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%.
