Question

A Common Measure of Student Achievement, The National Assessment of Educational Progress (NAEP) is the only assessment that measures what U.S. students know and can do in various subjects across the nation, states, and in some urban districts. Also known as The Nation's Report Card, NAEP has provided important information about how students are performing academically since 1969.

NAEP is a congressionally mandated project administered by the National Center for Education Statistics (NCES) within the U.S. Department of Education and the Institute of Education Sciences (IES).

NAEP is given to a representative sample of students across the country. Results are reported for groups of students with similar characteristics (e.g., gender, race and ethnicity, school location), not individual students. National results are available for all subjects assessed by NAEP.

In the most recent year, The NAEP sample of 1077 young women had mean quantitative score of 275. Individual NAEP scores have a Normal distribution with standard deviation of 60. (This is indicating that the population standard deviation σσ is 60)

Find a 99% Confidence Interval for the mean quantitative scores for young women.

a) Check that the normality assumptions are met. ?  

b) What is the 99% confidence interval for the mean quantitative scores for young women? ____ ≤μ≤ _____

c) Interpret the confidence interval obtained in previous question.

Answer 1

Solution :

Given that,

Point estimate = sample mean = = 275

Population standard deviation =    = 60

Sample size = n =1077

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576   ( Using z table )

Margin of error = E = Z_/2* ( /n)

= 2.576* (60 / 1077)

E= 4.7097

At 99% confidence interval estimate of the population mean is,

- E ≤μ≤ + E

275 - 4.7097 ≤μ≤275 + 4.7097

270.2903≤μ≤ 279.7097

At 99% confidence interval estimate of the population mean is,270.2903≤μ≤ 279.7097

Answer 2

Solution :

Given that,

Point estimate = sample mean = = 275

Population standard deviation =    = 60

Sample size = n =1077

At 99% confidence level the z is ,

  = 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z_/2 = Z_0.005 = 2.576   ( Using z table )

Margin of error = E = Z_/2* ( /n)

= 2.576* (60 / 1077)

E= 4.7097

At 99% confidence interval estimate of the population mean is,

- E ≤μ≤ + E

275 - 4.7097 ≤μ≤275 + 4.7097

270.2903≤μ≤ 279.7097

At 99% confidence interval estimate of the population mean is,270.2903≤μ≤ 279.7097