 1 and let jodis shifts 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 the population standard deviations $\sigma_1$ and $\sigma_2$ are known is $z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}$. Since we are testing the claim of a difference between the means, the null hypothesis $H_0:\mu_1=\mu_2$ (so $\mu_1 - \mu_2 = 0$).

Step2: Substitute the given values

We have $\bar{x}_1 = 5.3$, $n_1=34$, $\sigma_1 = 1.4$, $\bar{x}_2 = 4.7$, $n_2 = 41$, $\sigma_2=1.1$ and $\mu_1-\mu_2 = 0$.
\[

$$\begin{align*} z&=\frac{(5.3 - 4.7)-0}{\sqrt{\frac{1.4^{2}}{34}+\frac{1.1^{2}}{41}}}\\ &=\frac{0.6}{\sqrt{\frac{1.96}{34}+\frac{1.21}{41}}}\\ &=\frac{0.6}{\sqrt{0.057647+0.029512}}\\ &=\frac{0.6}{\sqrt{0.087159}}\\ &=\frac{0.6}{0.2952}\\ &\approx 2.03 \end{align*}$$

\]

Answer:

$2.03$