Answer to Question #169489 in Statistics and Probability for Sunera

Question

In a random sample of 500 observations, we found the proportion of successes to be 48%.

Estimate with 95% confidence the population proportion of successes.

a) Repeat part (a) with n=200.

b) Repeat part (a) with n = 1000.

c) Describe the effect on the confidence interval estimate of increasing the sample size.

Answer 1

Solution:

Given, $p=48\%=0.48, n=500$

So, $q=1-0.48=0.52$

(a) Standard error, $SE=\sqrt{\dfrac{pq}{n}}=\sqrt{\dfrac{0.48\times0.52}{500}}=0.022342$

z-value for 95% confidence interval is 1.96.

Now, confidence interval: $1.96\pm0.022342$

$=[1.937658,\ 1.982342]$

(b) n = 200

Standard error, $SE=\sqrt{\dfrac{pq}{n}}=\sqrt{\dfrac{0.48\times0.52}{200}}=0.035327$

z-value for 95% confidence interval is 1.96.

Now, confidence interval: $1.96\pm0.035327$

$=[1.924673,\ 1.995327]$

(c) n = 1000

Standard error, $SE=\sqrt{\dfrac{pq}{n}}=\sqrt{\dfrac{0.48\times0.52}{1000}}=0.015798$

z-value for 95% confidence interval is 1.96.

Now, confidence interval: $1.96\pm0.015798$

$=[1.944202,\ 1.975798]$

(d) Significant effect of increasing sample size on confidence interval is that it becomes precise or shorter than earlier.