1. A random sample of 64 cell phone users showed that they paid an average of $1

Question

1. A random sample of 64 cell phone users showed that they paid an average of $110 per month with a standard deviation of $12 for their cell phone service. Make a 95% confidence interval for the mean monthly cost of cell phone service for this population. For 2 and 3·A 95% confidence interval for the mean hourly wage paid to fast food workers is $7.45 to $7.75. The confidence interval was computed using a random sample of 100 workers. 2. Suppose that we used the same data to make a 90% confidence interval. Would the interval be wider or narrower (which) than the 95% confidence interval. 3. Suppose we took a random sample of 50 workers instead of 100 workers. Would we expect the 95% confidence interval to be wider or narrower (which ) than the interval based on a random sample of 100 workers?

Explanation / Answer

Kindly post 1 question per post as per forum rules, here answer to Q1

1. n = 64
Average = 110
Stdev = 12
Make a 95% CI ?
Ok, so a 95% CI has Z = 1.96 according to the Z-tables

So, CI is given by formula:
Xbar +/- Z*Stdev/sqrt(n)
= 110 +/- 1.96*12/sqrt(64)
= $107.06 to $112.94

So, the 95% CI for the population mean is ; $107.06 to $112.94

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