Question

In: Math

In looking at our class’s data as a sample of a larger population of students (who...

In looking at our class’s data as a sample of a larger population of students (who have taken, are taking, or may one day take this class), we find that the mean number of hours exercised per week during the summer is nearly 9 hours. We know that this is an estimate however. Is it likely that the true population mean is actually under 7 hours? Use a 95% confidence interval to determine this. If we’re willing to use a 99% confidence interval, does that change our findings? (Careful with your rounding!)

. mean exersum

Mean estimation                   Number of obs   =        215

--------------------------------------------------------------

             |       Mean   Std. Err.     [95% Conf. Interval]

-------------+------------------------------------------------

     exersum |   8.946512   .7143183     

Solutions

Expert Solution

Solution:

First of all we have to find 95% confidence interval for population mean.

Confidence interval = Xbar ± Z*SE

Where SE is standard error

We use Z confidence interval because sample size is given as n=215 which is larger as compared to n=30.

For 95% confidence level, Z = 1.96 (by using z-table)

Confidence interval = Xbar ± Z*SE

We are given Xbar = 8.946512, SE = 0.7143183

Confidence interval = 8.946512 ± 1.96*0.7143183

Confidence interval = 8.946512 ± 1.4001

Lower limit = 8.946512 - 1.4001 = 7.546412

Upper limit = 8.946512 + 1.4001 = 10.34661

Confidence interval = (7.546412, 10.34661)

Now, we have to find 99% confidence interval for population mean.

Confidence interval = Xbar ± Z*SE

Where SE is standard error

We use Z confidence interval because sample size is given as n=215 which is larger as compared to n=30.

For 99% confidence level, Z = 2.5758 (by using z-table)

Confidence interval = Xbar ± Z*SE

We are given Xbar = 8.946512, SE = 0.7143183

Confidence interval = 8.946512 ± 2.5758*0.7143183

Confidence interval = 8.946512 ± 1.8399

Lower limit = 8.946512 - 1.8399= 7.106612

Upper limit = 8.946512 + 1.8399= 10.78641

Confidence interval = (7.106612, 10.78641)

If we use the 99% confidence level instead of 95% confidence level, then it is observed that the width of the confidence interval is increases. The change in confidence level results in change in our findings.


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