Question

1. A primary health center health center reported that in a sample of 400 patients 80 had severe fever. The director of health has asked you to compute a 90% confidence interval for the population proportion with severe fever.

1. Compute the required confidence interval. Your answer should follow the logical sequence as done in class.

Incidentally the 95% confidence interval is (0.1608, 0.2392).

2. Explain why the 95% confidence interval will be larger than the 90% confidence interval.

The Director went and told the minister that the probability that the disease rate is between 16.08% and 23.92% is 0.95. Explain (briefly) why this is nonsensical!

Answer 1

(a)

Step 1:

n = 400

= 80/400 = 0.2

n = 400 X 0.2 = 80 > 10

n (1 - ) = 400 X 0.8 = 320 > 10

Both conditions are satisied.

So, the sample can be taken as large sample

Step 2:

= 0.10

From Table, critical values of Z = 1.645

Step 3:
Confidence Interval:

So,

Answer is:

(0.1671, 0.2329)

(b)

The 95% confidence interval (0.1608, 0.2392). will be larger than the 90% confidence interval (0.1671, 0.2329) because increasing the confidence level from 90% to 95% will increase the Margin of Error, resulting in a wider confidence interval.Alarger Margin of Error produces awider confidence interval that is more likely to contain the population proportion.

(c)

To interpret the 95% confidence interval (0.1608, 0.2392) as the probability that the disease rate is between 16.08% and 23.92% is 0.95 is nonsensical because the 95% confidence interval (0.1608, 0.2392). is a range of values we are 95% confident will contain the true unknown population proportion. If repeated samples are taken and the 95% confidence interval iscalculated foreach sample, 95% of these confidence intervals will contain the true unknown population proportion.