Question

Suppose that the sample proportion for the U.S. population aged between 25 and 34 years that has a Bachelors’s degree or higher is .35. Further suppose that this estimate came from a survery of 1282 people. (a) What are the end points of a 90% confidence interval estimate for the population proportion? (Round to 2 digits after the decimal place.) [ , ] (b) What are the end points of a 95% confidence interval estimate for the population proportion? (Round to 2 digits after the decimal place.) [ , ] (c) What are the end points of a 99% confidence interval estimate for the population proportion? (Round to 2 digits after the decimal place.) [ , ]

Answer 1

Solution:

Given,

n = 1282 ....... Sample size

Let denotes the sample proportion.

     = 0.35

a) 90% confidence interval estimate

c = 0.90

= 1- c = 1- 0.90 = 0.10

  /2 = 0.10 2 = 0.05 and 1- /2 = 0.950

= 1.645 (use z table)

E = /2 *  

= 1.645 * [0.35 *(1 - 0.35)/1282]

= 0.02

Now the confidence interval is given by

( - E)   ( + E)

(0.35 - 0.02)   (0.35 + 0.02)

0.33 0.37

90% confidence Interval is (0.33 , 0.37)

b)  95% confidence interval estimate

c = 0.95

= 1- c = 1- 0.95 = 0.05

  /2 = 0.05 2 = 0.025 and 1- /2 = 0.975

= 1.96

Now , the margin of error is given by

E = /2 *  

= 1.645 * [0.35 *(1 - 0.35)/1282]

= 0.03

Now the confidence interval is given by

( - E)   ( + E)

(0.35 - 0.03)   (0.35 + 0.03)

0.32 0.38

95% Confidence Interval is (0.32 , 0.38)

c)99% confidence interval

c = 0.99

= 1- c = 1- 0.99 = 0.01

  /2 = 0.005 and 1- /2 = 0.995

Search the probability 0.995 in the Z table and see corresponding z value

= 2.576

Now , the margin of error is given by

E = /2 *  

= 2.576* [0.35 *(1 - 0.35)/1282]

= 0.03

Now the confidence interval is given by

( - E)   ( + E)

(0.35 - 0.03)   (0.35 + 0.03)

0.32    0.38

99% confidence interval Interval is (0.32 , 0.38)