Question

You want to build an interval estimate for the proportion of all college-bound high school graduates in Indiana who plan to pursue a career in science.
										A survey a random sample 800 such students yields a point estimate of 22% from the survey.
13	The standard error of the sample proportion is,
a	0.0146
b	0.0159
c	0.0172
d	0.0186
14	The margin of error for a confidence interval with a confidence level of 90% is
a	0.034
b	0.029
c	0.024
d	0.019
15	The upper end of the 95% confidence interval for the population proportion is,
a	0.249
b	0.227
c	0.208
d	0.190

Answer 1

Question 13

We are given n = 800, p̂ = 0.22

Standard error = sqrt(p̂*(1 - p̂)/n)

Standard error = sqrt(0.22*(1 – 0.22)/800)

Standard error = sqrt(0.22*0.78/800)

Standard error = 0.014646

Answer: a. 0.0146

Question 14

Margin of error = Z*Standard error

We are given

Confidence level = 90%

So, Z = 1.6449 (by using z-table)

Margin of error = 1.6449*0.014646

Margin of error = 0.024091

Margin of error = 0.024

Answer: c. 0.024

Question 15

Upper end = p̂ + z*sqrt(p̂*(1 - p̂)/n)

Upper end = 0.22 + 1.6449*0.014646

Upper end = 0.22 + 0.024091

Upper end = 0.244091

Upper end = 0.244

Approximate answer: a. 0.249