Question

A hair dryer manufacturer claims that the mean life of its product is greater than 700 hours. A random sample of 40 of its motors have a mean life of 702 hours. It is known that the standard deviation of the lifetime of the motors is 15 hours. At  = 0.06, is there enough evidence to reject the company's claim?

Perform a thorough hypothesis test on your work including:

• Identification of the null and alternative hypotheses and indication of which is the claim
• Identification of the distribution to use and why
• Identification of the test to use and why
• Solving explicitly for the standardized test statistic and the P value
• Stating explicitly your basis for rejecting/failing to reject the null hypothesis AND the claim

Answer 1

Answer:

Given,

Ho : u = 700

Ha : u > 700

Here it is a right tailed test

i.e.,

Here we use z test

At 0.06 level of significance , Critical value z = 1.55

So reject Ho if z > 1.55

consider,

test statistic z = (x - u)/(s/sqrt(n))

substitute values

= (702 - 700)/(15/sqrt(40))

z = 0.843

P value = P(z > 0.843)

= 0.1996142 [since from z table]

= 0.1996

Here we observe that, p value > alpha, so we fail to reject Ho.

So there is no sufficient evidence to support the claim.