Question

- On March 28 at 1:30 pm, I randomly selected 10 states and found the number of coronavirus cases in each of them. This gave me an average of 939 cases with a standard deviation of 1543. Assume the number of cases in each state is normally distributed. Create a 90% confidence interval for the actual number of coronavirus cases in each state.

- On March 28 at 1:30 pm, the actual average number of cases in a U.S. state was 2102. Does this value fall in the interval you found in question 1. Discuss why or why not, and the reason you found the result you did.

- Using the scenario from question 1, how many states would need to be selected to be 90% confident that the sample mean number of cases would be within 500 of the actual mean?

Answer #1

A) df = 10 - 1 = 9

At 90% confidence level, the critical value is t0.05, 9 = 1.833

The 90% confidence interval is

No, 2102 doesn't lie in the confidence interval.

Margin of error = 500

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