Question

. A manager at a power company monitored the employee time required to process high. efficiency lamp bulb rebates. A random sample of 50 applications gave a sample mean time of 3.4 minutes and a standard deviation of 1.2 minutes. Could the manager conclude that the employee time required to process high-efficiency lamp bulb rebates is in fact lower than 3.6 ? Test the manager claim using uses a 5% signifieance level? Use any approach. A. Define both hypotheses. B. Calculate the test statistic. C. Calculate the p-value or use the classical approach. D. Do we reject the H.? Explain why. E. Write your conclusion in context with the problem

Answer 1

The manager can conclude that the employee time required to process high-efficiency lamp bulb rebates is lower than 3.6 minutes.

Step-by-step explanation:

To test the manager's claim, we will conduct a one-sample t-test. The null hypothesis (H0) is that the mean time required to process high-efficiency lamp bulb rebates is not lower than 3.6 minutes. The alternative hypothesis (Ha) is that the mean time is lower than 3.6 minutes.

To calculate the test statistic, we will use the formula: t = (x - μ) / (s / sqrt(n)), where x is the sample mean, μ is the population mean (3.6), s is the sample standard deviation, and n is the sample size. Plugging in the values, we get t = (3.4 - 3.6) / (1.2 / sqrt(50)) = -2.07.

The p-value can be calculated using the t-distribution table or software. At a significance level of 5%, the critical value for a one-tailed test is -1.677. Since the test statistic (-2.07) is less than the critical value, we can reject the null hypothesis. Therefore, we have enough evidence to conclude that the employee time required to process high-efficiency lamp bulb rebates is in fact lower than 3.6 minutes.