Analyze the defect rate metric, we can use hypothesis testing to compare the before and after defect rates conclude that the expected improvement has been achieved.

Use below data to perform hypothesis testing.

Perform hypothesis testing by defining H0 and H1. Mu1, Mu2 and Mud. Also draw a graphical representation showing the rejection.

Below is the answer from chatgtp:

we can use hypothesis testing to compare the before and after defect rates and determine if the expected improvement has been achieved. Let's use a significance level of 0.05.

Null hypothesis (H0): There is no significant difference between the before and after defect rates. Alternative hypothesis (H1): The after defect rate is significantly lower than the before defect rate.

We can use a two-sample proportion test to compare the before and after defect rates. Here are the steps:

Calculate the sample proportions of defective units for the before and after periods: Before: 8% = 0.08 After: 2% or less = 0.02 (assume 2% for this analysis)

Calculate the standard error of the difference between the two proportions: SE = sqrt(p1*(1-p1)1 + p2*(1-p2)2) where p1 and p2 are the sample proportions, and n1 and n2 are the sample sizes.

Assuming a sample size of 1000 units for both before and after periods, we get: SE = sqrt(0.08*(1-0.08)/1000 + 0.02*(1-0.02)/1000) = 0.010

Calculate the test statistic: z = (p1 - p2) / SE

z = (0.08 - 0.02) / 0.010 = 6.00

Calculate the p-value: p-value = P(Z > 6.00) = 0.000000000002

Compare the p-value to the significance level: p-value (0.000000000002)

Since the p-value is less than the significance level, we reject the null hypothesis and conclude that the after defect rate is significantly lower than the before defect rate. Therefore, the expected improvement has been achieved.

Answer from Chegg:

Here's a quicker summary of the solution:

Hypotheses:

Null hypothesis (H0): There is no significant difference between the before and after defect rates.

Alternative hypothesis (H1): The after defect rate is significantly lower than the before defect rate.

Steps:

Calculate the sample proportions of defective units for the before and after periods: Before: 8%, After: 2% or less (assume 2%).

Calculate the standard error of the difference between the two proportions: SE = sqrt(p1*(1-p1)1 + p2*(1-p2)2) where p1 and p2 are the sample proportions, and n1 and n2 are the sample sizes.

Assuming a sample size of 1000 units for both before and after periods, we get: SE = 0.010.

Calculate the test statistic: z = (p1 - p2) / SE = 6.00.

Calculate the p-value: p-value = 0.000000000002.

Compare the p-value to the significance level (0.05).

Since the p-value is less than the significance level, we reject the null hypothesis and conclude that the after defect rate is significantly lower than the before defect rate. Therefore, the expected improvement has been achieved.

Graphical representation: The rejection region is on the right side of the null distribution, with a critical value of z = 1.96. The observed test statistic (z = 6.00) falls in the rejection region, so we reject the null hypothesis

Explanation:

The hypothesis testing performed showed that the after defect rate is significantly lower than the before defect rate, with a p-value of 0.000000000002. Therefore, the expected improvement has been achieved, and the process improvement implemented was effective in reducing the defect rate.

What I need: Please attach excel results and histogram pictures for this answer.

\begin{tabular}{|l|l|l|l|} \hline Month & Number of units produced & Number of defective units & Proportion of defective units \\ \hline Jan & 1000 & 150 & 15% \\ \hline Feb & 1200 & 180 & 15% \\ \hline Mar & 1300 & 120 & 9.2% \\ \hline Apr & 1100 & 70 & 6.4% \\ \hline May & 1500 & 50 & 3.3% \\ \hline Jun & 1400 & 60 & 4.3% \\ \hline Jul & 1600 & 60 & 3.8% \\ \hline Aug & 1800 & 70 & 3.9% \\ \hline Sep & 2000 & 60 & 30 \\ \hline Oct & 1800 & 70 & 3.0% \\ \hline Nov & 2200 & 90 & 3.9% \\ \hline \end{tabular}