Statistics Homework

Question

The mean SAT score in mathematics is 541

. The standard deviation of these scores is 39

. A special preparation course claims that the mean SAT score, μ

, of its graduates is greater than 541

. An independent researcher tests this by taking a random sample of 100

 students who completed the course; the mean SAT score in mathematics for the sample was 545

. At the 0.05

 level of significance, can we conclude that the population mean SAT score for graduates of the course is greater than 541

? Assume that the population standard deviation of the scores of course graduates is also 39

State the null hypothesis H0

 and the alternative hypothesis H1

.

H0:

H1:

(b)Determine the type of test statistic to use.▼(Choose one)(c)Find the value of the test statistic. (Round to three or more decimal places.)

(d)Find the critical value. (Round to three or more decimal places.)

(e)Can we support the preparation course's claim that the population mean SAT score of its graduates is greater than 541

?

Yes 

No

.

Answer 1

a)

The prep course claims that its graduates have means larger than the SAT mean of 541.

State the null hypothesis H0

 and the alternative hypothesis H1

H0: µ = 541

H1: μ > 541

b) Since we know the standard deviation of the sample the researcher used to test this hypothesis (σ= 39), we can use the z-statistic. If we would not have known σ, we would have had to use the t-statistic.

c) z = (Xmean - µ) / (σ/ √n)

We will compute the z statistic and then subsequently find the probability of the sample mean being larger than µ = 541 under the null hypothesis.

z = (545- 541)/( 39/ 10) = 1.025

Because of the Central Limit Theorem which states that the distribution of the sample means will be approximately normally distributed:

P(Xmean > 541) = p( z > 1.025) ≈ 1- p(z <1.03) = 1-0.84849 = 0.1515

Thus, even under the null hypothesis in which mean scores post test prep is µ= 541, the probability of observing a sample Xmean = 545 is approx. 15%.

(d) For such an upper-tailed z-test, in order to be able to reject the null hypothesis at the α= 0.05 level of significance, we would need to have z = 1.645, which is the critical value for the z test. We can only reject H0 if z>= 1.645.

(e) We already have the answer for this under (c): We cannot support the preparation course's claim that the population mean SAT score of its graduates is greater than 541 at the level of significance α= 0.05. Even under the null hypothesis H0, the probability of observing a sample mean of 545 is approx. 15%.