A manager evaluates effectiveness of a major hardware upgrade by running a certain process 40 times before the upgrade and 40 times after it. Based on these data, the average running time is 8.6 minutes before the upgrade, 7.3 minutes after it. Historically, the standard deviation has been 1.8 minutes, and presumably it has not changed. Construct a 90% confidence interval showing how much the mean running time reduced due to the hardware upgrade.
Question
A manager evaluates effectiveness of a major hardware upgrade by running a certain process 40 times before the upgrade and 40 times after it. Based on these data, the average running time is 8.6 minutes before the upgrade, 7.3 minutes after it. Historically, the standard deviation has been 1.8 minutes, and presumably it has not changed. Construct a 90% confidence interval showing how much the mean running time reduced due to the hardware upgrade.
Solution
To construct a 90% confidence interval for the reduction in mean running time due to the hardware upgrade, we first need to calculate the mean difference in running time before and after the upgrade.
The mean running time before the upgrade is 8.6 minutes and after the upgrade is 7.3 minutes. So, the mean difference is 8.6 - 7.3 = 1.3 minutes.
Next, we need to calculate the standard error of the mean difference. The standard deviation is given as 1.8 minutes. The standard error (SE) is calculated as the standard deviation divided by the square root of the number of observations. Since the process was run 40 times before and 40 times after the upgrade, the total number of observations is 80.
SE = 1.8 / sqrt(80) = 0.201
Now, we can construct the 90% confidence interval. For a 90% confidence interval, the z-score is 1.645 (you can find this value in a standard z-table).
The confidence interval is calculated as the mean difference ± (z-score * SE).
Lower limit = 1.3 - (1.645 * 0.201) = 1.03 Upper limit = 1.3 + (1.645 * 0.201) = 1.57
So, the 90% confidence interval for the reduction in mean running time due to the hardware upgrade is between 1.03 minutes and 1.57 minutes.
Similar Questions
A tech company claims that their new software reduces the average time taken to process a specific task by 2 minutes compared to the existing software in the market. An independent tech reviewer wants to test this claim. They conduct an experiment with 40 users. Half of them use the new software, and the other half use the existing software. The differences in processing times (new software - existing software) are recorded for each user. The sample mean difference is found to be 1.5 minutes with a standard deviation of 1.2 minutes. Test the company's claim at a 1% significance level.
A university claims that their average graduation time for all their students is 4.7 years. The dean of the College of Business and Economics thinks that their students in that college graduate faster. They took a random sample of 150 students from their college and tracked them to see their graduation time. They found that the average graduation time for these students was 4.48 years with standard deviation of 0.83 year. Find the test statistic.
The manager of a computer shop is recording the time taken for customers to decide of which computer and accessories they buy from the time they enter the store. From the previous data, it is known that the average ‘decision’ time was 45 minutes. The manager assumes a normally distributed population with a standard deviation of 10 minutesa) What is the probability that a customer will take more than 60 minutes?
An analyst has timed a metal-cutting task for 50 cycles. The average time per cycle was 11.9 minutes, and the standard deviation was 1.32 minutes for a worker with a performance rating of 130 percent. Assume an allowance of 17 percent of task time. Calculate the standard time for this task. (Do not round intermediate calculations. Round the final answer to 2 decimal places.)
The manager of a grocery store has taken a random sample of 100 customers to determine whether or not the mean waiting time of all customers is significantly more than 3 minutes. She observes that the average length of time it took the customers in the sample to check out was 3.1 minutes. The population standard deviation is known to be 0.5 minutes.
Upgrade your grade with Knowee
Get personalized homework help. Review tough concepts in more detail, or go deeper into your topic by exploring other relevant questions.