Question

the manufacturer of a certain brand of auto batteries claims the mean life of these batteries is 45 months. A consumer protection agency that wants to check this claim took a random sample of 36 such batteries and found the mean life for this sample is 43.75 months with a standard deviation of 4.5 months. Find the p-value for the test hypothesis with the alternative hypothesis that the mean life of these batteries is less than 45 months.
Would you reject the manufactures claim at a 5% significance level? Explain

Answer 1

Z = (mean1 - mean2)/standard error (SE) of difference between means

SEdiff = √(SEmean1^2 + SEmean2^2)

SEm = SD/√n

If only one SD is provided, you can use just that to determine SEdiff.

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion related to that Z score. Is it less than .05?

Answer 2

standard deviation is 4months

Answer 3

Z=(43.75-45)/(4.5/sqrt36)=-1.67
p-value=0.0475

the manufacture claim is mu=45 months
p-value=2*0.0475=0.095
alpha=0.05
the p-value > alpha
therefore we fail to reject the null alternative