Respuesta :
Answer:
The P-value you would use to test the claim that the population mean of pencils produced in that factory have a mean length equal to 18.0 cm is 0.00736.
Step-by-step explanation:
We are given that a quality control specialist at a pencil manufacturer pulls a random sample of 45 pencils from the assembly line.
The pencils have a mean length of 17.9 cm. Given that the population standard deviation is 0.25 cm.
Let [tex]\mu[/tex] = population mean length of pencils produced in that factory.
So, Null Hypothesis, [tex]H_0[/tex] : [tex]\mu[/tex] = 18.0 cm {means that the population mean of pencils produced in that factory have a mean length equal to 18.0 cm}
Alternate Hypothesis, [tex]H_A[/tex] : [tex]\mu\neq[/tex] 18.0 cm {means that the population mean of pencils produced in that factory have a mean length different from 18.0 cm}
The test statistics that will be used here is One-sample z-test statistics because we know about the population standard deviation;
T.S. = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\bar X[/tex] = sample mean length of pencils = 17.9 cm
[tex]\sigma[/tex] = population standard deviation = 0.25 cm
n = sample of pencils = 45
So, the test statistics = [tex]\frac{17.9-18.0}{\frac{0.25}{\sqrt{45} } }[/tex]
= -2.68
The value of z-test statistics is -2.68.
Now, the P-value of the test statistics is given by;
P-value = P(Z < -2.68) = 1 - P(Z [tex]\leq[/tex] 2.68)
= 1- 0.99632 = 0.00368
For the two-tailed test, the P-value is calculated as = 2 [tex]\times[/tex] 0.00368 = 0.00736.
