Respuesta :
Solution: We are given:
[tex]\mu=0.75,\sigma=0.012,\bar{x}=0.73[/tex]
To find the P-value, we first need to find the value of test statistic.
[tex]z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
[tex]=\frac{0.73-0.75}{\frac{0.012}{\sqrt{50}}}[/tex]
[tex]=\frac{-0.02}{0.001697}[/tex]
[tex]=-11.79[/tex]
Now we can find the P-value.
[tex]P-value=P(z<-11.79) +P(z>11.79)[/tex]
[tex]=0.0000+0.0000[/tex]
[tex]=0.0000[/tex]
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.
Answer:
Step-by-step 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:
[tex]H_0: \mu = 0.75\\H_a:\mu \neq 0.75[/tex]
(two tailed test at 5% sign. level)
Mean difference = [tex]0.75-0.73=0.02[/tex]
Std error of sample = [tex]\frac{\sigma}{\sqrt{n}} =\frac{0.012}{\sqrt{50} } \\=0.00024[/tex]
Test statistic = Mean diff/std error
=[tex]\frac{0.02}{0.0024} \\=8.33[/tex]
Since population std dev is known and also sample size >30 z test can be used.
p value <0.0001
