Question

USING R : A random sample of 40 students took an SAT preparation course prior to taking the SAT. The sample mean of their quantitative SAT scores was 560 with a s.d. of 95, and the sample mean of their verbal SAT scores was 525 with a s.d. of 100. Suppose the mean scores for all students who took the SAT at that time was 535 for the quantitative and 512 for the verbal. Do the means for students who take this course differ from the corresponding means for all students at the 10% level of significance?

Answer 1

Solution:-

1)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 535
Alternative hypothesis: u 535

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 15.0208

DF = n - 1

D.F = 39
t = (x - u) / SE

t = - 1.664

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 39 degrees of freedom is less than -1.664 or greater than 1.664.

Thus, the P-value = 0.104

Interpret results. Since the P-value (0.104) is greater than the significance level (0.10), we cannot reject the null hypothesis.

2)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u = 512
Alternative hypothesis: u 512

Note that these hypotheses constitute a two-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.10. The test method is a one-sample t-test.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees of freedom (DF), and the t statistic test statistic (t).

SE = s / sqrt(n)

S.E = 15.8114

DF = n - 1

D.F = 39
t = (x - u) / SE

t = - 0.822

where s is the standard deviation of the sample, x is the sample mean, u is the hypothesized population mean, and n is the sample size.

Since we have a two-tailed test, the P-value is the probability that the t statistic having 39 degrees of freedom is less than -0.822 or greater than 0.822.

Thus, the P-value = 0.416.

Interpret results. Since the P-value (0.416) is greater than the significance level (0.10), we cannot reject the null hypothesis.