Question

The College Board reported the following mean scores for the three parts of the Scholastic Aptitude Test (SAT) (The World Almanac, 2009): Critical Reading 502 Mathematics515 Writing 494 Assume that the population standard deviation on each part of the test is 10 a) What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 502 on the Critical Reading part of the test? b) What is the probability a sample of 90 test takers will provide a sample mean test score within 10 points of the population mean of 515 on the Mathematics part of the test? Compare this probability to the value computed in part (a). c) What is the probability a sample of 100 test takers will provide a sample mean test score within 10 of the population mean of 494 on the writing part of the test? Comment on the differences between this probability and the values computed in parts (al and (b).

Answer 1

(a)

Critical reading:

= 502

= 100

n = 90

SE = /

= 100/ = 10.5409

Z= 10/10.5409 = 0.9487

Table of Area Under Standard Normal Curve gives area = 0.3289

So,

Probability that sample mean test score within 10 pointsof population mean = 2 X 0.3289 = 0.6578

So,

Answer is:

0.6578

(b)

Mathematics:

= 515

= 100

n = 90

SE = /

= 100/ = 10.5409

Z= 10/10.5409 = 0.9487

Table of Area Under Standard Normal Curve gives area = 0.3289

So,

Probability that sample mean test score within 10 points of population mean = 2 X 0.3289 = 0.6578

So,

Answer is:

0.6578

The probability value = 0.6578 is same as the value computed in (a)

(c)

Writing:

= 494

= 100

n = 100

SE = /

= 100/ = 10

V100

Z= 10/10 = 1

Table of Area Under Standard Normal Curve gives area = 0.3413

So,

Probability that sample mean test score within 10 points of population mean = 2 X 0.3413= 0.6826

So,

Answer is:

0.6826

The probability value obtained in (c) = 0.6826 is greater than that obtained in (a) & (b) = 0.6578. This increase is due to increase in Sample Size from n = 90 to n = 100.