Question

Two hundred patients who had symptoms were also tested. Of these patients,9 percent was infected with the virus. Find the 95% confidence interval for the percent of all paitents who have symptoms at that hospital and who are infected with the virus

Answer 1

Confidence interval for Population Proportion is given as below:

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We are given

n = 200

P = x/n = 0.09

Confidence level = 95%

Critical Z value = 1.96

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.09 ± 1.96* sqrt(0.09*(1 – 0.09)/ 200)

Confidence Interval = 0.09 ± 1.96* 0.0202

Confidence Interval = 0.09 ± 0.0397

Lower limit = 0.09 - 0.0397 = 0.0503

Upper limit = 0.09 + 0.0397 = 0.1297

Confidence interval = (0.0503, 0.1297)

Confidence interval = (5.03%, 12.97%)