Question

“A sample of 60 individuals in the Midwest showed the sample average per person annual expenditure for prescription drugs of $745. Suppose the population standard deviation expenditure for prescription drugs is $300.”

a. Construct a 95% confidence interval estimate for the population mean annual expenditure for prescription drugs in the Midwest.

b. Is $838 in the interval from part a? What would this imply about the population mean being $838?

c. what would happen to the width of the confidence interval estimate if the confidence level changed to 99%? Explain.

d. Determine the minimum sample size to ensure that a 99% confidence interval has a margin of error of $50. Assume the population standard deviation is $300.

Answer 1

The formula for confidence interval of population mean is

where

= sample mean

= population standard deviation

n = sample size

=  critical z-value

The "Z" values for Confidence Interval are given below:

Confidence Interval	Z
80%	1.282
85%	1.440
90%	1.645
95%	1.960
99%	2.576
99.5%	2.807
99.9%	3.291

Given

n = 60

From the table, the Z-value for 95% confidence level is 1.96

The 95% confidence interval estimate for the population mean is

  

( 669.09 , 820.91)

Therefore, the 95% confidence interval estimate for the population mean annual expenditure for prescription drugs in the Midwest is $669.09< < $820.91

b)

The given $838 is outside the interval $669.09< < $820.91 .

95% of the time, when we calculate a confidence interval in this way, the true mean will be between the two values. 5% of the time, it will not.

c)

Using confidence level = 99%

From the table, the Z-value for 99% confidence level is 2.576

The 99% confidence interval estimate for the population mean is

  

( 645.23 , 844.77)

Therefore, the 99% confidence interval estimate for the population mean annual expenditure for prescription drugs in the Midwest is $645.23< < $844.77.

So, the width of the confidence interval estimate would increase if the confidence level changed to 99%.

d)

Given

Confidence level = 99%

Margin of error = $50.

Population standard deviation ()= $300

From the table, the Z-value for 99% confidence level is 2.576

Now

Margin of error = critical value * standard error

The minimum sample size to ensure that a 99% confidence interval has a margin of error of $50 is 239.