I. Methodology:

For this hypothesis test, we will use a one-sample t-test to determine if the number of defective flash drives produced by the company per week is less than seven.

Null Hypothesis: The number of defective flash drives produced by the company per week is equal to or greater than seven. (µ ≥ 7)

Alternative Hypothesis: The number of defective flash drives produced by the company per week is less than seven. (µ < 7)

Significance level: 0.05

II. Analysis:

First, we will need to calculate the sample mean and standard deviation of the number of defective flash drives produced by the company over the last 30 weeks. The sample data is given as:

4, 5, 3, 4, 6, 5, 3, 4, 6, 5, 4, 3, 6, 5, 4, 3, 6, 5, 4, 3, 6, 5, 4, 3, 6, 5, 4, 3, 6, 5

Sample Mean: 4.5 Sample Standard Deviation: 0.94 Sample Size: 30

Next, we will calculate the t-test statistic using the sample mean, sample standard deviation, and the hypothesized population mean (µ = 7).

t = (4.5 – 7) / (0.94 / √30) = -6.13

The p-value can be determined using a t-distribution table with 29 degrees of freedom and a two-tailed test (since we are testing if the mean is less than or equal to 7, and not just less than 7).

p-value = 0.0001

Given the small p-value (less than 0.05), we can reject the null hypothesis and conclude that the number of defective flash drives produced by the company per week is less than seven.

Decision: Reject the Null Hypothesis

Conclusion: Based on the sample data and hypothesis test, we can conclude that the number of defective flash drives produced by the company per week is less than seven with a significance level of 0.05. This supports the claim of the operations manager that the number of defects produced by the process is less than seven defective flash drives per week.