Assume that a sample is used to estimate a population proportion .

Question

Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 101 with 48 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

90% C.I. = 

Answer should be obtained without any preliminary rounding. However, the critical value may be rounded to 3 decimal places.

Answer 1

In general, CI will look like p̂-E<p<p̂+E, where p̂ is the point estimate of population proportion p i.e. sample proportion and E is the margin of error ( E= z_α/2*√( (p̂*q̂)/n) )

1.  Identify the sample statistics n and x.          n=101, x=48

2.  Find the point estimate p̂ (which is sample proportion). p̂=x/n=48/101=0.475

3.  Verify that the sampling distribution of p̂ can be approximated by a normal distribution (np̂≥5 and nq̂≥5).       101*0.475=47.975≥5 and 101*(1-0.475)=53.025≥5

4.  Find the critical value z_α/2 (using z-Table or any stat app) that corresponds to the given level of confidence.   In our case confidence level (1-α)%=90% so α=10%=0.1. Hence, z_0.1/2=1.645

5.  Find the margin of error E= z_α/2*√( (p̂*q̂)/n) = 1.645*√((0.4752*0.5248)/101)=0.0817

6.  Form the Interval. 90% CI for population proportion p is from p - E to p + E or p=p̂±E=0.4752±0.0817. Hence, the 90% CI is (0.3935, 0.5570)