Question

in a random sample of 28 people, the mean commute time to work was 31.2 minutes and the standard deviation was 7.3 minutes. assume the population is normally distributed and use a t-distribution to construct a 99% confidence interval for the population mean u. What is the margin of error of u

Answer 1

Answer:

The margin of error of u is of 3.8.

The 99% confidence interval for the population mean u is between 27.4 minutes and 35 minutes.

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 = 28 - 1 = 27

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 27 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$. So we have T = 2.7707

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.7707\frac{7.3}{\sqrt{28}} = 3.8$

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

The margin of error of u is of 3.8.

The lower end of the interval is the sample mean subtracted by M. So it is 31.2 - 3.8 = 27.4 minutes

The upper end of the interval is the sample mean added to M. So it is 31.2 + 3.8 = 35 minutes

The 99% confidence interval for the population mean u is between 27.4 minutes and 35 minutes.