A study of stress on the campus of your university reported a mean stress level of 78 (on a 0 to 100 scale with a higher score indicating more stress) with a margin of error of …

Question

asked May 18, 2021 202k views

1 Answer

Yassir Ennazk · Answer 1 · 2021-05-22T08:08:31+0000

Answer:

The 90% confidence interval would be between 73.80 and 82.20.

Explanation:

95% confidence interval:

We have that to find our
$\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = (1-0.95)/(2) = 0.025$

Now, we have to find z in the Ztable as such z has a pvalue of
$1-\alpha$ .

So it is z with a pvalue of
1-0.025 = 0.975 , so
z = 1.96

Now, find the margin of errorM as such

$M = z*(\sigma)/(√(n))$

In which
$\sigma$ is the standard deviation of the population and n is the size of the sample.

We have that M = 5, n = 64. We have to find the standard deviation of the population. So

$M = z*(\sigma)/(√(n))$

$5 = 1.96*(\sigma)/(√(64))$

$1.96\sigma = 5*8$

$\sigma = (5*8)/(1.96)$

$\sigma = 20.41$

90% confidence interval.

We have that to find our
$\alpha$ level, that is the subtraction of 1 by the confidence interval divided by 2. So:

$\alpha = (1-0.9)/(2) = 0.05$

Now, we have to find z in the Ztable as such z has a pvalue of
$1-\alpha$ .

So it is z with a pvalue of
1-0.05 = 0.95 , so
z = 1.645

Now, find M as such

$M = z*(\sigma)/(√(n))$

In which
$\sigma$ is the standard deviation of the population and n is the size of the sample.

M = 1.645*(20.41)/(√(64)) = 4.20

The lower end of the interval is the sample mean subtracted by M. So it is 78 - 4.20 = 73.80.

The upper end of the interval is the sample mean added to M. So it is 78 + 4.20 = 82.20

The 90% confidence interval would be between 73.80 and 82.20.