Answered

Welcome to Westonci.ca, the ultimate question and answer platform. Get expert answers to your questions quickly and accurately. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.

A stationery store wants to estimate the mean retail value of greeting cards that it has in its inventory. A random sample of 15 greeting cards indicates an average value of $3.00 and a s of $0.40. The critical value for a 99% confidence interval is

Sagot :

Answer:

The critical value for a 99% confidence interval is T = 2.9768

Step-by-step explanation:

We have the standard deviation of the sample, which means that the t-distribution is used to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 15 - 1 = 14

Critical value

Now, we have to find a value of T, which is found looking at the t table, with 14 degrees of freedom(y-axis) and a confidence level of [tex]1 - \frac{1 - 0.99}{2} = 0.995[/tex]. So we have T = 2.9768

The critical value for a 99% confidence interval is T = 2.9768

Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.