Question

X6.2.9-T

Construct the indicated confidence interval for the population mean μ using the t-distribution. Assume the population is normally distributed 

c = 0.90, x̅ = 12.9, s = 4.0, n = 9 

The 90% confidence interval using a t-distribution is 

6.2.17-T 

In a random sample of 26 people, the mean commute time to work was 34.8 minutes and the standard deviation was 7.2 minutes. Assume the population is normally distributed and use a t-distribution to construct a 98% confidence interval for the population mean μ What is the margin of error of μ? Interpret the results. 

The confidence interval for the population mean μ is 

Answer 1

1)

X bar = 12.9

Std Dev (s) = 4

n = 9

Degrees of Freedom(df) = n -1 = 8

Alpha = 0.1

T Critical using the alpha = 0.1 and df = 1.859

Confidence Interval using t distribution is calculated using the below formulae:

= 12.9 +/- 1.859 * (4/3) = {10.42,15.38}

The 90% confidence interval using a t distribution is (10.4,15.4)

2)

X bar = 34.8

Std Dev (s) = 7.2

n = 26

Degrees of Freedom(df) = n -1 = 25

Alpha = 0.02

T Critical using the alpha = 0.02 and df = 2.478

Confidence Interval using t distribution is calculated using the below formulae:

= 34.8 +/- 2.478 * (7.2/sqrt(26)) = {31.3,38.3}

The 98% confidence interval using a t distribution is {31.3,38.3}

Margin of Error = t*s/n^1/2 = 3.5