Question

• From your data, what is the point estimate, p̂ of the population proportion?
• Write down the confidence interval that you obtained. Interpret the result.
• What is the margin of error?
• Using the same data, construct a 98% confidence interval for the population proportion. Then, answer the following three questions:

(i) What happens to the length of the interval as the confidence level is increased?

(ii) How has the margin of error changed?

(iii) If the sample size is increased, what do you think will happen to the margin of error? Why?

One sample proportion confidence interval:

Outcomes in : Sex
Success : M
p : Proportion of successes
Method: Standard-Wald

90% confidence interval results:

Variable	Count	Total	Sample Prop.	Std. Err.	L. Limit	U. Limit
Sex	4791	5373	0.89168063	0.0042398385	0.88470671	0.89865454

Answer 1

A) = 0.892

The 90% confidence interval is

0.8847 < p < 0.8987

We are 90% confident that the true population proportion lies in the above confidence interval.

Margin of error = (0.8987 - 0.8847)/2 = 0.007

B) At 98% confidence interval the critical value is z_0.01 = 2.33

The 98% confidence interval for the population proportion is

+/- z_0.01 * sqrt((1 - )/n)

= 0.892 +/- 2.33 * sqrt(0.892 * (1 - 0.892)/5373)

= 0.892 +/- 0.0099

= 0.8821, 0.9019

i) As the confidence level increases, the length of the confidence interval also increases.

ii) The margin of error has increased.

iii) As the sample size increases, the margin of error decreases.