Question

Scenario:

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.

Question:

3. 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

Solution :

Given that,

1) Point estimate = sample mean = = 939

sample standard deviation = s = 1543

sample size = n = 10

Degrees of freedom = df = n - 1 = 10 - 1 = 9

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

t/2,df = t0.05,9 = 1.833

Margin of error = E = t/2,df * (s /n)

= 1.833 * ( 1543 / 10)

Margin of error = E = 894.39

The 90% confidence interval estimate of the population mean is,

  ± E  

= 939 ± 894.39

= ( 44.61, 1833.39 )

2) Margin of error = E = 500

sample size = n = [t/2,df * s / E] 2

n = [1.833 * 1543 / 500]2

n = 31.99

Sample size = n = 32