Question

A computer literacy survey conducted by the ministry of health among a random sample of 1000 skilled employees in Kenya showed that 40% of them are computer literate. Based on this information, determine the 95% confidence interval for computer literacy among employees in Kenya.

Answer 1

The formula for confidence interval is given by,

$p\ \pm \ Z\alpha*\sqrt{\frac{p(1-p)}{n}}$ .............(Eq. 1)

where,

p = probability of employees who are computer literate from random selection of samples

n = sample size

In our case,

p = 40% = 0.4

n = 1000

$\alpha$ = 1-95%= 5%

Since it is a 2 tailed test the value of Z = Z$\alpha$_/2=Z_0.025

From the standard normal distribution table we get the value of Z_0.025 = 1.96

Substituting the values in equation 1 we get,

$0.4\ \pm\ 1.96\sqrt{\frac {0.4(1-0.4)}{1000}} = 0.4\pm0.0155$

Hence the confidence interval for literacy rate is (0.3845, 0.4155)