Question

As part of a study on pulse rates, a random sample of 10 study participants had their pulse rate taken and asked to sit quietly for 5 minutes. At the end of 5 minutes, their pulse rates were taken again. The mean difference between the two pulse rates for the 10 participants is -1.2 beats per minute with standard deviation of 2.97. Researchers are interested in testing whether or not the mean difference in pulse rates before and after the 5 minute waiting period is 0. Based on the t-test statistic and p-value, researchers found little to no evidence against the null hypothesis. Give the final statement of the hypothesis test.
a. There is enough evidence to conclude the mean difference in pulse rates before and after the 5 minute waiting period is 0.
b. There is enough evidence to conclude the mean difference in pulse rates before and after the 5 minute waiting period is different from 0.
c. There is not enough evidence to conclude the mean difference in pulse rates before and after the 5 minute waiting period is 0.
d. There is not enough evidence to conclude the mean difference in pulse rates before and after the 5 minute waiting period is different from 0.

Answer 1

Answer:

$t = \frac{-1.2-0}{\frac{2.97}{\sqrt{10}}}= -1.278$

The degrees of freedom are :

$df=n-1=10-1=9$

$p_v =2*P(t_{9}$

Since the p value is a very high value is higher than the typical values for the significance level (0.05,0.1) so then we have enough evidence to FAIL to reject the null hypothesis and there is no enough evidence to conclude that the true difference in the means is different from 0. And the best conclusion would be:

d. There is not enough evidence to conclude the mean difference in pulse rates before and after the 5 minute waiting period is different from 0.

Step-by-step explanation:

We are interested to check if the true difference in pulse rates before and after after 5 minute waiting period is 0 or no so then the system of hypothesis are:

Null hypothesis: $\mu_d =0$

Alternative hypothesis:$\mu_d \neq 0$

We have the following data given:

$\bar X_d = -1.2$ sample mean for the difference

$n =10$ the sample size

$s_d = 2.97$ sample standard deviation

The statistic is given by:

$t= \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}$

And replacing we got:

$t = \frac{-1.2-0}{\frac{2.97}{\sqrt{10}}}= -1.278$

The degrees of freedom are :

$df=n-1=10-1=9$

Since we are conducting a bilateral test te p value would be given by:

$p_v =2*P(t_{9}$

Since the p value is a very high value is higher than the typical values for the significance level (0.05,0.1) so then we have enough evidence to FAIL to reject the null hypothesis and there is no enough evidence to conclude that the true difference in the means is different from 0. And the best conclusion would be:

d. There is not enough evidence to conclude the mean difference in pulse rates before and after the 5 minute waiting period is different from 0.