Question

In: Statistics and Probability

“A sample of 60 individuals in the Midwest showed the sample average per person annual expenditure...

“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.

Solutions

Expert Solution

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.


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