Question

Problems 1-5: We want to know the true proportion of students that are ok with online courses. We take a sample of 120 students, and 72 said they are ok with online courses. We want to create a 95% confidence interval. Answer the questions below:

1) What is the point estimate for the population proportion?

2) What is the standard error?

3) What is the z score for 95% confidence?

4) What is a 95% confidence interval for this data?

5) What is the margin of error for this data?

Answer 1

1)

Point estimate for proportion = 72 / 120 = 0.60

2)

Standard error = sqrt [ ( 1 - ) / n ]

= sqrt [ 0.60 ( 1 - 0.60) / 120 ]

= 0.0447

3)

From Z table,

z-score = 1.96

4)

95% confidence interval is

- Z * SE < P < + Z * SE

0.60 - 1.96 * 0.0447 < p < 0.60 + 1.96 * 0.0447

0.512 < p < 0.688

95% CI is ( 0.512 , 0.688 )

5)

Margin of error = Z * SE  

= 1.96 * 0.0447  

= 0.088