Question

. In a random sample of five people, the mean driving distance to work was 25.4 miles and the standard deviation was 5.2 miles. Assume the random variable is normally distributed and use a tdistribution to find the margin of error and construct confidence intervals for the population mean at the following confidence levels: (4 points each) a. 85% b. 90% c. 99%

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 25.4

sample standard deviation = s = 5.2

sample size = n = 5

Degrees of freedom = df = n - 1 = 5 - 1 = 4

a. At 85% confidence level

= 1 - 85%

=1 - 0.85 =0.15

/2 = 0.075

t/2,df = 1.778

Margin of error = E = t/2,df * (s /n)

= 1.778 * (5.2 / 5)

Margin of error = E = 4.13

The 85% confidence interval estimate of the population mean is,

- E < < + E

25.4 - 4.13 < < 25.4 + 4.13

21.27 < < 29.53

b. At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = 2.132

Margin of error = E = t/2,df * (s /n)

= 2.132 * (5.2 / 5)

Margin of error = E = 4.96

The 90% confidence interval estimate of the population mean is,

- E < < + E

25.4 - 4.96 < < 25.4 + 4.96

20.44 < < 30.36

c. At 99% confidence level

= 1 - 99%

=1 - 0.99 =0.01

/2 = 0.005

t/2,df = 4.604

Margin of error = E = t/2,df * (s /n)

= 4.604 * (5.2 / 5)

Margin of error = E = 10.71

The 99% confidence interval estimate of the population mean is,

- E < < + E

25.4 - 10.71 < < 25.4 + 10.71

14.69 < < 36.11