A statistics professor created a course to assess its effect on GRE-Quantitative

Question

A statistics professor created a course to assess its effect on GRE-Quantitative scores. GRE-Q has a population mean of 500 and a population standard deviation of 100. This statistics professor took a SRS of 16 students and had them go through his course. The 16 students had an average (mean) GRE-Q score of 560. Assume the sample of 16 scores are normal in shape. Use this information to answer questions (18 - 25).

18.       State the null and alternative/research hypotheses for the above research scenario.

19.       Given the above null and alternative/research hypotheses and with alpha = .05, what is     Zcrit equal to?

20.       What is the value of Zobt?

21.       What is the p-value for the outcome (what is the probability of getting an outcome this extreme (one or two tail test?)?

22.       Was there are relationship between the course (course vs. no course) and the GRE-Q?

23.       If appropriate, determine a 95% confidence interval for the sample mean = 560.

24.       If appropriate, determine Cohen’s d.

25.       What is the probability of the error that the statistics professor might be making?

Explanation / Answer

18) H0: mu = 500

    H1: mu not equal to 500

19) At alpha = 0.05, the critical value is z0.025 = 1.96

20) The test statistic z = (bar x - mu)/(sigma/sqrt(n))

                                   = (560 - 500)/(100/sqrt(16))

                                    = 2.4

21) P-value = 2 * P(Z > 2.4)

                    = 2 * 1 - P(Z < 2.4)

                     = 2 * (1 - 0.9918)

                     = 2 * 0.0082 = 0.0164

As the P-value is less than the significance level (0.0164 < 0.05), so the null hypothesis is rejected.

So we can conclude that there is a difference between the course and the GRE-Q.

23) At 95% confidence interval the critical value is z0.025 = 1.96

So the 95% confidence interval is

bar x +/- z0.025 * sd/sqrt(n)

= 560 +/- 1.96 * 100/sqrt(16)

= 560 +/- 49

= 511, 609

24) Cohen's d = (bar x - mu)/sd

                       = (560 - 500)/100 = 0.6

25) The probability of error that the statistics professor might be making = 0.05

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