Question

1. SAT scores are distributed with a mean of 1,500 and a standard deviation of 300. You are interested in estimating the average SAT score of first year students at your college. If you would like to limit the margin of error of your 95% confidence interval to 25 points, how many students should you sample Make sure to give a whole number answer.

2. You measure 26 textbooks' weights, and find they have a mean weight of 69 ounces. Assume the population standard deviation is 6.1 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight. Give your answers as decimals, to two places __ < μ __

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=30.6σ=30.6. You would like to be 99% confident that your estimate is within 3 of the true population mean. How large of a sample size is required?

Answer 1

Solution :

1) Given that,

Population standard deviation = = 300

Margin of error = E = 25

At 95% confidence level the z is,

= 1 - 95%

= 1 - 0.95 = 0.05

/2 = 0.025

Z_/2 = 1.96

sample size = n = [Z_/2* / E] ²

n = [1.96 *300 / 25 ]²

n = 553.19

Sample size = n = 554

2) Given that,

Point estimate = sample mean = = 69

Population standard deviation =    = 6.1

Sample size = n = 26

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 * ( /n)

= 1.645 * ( 6.1/  26 )

= 1.97

At 90% confidence interval estimate of the population mean is,

- E < < + E

69 - 1.97 <   < 69 + 1.97

( 67.03 <   < 70.97 )

3) Given that,

Population standard deviation = = 30.6

Margin of error = E = 3

At 99% confidence level the z is,

= 1 - 99%

= 1 - 0.99 = 0.01

/2 = 0.005

Z_/2 = 2.576

sample size = n = [Z_/2* / E] ²

n = [ 2.576 * 30.6 / 3]²

n = 690.39

Sample size = n = 691