Question

The mean SAT score in mathematics,

μ

, is

559

. The standard deviation of these scores is

39

. A special preparation course claims that its graduates will score higher, on average, than the mean score

559

. A random sample of

50

students completed the course, and their mean SAT score in mathematics was

561

. At the

0.05

level of significance, can we conclude that the preparation course does what it claims? Assume that the standard deviation of the scores of course graduates is also

39

.

Perform a one-tailed test. Then fill in the table below.

Carry your intermediate computations to at least three decimal places, and round your responses as specified in the table. (If necessary, consult a list of formulas.)

The null hypothesis: H0: The alternative hypothesis: H1: The type of test statistic: The value of the test statistic: (Round to at least three decimal places.) The p-value: (Round to at least three decimal places.) Can we support the preparation course's claim that its graduates score higher in SAT? Yes No

Answer 1

Solution:

Given:

Claim: Graduates will score higher, on average, than the mean score 559.

Sample size = n = 50

Sample mean =

Level of significance = 0.05

Part 1) The null hypothesis:

Part 2) The alternative hypothesis:

Part 3) The type of test statistic:

Since sample size is large and population standard deviation is known we use z test.

Part 4) The value of the test statistic:

Part 5) The p-value:

p-value = P( Z> z test statistic)

p-value = P( Z> 0.363)

p-value = 1 - P( Z< 0.363)

Use following Excel command:

=1-NORM.S.DIST(z,cumulative)

=1-NORM.S.DIST(0.363,TRUE)

=0.358

Thus p-value = 0.358

Part 6) Can we support the preparation course's claim that its graduates score higher in SAT?

Since p-value = 0.358 > 0.05 significance level, we fail to reject H0, thus we can not support the preparation course's claim that its graduates score higher in SAT.

Thus answer is: No