Question

1. A medical institute needs information concerning cough and fever symptoms of the surrounding metropolitan population. A random sample of 420 people found that 98 had at least one of the above ailments in the past month. Find a 99% confidence interval for the population proportion that may be at risk of exhibiting similar symptoms (4 Points).

Conditions:

Work:

Interval:

Conclusion:

Answer 1

Solution :

Given that,

n = 420

x = 98

Point estimate = sample proportion = = x / n = 98/420=0.233

1 -   = 1- 0.233 =0.767

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 ( Using z table )

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

= 2.576* (((0.233*0.767) /420 )

= 0.0531

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.233- 0.0531< p < 0.233+0.0531

0.1799< p < 0.2861

The 99% confidence interval for the population proportion p is : 0.1799, 0.2861