Question

You are the Colonel of the Maryland State Police, and you are concerned about drinking among state troopers.  You take a random sample of 150 police offers and find that the median number of drinks per week is 2 and the mean number of drinks per week is 4.7 with a standard deviation of 8.

1. Construct a 90% confidence interval around the appropriate point estimate. Interpret your results.

2. Construct a 93% confidence interval around your point estimate.  Interpret your results.  Why did the interval change?

                  

3. What would happen if we increased our confidence level to 99% (You do not need to recalculate the CI)?

Answer 1

Solution:

Part a

We are given

Xbar = 4.7

S = 8

n = 150

df = n – 1 = 149

Confidence level = 90%

Critical t value = 1.6551

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 4.7 ± 1.6551*8/sqrt(150)

Confidence interval = 4.7 ± 1.6551*0.653197265

Confidence interval = 4.7 ± 1.0811

Lower limit = 4.7 - 1.0811 = 3.62

Upper limit = 4.7 + 1.0811 = 5.78

Confidence interval = (3.62, 5.78)

Part b

We are given

Xbar = 4.7

S = 8

n = 150

df = n – 1 = 149

Confidence level = 93%

Critical t value = 1.8250

(by using t-table)

Confidence interval = Xbar ± t*S/sqrt(n)

Confidence interval = 4.7 ± 1.8250*8/sqrt(150)

Confidence interval = 4.7 ± 1.8250*0.653197265

Confidence interval = 4.7 ± 1.1921

Lower limit = 4.7 - 1.1921= 3.51

Upper limit = 4.7 + 1.1921= 5.89

Confidence interval = (3.51, 5.89)

3. What would happen if we increased our confidence level to 99% (You do not need to recalculate the CI)?

If we increase our confidence level to 99%, then the width of the confidence interval will be increase. We already observed in above two parts, if we increase confidence level, then the width of the confidence interval will be increase.