Question

A fire insurance company thought that the mean distance from a home to the nearest fire department in a suburb of Chicago was at least 5.9 miles. It set its fire insurance rates accordingly. Members of the community set out to show that the mean distance was less than 5.9 miles. This, they thought, would convince the insurance company to lower its rates. They randomly indentified 62 homes and measured the distance to the nearest fire department from each. The resulting sample mean was 5.3. If σ = 2 miles, does the sample show sufficient evidence to support the community's claim at the α = .05 level of significance?

(a) Find z. (Give your answer correct to two decimal places.)


(ii) Find the p-value. (Give your answer correct to four decimal places.)

(b) State the appropriate conclusion.

Reject the null hypothesis, there is not significant evidence that the mean distance is less than 5.9 miles. Reject the null hypothesis, there is significant evidence that the mean distance is less than 5.9 miles.     Fail to reject the null hypothesis, there is significant evidence that the mean distance is less than 5.9 miles. Fail to reject the null hypothesis, there is not significant evidence that the mean distance is less than 5.9 miles.

2.From candy to jewelry to flowers, the average consumer was expected to spend $104.21 for Mother's Day in 2005, according to the Democrat & Chronicle article "Mom's getting more this year" (May 7, 2005). Local merchants thought this average was too high for their area. They contracted an agency to conduct a study. A random sample of 62 consumers was taken at a local shopping mall the Saturday before Mother's Day and produced a sample mean amount of $94.33. If σ = $29.93, does the sample provide sufficient evidence to support the merchants' claim at the .05 significance level?

(a) Find z. (Give your answer correct to two decimal places.)


(ii) Find the p-value. (Give your answer correct to four decimal places.)

(b) State the appropriate conclusion.

Reject the null hypothesis, there is not significant evidence to support the merchants' claim. Reject the null hypothesis, there is significant evidence to support the merchants' claim.     Fail to reject the null hypothesis, there is significant evidence to support the merchants' claim. Fail to reject the null hypothesis, there is not significant evidence to support the merchants' claim.

Answer 1

Solution :

1 ) Given that

= 5.9

= 5.3

= 2

n = 62

This is the right tailed test .

The null and alternative hypothesis is ,

H₀ :    = 5.9

H_a : > 5.9

Test statistic = z

= ( - ) / / n

= (5.3 - 5.9) / 2 / 62

= −2.362

Test statistic = z = −2.36

P(z > −2.36) = 1 - P(z < −2.36 ) = 1 -0.0091

P-value = 0.9909

= 0.05  

P-value ≥

0.9909 ≥ 0.05

Do not reject the null hypothesis .

Therefore, there is not enough evidence to claim that the population mean μ is greater than 5.9, at the 0.05 significance level  

2 ) Given that

= 104.21

= 94.33

= 29.93

n = 62

This is the two tailed test .

The null and alternative hypothesis is ,

H₀ :    = 104.21

H_a : 104.21

Test statistic = z

= ( - ) / / n

= (94.33- 104.21) / 29.93 / 62

=−2.599

Test statistic = z = −2.60

P-value = 0.0093

= 0.05  

P-value ,

0.0093 < 0.05

Reject the null hypothesis

Therefore, there is not enough evidence to claim that the population mean μ is different than 104.21, at the 0.05 significance level