Question

5. A professor theorizes that science students perform differently on multiple choice tests than on essay tests. The professor recruited 6 science students and had them take both types of test. Their test scores were recorded. The data are given in the following table. Based on the data, can you support the professor’s theory? Perform an appropriate test at α = 0.01.

Student	1	2	3	4	5	6
Multiple choice	8	7	9	4	6	5
Essay	5	8	7	3	9	4

• Which hypothesis test should be used?

 one-sample Z-test  one-sample t-test  independent-samples t-test  paired-samples t-test

• Step 1. State the hypotheses (both in words and in symbols)

• H₀:

• H₁:

• Step 2: α = 0.01

• Step 3: Use SAS to calculate a test statistic (write a SAS code for reading in the data set and analyzing it, and upload your SAS code on the Cyber campus website). Copy and paste the relevant SAS output. Report the obtained test statistic and calculate effect size (Cohen’s d).

• Step 4: Make a decision.

• Step 5: State a conclusion (You should include the test statistic, p-value, effect size, and a verbal interpretation of the result).

Answer 1

Solution:-

Paired samples t-test.

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

Null hypothesis: u_d = 0

Alternative hypothesis: u_d ≠ 0

Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the difference between sample means is too big or if it is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.01. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

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

Student	Multiple choice	Essay	d = A - B	(d - dbar)^2
1	8	5	3	6.25
2	7	8	-1	2.25
3	9	7	2	2.25
4	4	3	1	0.25
5	6	9	-3	12.25
6	5	4	1	0.25
Sum	39	36	3	23.5
Mean	6.5	6	0.5	3.916666667

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 2.168

SE = s / sqrt(n)

S.E = 0.8851

DF = n - 1 = 6 -1

D.F = 5

t = [ (x₁ - x₂) - D ] / SE

t = 0.565

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Since we have a two-tailed test, the P-value is the probability that a t statistic having 5 degrees of freedom is more extreme than 0.565; that is, less than - 0.565 or greater than 0.565.

Thus, the P-value = 0.596

Interpret results. Since the P-value (0.596) is greater than the significance level (0.01), hence we failed to reject the null hypothesis.

Do not reject H₀. From the above test we have sufficient evidence in the favor of the claim that science students perform differently on multiple choice tests than on essay tests.