Question

A simple random sample of 50 stainless steel metal screws is obtained. The screws have a mean length of 0.73 inches. Assume the population standard deviation is 0.012 inches and the confidence level is α = 0.05. Find the P-value you would use to test the claim that the screws have a mean length equal to 0.75 inches as indicated on the package label.

0.9999


0.0002


1.9998


0.0001

Answer 1

Explanation:

A simple random sample of 50(n) stainless steel metal screws is obtained. The screws have a mean length of 0.73(mu) inches. Assume the population standard deviation is 0.012(sigma) inches and the confidence level is α = 0.05.

Set up hypotheses as:

`H_0: \mu = 0.75\\H_a:\mu \\eq 0.75`

(two tailed test at 5% sign. level)

Mean difference =
`0.75-0.73=0.02`

Std error of sample =
`(\sigma)/(√(n)) =(0.012)/(√(50) ) \\=0.00024`

Test statistic = Mean diff/std error

=
`(0.02)/(0.0024) \\=8.33`

Since population std dev is known and also sample size >30 z test can be used.

p value <0.0001

Answer 2

Solution: We are given:

`\mu=0.75,\sigma=0.012,\bar{x}=0.73`

To find the P-value, we first need to find the value of test statistic.

`z=\frac{\bar{x}-\mu}{(\sigma)/(√(n))}`

`=(0.73-0.75)/((0.012)/(√(50)))`

`=(-0.02)/(0.001697)`

`=-11.79`

Now we can find the P-value.

`P-value=P(z<-11.79) +P(z>11.79)`

`=0.0000+0.0000`

`=0.0000`

Therefore, the P-value = 0.0000

But we have no such option given. I would suggest to use 0.0001 as it is closest to 0.0000.