Question

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, find 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.​

Answer 1

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 $\mu$ = population mean length of  pencils produced in that factory.

So, Null Hypothesis, $H_0$ : $\mu$ = 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, $H_A$ : $\mu\neq$ 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.  =  $\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }$  ~ N(0,1)

where, $\bar X$ = sample mean length of pencils = 17.9 cm

           $\sigma$ = population standard deviation = 0.25 cm

           n = sample of pencils = 45

So, the test statistics =  $\frac{17.9-18.0}{\frac{0.25}{\sqrt{45} } }$  

                                    =  -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 $\leq$ 2.68)

                      = 1- 0.99632 = 0.00368

For the two-tailed test, the P-value is calculated as = 2 $\times$ 0.00368 = 0.00736.