Answer to Question #263105 in Statistics and Probability for Cameron

Question

In a random sample of 200 observations, we found the proportion of successes to be 48%. Estimate with 95% confidence the population proportion of successes. (b) Repeat part (a) with n = 500. (c) Repeat part (a) with n = 1000. (d) Describe the effect on the confidence interval estimate of increasing the sample size.

Answer 1

The confidence interval for proportion can be estimated by the next formula:

$P(u \in (p-Cr({\frac {1+a} 2})*\sqrt{{\frac {p(1-p)} n}};p+Cr({\frac a 2})*\sqrt{{\frac {p(1-p)} n}}))=a$ , where p is the sample proportion mean, n is a sample size, Cr is a critical value, a is a confidence level.

Since we have a big sample size in each situation, then we can use a normal-distributed critical value.

$P(N(0,1)>Cr)={\frac {1+a} 2}=0.975\implies Cr=1.96$

By substitution in the formula values Cr = 1.96, p =0.48, we obtain:

$P(u \in (0.48-1.96*\sqrt{{\frac {0.2496} n}};0.48+1.96*\sqrt{{\frac {0.2496} n}}))=0.95$

(a) for n = 200 we have

$P(u \in (0.48-1.96*0.0353;0.48+1.96*0.0353))=0.95$

the 95% confidence interval is $u \in (0.411;0.549)$

(b) for n = 500 we have

$P(u \in (0.48-1.96*0.0223;0.48+1.96*0.0223))=0.95$

the 95% confidence interval is $u \in (0.436;0.524)$

(c)for n = 1000 we have

$P(u \in (0.48-1.96*0.0158;0.48+1.96*0.0158))=0.95$

the 95% confidence interval is $u \in (0.449;0.511)$

(d) As we can see, the increasing of the sample size makes confidence interval more narrow, which is quite logical, cause the more data you have, the more accurate conclusion you can make