Question

Sherries Berries store claims that the batteries of their competitors have a lifetime of less than 600 hours. The Mall Inspectors test a sample of 26 batteries, and found: X = 602 hours   s = 40 hours

a) Test at alpha =.01

b) Construct a 2-tailed 95% C.I.E of μ. The upper limit of the confidence interval is ______

Steps to be covered:

1. state the hypothesis and identify the claim
2. find the critical value from the table and mention the acceptance range
3. compute the test valuemake the decision to reject or not the null hypothesis

Answer 1

Solution :

Given that,

Population mean = = 600

Sample mean = = 602

Sample standard deviation = s = 40

Sample size = n = 26

a)

Level of significance = = 0.01

This is a left tailed test.

The null and alternative hypothesis is,

Ho: 600

Ha: 600

The test statistics,

t = ( - )/ (s/)

= ( 602 - 600) / ( 40 / 26)

= 0.255

Based on the information provided, the significance level is = 0.01, and the critical value for a left-tailed test is ​=−2.485.

Since it is observed that t = 0.255 ≥ =−2.485, it is then concluded that the null hypothesis is not rejected.

Conclusion:

It is concluded that the null hypothesis Ho is not rejected. Therefore, there is not enough evidence to claim that the population mean μ is less than 600, at the 0.01 significance level.

b)

At 95% confidence level the t is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

t _/2,df = t_0.025,24 = 2.064

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

= 2.064 * ( 40 / 26)

= 16.191

The 95% confidence interval estimate of the population mean is,

- E < < + E

602 - 16.191 < < 602 + 16.191

585.809 < < 618.191

The upper limit confidence interval = 618.191