Question

Assume the population has a normal distribution. A sample of 25 randomly selected students Has a mean test score of 81.5 With a standard deviation of 10.2. Construct a 90% confidence interval for the mean test score.

Answer 1

Answer:

The 90% confidence interval for the mean test score is between 77.29 and 85.71.

Step-by-step explanation:

We have the standard deviation for the sample, so we use the t-distribution 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 = 25 - 1 = 24

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 24 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$. So we have T = 2.064

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.064\frac{10.2}{\sqrt{25}} = 4.21$

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 81.5 - 4.21 = 77.29

The upper end of the interval is the sample mean added to M. So it is 81.5 + 4.21 = 85.71.

The 90% confidence interval for the mean test score is between 77.29 and 85.71.