Question

1. The standard deviation of sample size of n = 15 is 345 and the mean is 6784. Given that the

population follows a normal distribution, construct a 90% confidence interval estimate of the

mean of the population.

Answer 1

Answer:

The 90% confidence interval estimate of the mean of the population is between 6627 and 6941.

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 = 15 - 1 = 14

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$. So we have T = 1.761

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 1.761\frac{345}{\sqrt{15}} = 157$

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 6784 - 157 = 6627

The upper end of the interval is the sample mean added to M. So it is 6784 + 157 = 6941

The 90% confidence interval estimate of the mean of the population is between 6627 and 6941.