Question

1. Using the scenario from question 1 (Question 1; 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.) now,  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

Solution :

Given that,

standard deviation =s =   =1543

Margin of error = E = 500

At 90% confidence level

= 1 - 90%  

= 1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z0.05 = 1.65 ( Using z table ( see the 0.05 value in standard normal (z) table corresponding z value is 1.65 )  

sample size = n = [Z/2* / E] 2

n = ( 1.65 * 1543/ 500)2

n =25.92

Sample size = n =26