Confidence Interval for Proportions Og = Standard Error for the sample proportio

Question

Confidence Interval for Proportions Og = Standard Error for the sample proportion rt Finding the sample proportion and the standard error for P Example 3 In a recent survey of 1 50 students, |25 had a student loan. Find and a confidence interval for a population proportion Constructing Example 4. Find the 95% confidence interval of the true proportion of students who have student loan. Margin Error is The confidence interval is Hence, we can be 95% confident that the percentage of students who study wit and based on a sample or size loan is between

Explanation / Answer

Ans:

3)sample proportion=125/150=0.833

standard error of sample proportion=sqrt(0.833*(1-0.833)/150)=0.030

4)

z value for 95% CI is 1.96

Margin of error=1.96*0.030=0.059

95% Confidence interval of p

=0.833+/-0.059

=(0.774, 0.892)

We are 95% confident the percentage of students who study with loan is between 77.4% and 89.2% based on sample size of 150.

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