Question

A study of stress on the campus of your university reported a mean stress level of 76 (on a 0 to 100 scale with a higher score indicating more stress) with a margin of error of 4 for 95% confidence. The study was based on a random sample of 49 undergraduates.Give the 95% confidence interval.

Answer 1

Answer:

The 95% confidence interval for population mean is (72, 80).

Step-by-step explanation:

The (1 - α) % confidence interval for population mean (μ) is:

$CI=\bar x\pm z_{\alpha /2}\frac{\sigma}{\sqrt{n}}$

The Margin of error for this confidence interval is:

$MOE=z_{\alpha /2}\frac{\sigma}{\sqrt{n}}$

The confidence interval for μ can also be written as:

$CI=\bar x\pm MOE$

Given:

$\bar x=76\\MOE=4$

Compute the 95% confidence interval for population mean as follows:

$CI=\bar x\pm MOE\\=76\pm4\\=(72, 80)$

Thus, the 95% confidence interval for population mean is (72, 80).

Answer 2

Answer:

$\bar X \pm ME$

And if we find the limits we got:

$76-4 = 72$

$76 + 4 = 80$

So then the 95% confidence interval would be (72.0,80.0)

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$\bar X = 76$ represent the sample mean for the sample  

$\mu$ population mean (variable of interest)  

s represent the sample standard deviation  

n=49 represent the original sample size  

Confidence =95% or 0.95

ME=4 represent the margin of error.

Solution to the problem

The confidence interval for the mean is given by the following formula:  

$\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}$   (1)  

The margin of error is defined as:

$ME= t_{\alpha/2}\frac{s}{\sqrt{n}}$

The formula for the confidence interval is equivalent to:

$\bar X \pm ME$

And if we find the limits we got:

$76-4 = 72$

$76 + 4 = 80$

So then the 95% confidence interval would be (72.0,80.0)