Question

Manufacturer's representative argues that the average battery life is 585 hours. For testing were selected 42 batteries. The average lifetime was 620 hours with standard deviation s = 120 h.

A) Find the 95% confidence interval.

B) Whether there is reason to believe that the term is overvalued? Find p-value

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 620

sample standard deviation = s = 120

sample size = n = 42

Degrees of freedom = df = n - 1 = 42 - 1 = 41

At 95% confidence level

= 1 - 95%

=1 - 0.95 = 0.05

/2 = 0.025

t/2,df = t 0.025 , 41  = 2.020

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

= 2.020 * ( 120 / 42 )

Margin of error = E = 37.40

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

- E < < + E

620 - 37.40 < < 620 - 37.40

582.60 < < 657.40

( 582.60 , 657.40 )

( B )

= 585

= 620

s = 120

n = 42

This is the right tailed test .

The null and alternative hypothesis is ,

H0 :   = 585

Ha : > 585

Test statistic = T

= ( - ) /s / n

= ( 620 - 585 ) / 120 / 42

= 1.89

The test statistic = 1.89

df = n - 1 = 42 - 1 = 41

P- value = 1 - ( t 1.89 , 41 )

= 1 - 0. 9671

= 0.0329

P-value = 0.0329