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 5 for 95% confidence. The study was based on a random sample of 64 undergraduates. If you wanted 90% confidence for the same study, what would be the confidence interval

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

Step-by-step explanation:

95% confidence interval:

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

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find the margin of errorM as such

In which 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

90% confidence interval.

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

Now, we have to find z in the Ztable as such z has a pvalue of .

So it is z with a pvalue of , so

Now, find M as such

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

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.

won't that be plagrisim thought?

step-by-step explanation:

2y+y=5

add 2y and y