Question

An automobile assembly line operation has a scheduled mean completion time, μ, of 12 minutes. The standard deviation of completion times is 1.6 minutes. It is claimed that, under new management, the mean completion time has decreased. To test this claim, a random sample of 33 completion times under new management was taken. The sample had a mean of 11.2 minutes. Assume that the population is normally distributed. Can we support, at the 0.05 level of significance, the claim that the mean completion time has decreased under new management? Assume that the standard deviation of completion times has not changed.

Answer 1

Answer with explanation:

Let
`\mu` be the population mean.

Null hypothesis :
`\mu=12`

Alternative hypothesis :
`\mu<12`

Since Alternative hypothesis is left tailed so , the test is a left tailed test.

Given : n=33 > 30 , so we use z-test.

Test statistic :
`z=\frac{\overline{x}-\mu}{(\sigma)/(√(n))}`

i.e.
`z=(11.2-12)/((1.6)/(√(33)))\approx-2.87`

Using z-value table,

P-value for left tailed test =
`P(z<-2.87)=0.0020524`

Since , the p-value (0.0020524) is less than the 0.05 level of significance, it means we reject the null hypothesis.

Therefore, we have enough evidence to support the claim that the mean completion time has decreased under new management.