Question

A production line operates with a mean filling weight of 16 ounces per container. Overfilling or underfilling presents a serious problem and when detected requires the operator to shut down the production line to readjust the filling mechanism. From past data, a population standard deviation S=.8 ounces is assumed. A quality control in inspector selects a sample of 30 items every hour and at that time makes the decision of whether to shut down the line for readjustment. The level of significance is a = .05.- State the hypothesis test for this quality control application.- If a sample mean of xbar=16.32 ounces were found, what is the p-value ? What action would you recommend ?- If a sample mean of xbar=15.82 ounces were found, what is the p-value ? What action would you recommend ?- Use the critical value approach. What is the rejection rule for the preceding hypothesis testing procedure ? Repeat the last two parts : Do you reach the same conclusion ?

Answer 1

Answer:

Part1) $z=\frac{16.32-16}{\frac{0.8}{\sqrt{30}}}=2.191$    

$p_v =2*P(Z>2.191)=0.0284$

Readjustment is needed

Part 2)  $z=\frac{15.82-16}{\frac{0.8}{\sqrt{30}}}=-1.232$    

$z=\frac{15.82-16}{\frac{0.8}{\sqrt{30}}}=-1.232$    

$p_v =2*P(Z<-1.232)=0.218$

No readjustment is needed

Step-by-step explanation:

Data given and notation

Part 1    

$\bar X=16.32$ represent the sample mean    

$\sigma=0.8$ represent the population standard deviation    

$n=30$ sample size    

$\mu_o =16$ represent the value that we want to test    

$\alpha=0.05$ represent the significance level for the hypothesis test.    

z would represent the statistic (variable of interest)    

$p_v$ represent the p value for the test (variable of interest)    

State the null and alternative hypotheses.    

We need to apply a two tailed test.    

What are H0 and Ha for this study?    

Null hypothesis:  $\mu =16$    

Alternative hypothesis :$\mu \neq 16$    

Compute the test statistic  

The statistic for this case is given by:    

$z=\frac{\bar X-\mu_o}{\frac{\sigma}{\sqrt{n}}}$ (1)    

z-test: "Is used to compare group means. Is one of the most common tests and is used to determine if the mean is (higher, less or not equal) to an specified value".    

Calculate the statistic    

We can replace in formula (1) the info given like this:    

$z=\frac{16.32-16}{\frac{0.8}{\sqrt{30}}}=2.191$    

Give the appropriate conclusion for the test  

Since is a two tailed test the p value would be:    

$p_v =2*P(Z>2.191)=0.0284$

Using the value of $\alpha =0.05$ we see that $pv<\alpha$ and we have enough evidence to reject the null hypothesis.

Readjustment is needed

The rejection zone is given by:

$(-\infty ;-1.96) , (1.96;\infty)$    

Part 2

$\bar X=15.82$ represent the sample mean    

We can replace in formula (1) the info given like this:    

$z=\frac{15.82-16}{\frac{0.8}{\sqrt{30}}}=-1.232$    

Give the appropriate conclusion for the test  

Since is a two tailed test the p value would be:    

$p_v =2*P(Z<-1.232)=0.218$

Using the value of $\alpha =0.05$ we see that $pv>\alpha$ and we have enough evidence to FAIL to reject the null hypothesis.

No readjustment is needed

The rejection zone is given by:

$(-\infty ;-1.96) , (1.96;\infty)$  

What action would you recommend ?

Review the procedure.

Do you reach the same conclusion ?

No we got different conclusions