Mathematics: A researcher was interested in comparing the resting heart rate of people who exercise regularly and people who do not exercise regularly. Independent simple random samples of 16 people ages who do not exercise regularly and 12 people ages two who do exercise regular were selected and the resting heart rate of each person was measured. The summary statistics are as follows. Do not exercise Do exercise x_1 = 73.5 x_2 = 69.3 s_1 = 10.2 s_2 = 8.7 n_1 = 16 n_2 = 12 (a) Use a 0.025 significance level

Check the attached for the solutionStep-by-step explanation:...

A researcher was interested in comparing the resting

Ответ:

Loowkeychrissy7227

28.01.2022

$t=\frac{(73.5-69.3)-0}{\sqrt{\frac{10.2^2}{16}+\frac{8.7^2}{12}}}}=1.173$

Using the critical value method we need to find a critical value in the t distribution who accumulates 0.025 of the area on the right and we got:

$t_{cric}= 2.06$

So then since our calculated value is lower then the critical value we fail to reject the null hypothesis and we can't conclude that the true mean resting pulse rate of people who do not exercise regularly is larger than the mean resting pulse rate of people who exercise regularly.

Step-by-step explanation:

Information given

$\bar X_{1}=73.5$ represent the mean for sample of people who do not exercise

$\bar X_{2}=69.3$ represent the mean for sample of people who do exercise

$s_{1}=10.2$ represent the sample standard deviation for 1

$s_{2}=8.7$ represent the sample standard deviation for 2

$n_{1}=16$ sample size for the group 1

$n_{2}=12$ sample size for the group 2

$\alpha=0.01$ Significance level

t would represent the statistic

System of hypothesis

We want to verify if the mean resting pulse rate of people who do not exercise regularly is larger than the mean resting pulse rate of people who exercise regularly, the system of hypothesis would be:

Null hypothesis: $\mu_{1}-\mu_{2} \leq 0$