Question

A report regarding the status of corona virus spread in California published Monday (March 23, 2020) showed 2133 positive cases out of a total 25,200 tests. Using the 95% confidence level, what is the confidence interval for the proportion of California residents having the corona virus.

BONUS: A population has a standard deviation of 50. A randomly selected sample of 51 items is taken from the population. The sample has a mean of 52. What is the margin of error at the 90% confidence interval? 95% confidence interval? 99% confidence interval?

Answer 1

Solution :

n = 25,200

x = 2133

= x / n = 2133 / 25,200 = 0.085

1 - = 1 - 0.085 = 0.915

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 1.960 * (((0.085 * 0.915) /25,200)

= 0.003

A 95% confidence interval for population proportion p is ,

- E < P < + E

0.085 - 0.024 < p < 0.085 + 0.003

0.081 < p < 0.088

Given that,

= 52

= 50

n = 51

a ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z/2 = Z0.05 = 1.645

Margin of error = E = Z/2* (/n)

= 1.645 * (50 / 51 )

= 11.5

Margin of error = 11.5

b ) At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.960

Margin of error = E = Z/2* (/n)

= 1.960 * (50 / 51 )

= 13.7

Margin of error =13.7

c ) At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576

Margin of error = E = Z/2* (/n)

=2.576 * (50 / 51 )

= 18.0

Margin of error =18.0