Question

5. Mean of 1500, standard deviation of 300. Estimating the average SAT score, limit the margin of error to 95% confidence interval to 25 points, how many students to sample?

6. Population proportion is 43%, would like to be 95% confident that your estimate is within 4.5% of the true population proportion. How large of a sample is required?

7. P(z<1.34) (four decimal places)

8. Candidate only wants a 2.5% margin error at a 97.5% confidence level, what size of sample is needed?

9. Estimate this proportion to within 4% at the 95% confidence level, how many randomly selected college students must we survey?

10. 420 people were asked if they like dogs, 22% said they did. Find the margin of error for the poll at 95% confidence level. (four decimals

Answer 1

5. Mean of 1500, standard deviation of 300. Estimating the average SAT score, limit the margin of error to 95% confidence interval to 25 points, how many students to sample?

Answer)

As the population s.d is mentioned here we can use standard normal z table to estimate the sample size

Margin of error (MOE) = Z*S D/√N

critical value for 95% confidence level from z table is 1.96

Margin of error = 25

25 = 1.96*300/√n

N = 554

6. Population proportion is 43%, would like to be 95% confident that your estimate is within 4.5% of the true population proportion. How large of a sample is required?

Answer)

Margin of error (MOE) = Z*√{p*(1-p)}/√n

0.045 = 1.96*√{0.43*(1-0.43)}/√n

N = 465