Question

1. A recent study of 28 city residents showed that the mean of the time they had lived at their present address was 9.3 years and the standard deviation was 2 years. Assuming a normal population, find the 90% confidence interval of the true mean. Interpret the interval estimate.

2. A US Travel Data Center survey conducted for Better Homes and Gardens of 1500 adults found that 39% said that they would take more vacations this year than last year. Find the 95% confidence interval for the true proportion of adults who said that they will travel more this year. Interpret the interval estimate.

Answer 1

Question 1

A recent study of 28 city residents showed that the mean of the time they had lived at their present address was 9.3 years and the standard deviation was 2 years. Assuming a normal population, find the 90% confidence interval of the true mean. Interpret the interval estimate.

Solution:

Confidence interval for Population mean is given as below:

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

From given data, we have

Xbar = 9.3

S = 2

n = 28

df = n – 1 = 27

Confidence level = 90%

Critical t value = 1.7033

(by using t-table)

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

Confidence interval = 9.3± 1.7033*2/sqrt(28)

Confidence interval = 9.3± 0.6438

Lower limit = 9.3 - 0.6438 = 8.6562

Upper limit = 9.3 + 0.6438 = 9.9438

Confidence interval = (8.6562, 9.9438)

We are 90% confident that the mean time lived at present address will lies between 8.7 years and 9.9 years.

Question 2

A US Travel Data Center survey conducted for Better Homes and Gardens of 1500 adults found that 39% said that they would take more vacations this year than last year. Find the 95% confidence interval for the true proportion of adults who said that they will travel more this year. Interpret the interval estimate.

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

n = 1500

P = x/n = 0.39

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.39 ± 1.96* sqrt(0.39*(1 – 0.39)/1500)

Confidence Interval = 0.39 ± 1.96* 0.0126

Confidence Interval = 0.39 ± 0.0247

Lower limit = 0.39 - 0.0247 =0.3653

Upper limit = 0.39 + 0.0247 =0.4147

Confidence interval = (0.3653, 0.4147)

We are 95% confident that the population proportion of adults who said that they would take more vacations this year than last year will lies between 36.5% and 41.5%.