Question

Find the margin of error for a 95% confidence interval (nearest hundredths) of the mean test score of Statistics students at Clayton State University given the following sample data scores: 87 85 56 82 67 77 73 71 90 74 76 79 81 84

Find a 99% confidence interval estimate (nearest hundredth) using the data in #12.

The height (in inches) of males in the United States is believed to be Normally distributed, with mean µ. The average height of a random sample of 50 American adult males is x bar = 69.72 inches, and the sample standard deviation of the 50 heights is s = 4.23. For a 95% confidence interval, find the margin of error.

Answer 1

Find the margin of error for a 95% confidence interval (nearest hundredths) of the mean test score of Statistics students at Clayton State University given the following sample data scores: 87 85 56 82 67 77 73 71 90 74 76 79 81 84

Confidence interval for Population mean is given as below:

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

From given data, we have

Xbar = 77.28571429

S = 8.89622713

n = 14

df = n – 1 = 13

Confidence level = 95%

Critical t value = 2.1604

(by using t-table)

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

Confidence interval = 77.28571429 ± 2.1604*8.89622713/sqrt(14)

Confidence interval = 77.28571429 ± 2.1604*2.377616711

Confidence interval = 77.28571429 ± 5.1365

Lower limit = 77.28571429 - 5.1365 = 72.15           

Upper limit = 77.28571429 + 5.1365 = 82.42

Find a 99% confidence interval estimate (nearest hundredth) using the data in #12.

Confidence interval for Population mean is given as below:

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

From given data, we have

Xbar = 77.28571429

S = 8.89622713

n = 14

df = n – 1 = 13

Confidence level = 99%

Critical t value = 3.0123

(by using t-table)

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

Confidence interval = 77.28571429 ± 3.0123*8.89622713/sqrt(14)

Confidence interval = 77.28571429 ± 3.0123*2.377616711

Confidence interval = 77.28571429 ± 7.1620

Lower limit = 77.28571429 - 7.1620= 70.12            

Upper limit = 77.28571429 + 7.1620= 84.45

The height (in inches) of males in the United States is believed to be Normally distributed, with mean µ. The average height of a random sample of 50 American adult males is x bar = 69.72 inches, and the sample standard deviation of the 50 heights is s = 4.23. For a 95% confidence interval, find the margin of error.

Margin of error = t*S/sqrt(n)

Confidence level = 95%

S = 4.23

n = 50

df = n – 1 = 49

Critical t value = 2.0096

(by using t-table)

Margin of error = 2.0096*4.23/sqrt(50)

Margin of error = 1.2022