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 estimating the mean difference in pulse rates before and after the 5 minute waiting period for the population of study participants. Calculate the 90% confidence interval for the mean difference in pulse rates for the population of study participants. Use t*

Answer 1

Answer:

The 90% confidence interval for the mean difference in pulse rates for the population of study participants is between -2.9 beats per minute and 0.5 beats per minute.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 10 - 1 = 9

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 9 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$. So we have T = 1.8331

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.8331\frac{2.97}{\sqrt{10}} = 1.7$

In which s is the standard deviation of the sample and n is the size of the sample.

The lower end of the interval is the sample mean subtracted by M. So it is -1.2 - 1.7 = -2.9 beats per minute

The upper end of the interval is the sample mean added to M. So it is -1.2 + 1.7 = 0.5 beats per minute.

The 90% confidence interval for the mean difference in pulse rates for the population of study participants is between -2.9 beats per minute and 0.5 beats per minute.