Question

1. Assume that a sample is used to estimate a population proportion p. Find the 90% confidence interval for a sample of size 112 with 28% successes. Enter your answer as a tri-linear inequality using decimals (not percents) accurate to three decimal places.

2.  You want to obtain a sample to estimate a population proportion. Based on previous evidence, you believe the population proportion is approximately 88%. You would like to be 98% confident that your estimate is within 3% of the true population proportion. How large of a sample size is required? n= ________

3.  You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ=43.7σ=43.7. You would like to be 90% confident that your estimate is within 3 of the true population mean. How large of a sample size is required? n= _____

Answer 1

Solution :

Given that,

1) n = 112

Point estimate = sample proportion = = 28% = 0.28

1 - = 1 - 0.28 = 0.72

At 90% confidence level

= 1 - 90%

=1 - 0.90 =0.10

/2 = 0.05

Z/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 (((0.28 * 0.72) / 112)

= 0.070

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.28 - 0.070 < p < 0.28 + 0.070

(0.210 < p < 0.350)

2) Given that,

= 88% = 0.88

1 - = 1 - 0.88 = 0.12

margin of error = E = 3% = 0.03

At 98% confidence level

= 1 - 98%

=1 - 0.98 =0.02

/2 = 0.01

Z/2 = 2.326

sample size = n = (Z _{/ 2} / E )² * * (1 - )

= (2.326 / 0.03 )² * 0.88 * 0.12

= 634.80

sample size = n = 635

3) Given that,

Population standard deviation = = 43.7

Margin of error = E = 3

At 90% confidence level the z is,

= 1 - 90%

= 1 - 0.90 = 0.10

/2 = 0.05

Z_/2 = 1.645

sample size = n = [Z_/2* / E] ²

n = [1.645 * 43.7 / 3 ]²

n = 574.18

Sample size = n = 575