Question

The mean SAT score in mathematics, u, is 512. The standard deviation of these scores is 25. A special preparation course claims that its graduates will score higher, on average, than the mean score 512. A random sample of 25 students completed the course, and their mean SAT score in mathematics was 520. Assume that the population is normally distributed. At the 0.1 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 25. 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. The null hypothesis: The alternative hypothesis: H :| The type of test statistic: (Choose one) D+O O_O > 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

The given details are:

$\bar{X}=520$ sample size n = 25 and the standard deviation $\sigma$ = 25.

For the given details and based on the claim the hypotheses are:

Null Hypothesis

$Ho: \mu =512$

Alternate hypothesis

$Ha: \mu > 512$

Based on the hypothesis it will be a One-tailed test.

Test Statistic:

Since the population standard deviation is known hence Z -distribution is applicable for hypothesis testing.

$Z = \frac{\bar{X}-\mu}{\frac{\sigma }{\sqrt{n}}} = \frac{520-512}{\frac{25}{\sqrt{25}}}=1.6$

Rejection region:

Based on the significance level 0.10 the critical value for the rejection region is computed using excel formula which is =NORM.S.INV(1-0.10), thus the Zc computed as 1.282.

So, reject Ho if Z > Zc and P-value is less than 0.10.

P-value:

The P-value is computed based on the Test statistic calculated above using excel formula for normal distribution which is =1-NORM.S.DIST(1.6, TRUE), this results in P-value = 0.055.

Conclusion:

Since the P-value is less than 0.10 and Z is greater than Zc hence we can reject the null hypothesis and hence conclude that there is sufficient evidence to support the claim.

Yes.