Class Project: Quality Assurance. This exercise can be don individually or, better yet, as a class project. For the pretzel-packagin hypothesis test in Example 9.1 on page 352, the null and alternativ hypotheses are, respectively,
$H_0:\mu =454g$ (machine is working properly)
$H_0:\mu \ne 454g$ (machine is not working properly).
where $\mu$ is the mean net weight of all bags of pretzels packaged. The net weights are normally distributed with a standard deviation of 7.8g.
a. Assuming that the null hypothesis is true, simulate 100 samples of 25 net weights each.
b. Suppose that the hypothesis test is performed at the $5\%$significance level. Of the 100 samples obtained in part (a), roughly how many would you expect to lead to rejection of the null hypothesis? Explain your answer.

c. Of the 100 samples obtained in part (a), determine the number that lead to rejection of the null hypothesis.
d. Compare your answers from parts (b) and (c), and comment on any observed difference.

It is given that the null and alternative hypotheses are stated as follows:
$H_0:\mu =454g$(Machine is working properly.)
Versus
$H_0:\mu \ne 454g$ (Machine is not working properly)
Here the mean net weight of all bags of pretzels package is given as$\mu$ .
Also it is given that the net weights are normally distributed with a standard deviation of$7.8g$.
That means $\sigma =7.8g$

Using MINITAB, we simulate 100 samples of 25 net weights each by assuming$H_0$is true as follows:
Step 1: Calc$\to$Random data$\to$Normal
Step 2: enter the given data as follows:
Step 3: Click OK.
From the above three steps, we simulate 100 samples of size in the columns, C1 to C100.

It is given that the null and alternative hypotheses are stated as follows:
$H_0:\mu =454g$(Machine is working properly.)
Versus
$H_0:\mu \ne 454g$(Machine is not working properly)
Here the mean net weight of all bags of pretzels package is given as$\mu$ .
Also it is given that the net weights are normally distributed with a standard deviation of$7.8g$.
That means $\sigma =7.8g$

Here we have informed that the hypothesis test is performed at$5\%$ significance level. That means, we have
Thus, the probability of rejecting the null hypothesis is$0.05$. That means,
Therefore for 100 samples,the expected number of samples are rejected, is$0.05$

It is given that the null and alternative hypotheses are stated as follows:
$H_0:\mu =454g$(Machine is working properly.)
Versus
$H_0:\mu \ne 454g$(Machine is not working properly)
Here the mean net weight of all bags of pretzels package is given as$\mu$.
Also it is given that the net weights are normally distributed with a standard deviation of$7.8g$.
That means $\sigma =7.8g$

Using MINITAB, we obtain the$95\%$ confidence interval for the 100 samples simulated in part
(a), as follows:
1 Store the data in a column named Cadmium level.
2 Choose Stat ? Basic Statistics ?1-Sample Z. .
3 Select the Samples in columns option button.
4 Click in the Samples in columns text box and specify C1-C2.
5 Click in the Standard deviation text box and type 78
6 Click the Options ... button.
7 Type 95 in the Confidence level text box.
8 Click the arrow button at the right of the Alternative drop-down list box and select greater than.
9 Click OK twice.

As per the requirement, we observe the number of samples for which the $95\%$confidence interval does not contain the value of mean

It is given that the null and alternative hypotheses are stated as follows:
$H_0:\mu =454g$(Machine is working properly.)
Versus
role="math" localid="1651256756555" $H_0:\mu \ne 454g$(Machine is not working properly)
Here the mean net weight of all bags of pretzels package is given as$\mu$ .
Also it is given that the net weights are normally distributed with a standard deviation of$7.8g$.
That means $\sigma =7.8g$

From the parts (b) and (c), we observe that the expected value of the number of samples that lead to reject the null hypothesis is greater as compared to the observed value.
Therefore, we fail to reject the null hypothesis and hence we conclude that the machine is working properly.