Question

fran is training for her first marathon, and she wants to know if there is a significant difference between the mean number of miles run each week by group runners and individual runners who are training for marathons. she interviews 34 randomly selected people who train in groups, and finds that they run a mean of 41.4 miles per week. assume that the population standard deviation for group runners is known to be 2.6 miles per week. she also interviews a random sample of 42 people who train on their own and finds that they run a mean of 42.4 miles per week. assume that the population standard deviation for people who run by themselves is 1.4 miles per week. test the claim at the 0.10 level of significance. let group runners training for marathons be population 1 and let individual runners training for marathons 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 $z=\frac{(\bar{x}_1-\bar{x}_2)-(\mu_1 - \mu_2)}{\sqrt{\frac{\sigma_1^{2}}{n_1}+\frac{\sigma_2^{2}}{n_2}}}$. Since the claim is to test if there is 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 are given $\bar{x}_1 = 41.4$, $\sigma_1=2.6$, $n_1 = 34$, $\bar{x}_2=42.4$, $\sigma_2 = 1.4$, $n_2=42$. Substituting into the formula:
\[

$$\begin{align*} z&=\frac{(41.4 - 42.4)-0}{\sqrt{\frac{2.6^{2}}{34}+\frac{1.4^{2}}{42}}}\\ &=\frac{- 1}{\sqrt{\frac{6.76}{34}+\frac{1.96}{42}}}\\ &=\frac{-1}{\sqrt{0.1988 + 0.0467}}\\ &=\frac{-1}{\sqrt{0.2455}}\\ &=\frac{-1}{0.4955}\\ &\approx - 2.02 \end{align*}$$

\]

Answer:

$-2.02$