Question

A simple random sample with n =54 provided a sample mean of 24.0 and a sample standard deviation of 4.5. a. develop a 90% CI for the pop mean b. develop a 95% CI for the pop mean c. develop a 99% CI for the pop mean d. what happens to the margin of error and the CI as the confidence level increases?

Answer 1

Answer

(A) we have n = 5, sample mean = 24, sample standard deviation = 4.5

we have to find the 90% confidence interval for the population mean.

Formula for the confidence interval is given as

We are using t distribution because population standard deviation is unknown.

so, degree of freedom = n-1 = 54-1 = 53

using the degree of freedom 53 and level of confidence = 1-0.90 = 0.10

we get, t critical value = 1.67

setting all the given values in the confidence formula, we get

CI-24 ± 1.67 * (4.5/V54)-24 ± (1.67 * 0.6124)

this gives us

CI = (22.9773, 25.0227)

This is the required 90% confidence interval for the population mean.

(B) we have n = 5, sample mean = 24, sample standard deviation = 4.5

we have to find the 95% confidence interval for the population mean.

Formula for the confidence interval is given as

We are using t distribution because population standard deviation is unknown.

so, degree of freedom = n-1 = 54-1 = 53

using the degree of freedom 53 and level of confidence = 1-0.95 = 0.05

we get, t critical value = 2.01

setting all the given values in the confidence formula, we get

this gives us

CI = (22.7691,25.2309)

This is the required 95% confidence interval for the population mean.

(C) we have n = 5, sample mean = 24, sample standard deviation = 4.5

we have to find the 99% confidence interval for the population mean.

Formula for the confidence interval is given as

We are using t distribution because population standard deviation is unknown.

so, degree of freedom = n-1 = 54-1 = 53

using the degree of freedom 53 and level of confidence = 1-0.99 = 0.01

we get, t critical value = 2.67

setting all the given values in the confidence formula, we get

this gives us

CI = (22.3650,25.6350)

This is the required 99% confidence interval for the population mean.

(d) As we increase the confidence level, then length of confidence interval is increasing, i.e. the width of confidence interval is increasing due to increased margin of error.

Similarly, margin of error is given as

Since t value is increasing with increase in confidence level, so margin of error is also increasing with increasing confidence level.