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: 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

here we use independent-samples t-test with

(step1)

null hypothesis H0:1=2 (science students perform same (or not different)  on multiple choice tests(1) than on essay tests (2))and

alternate hypothesis H1:1≠2 (science students perform differently on multiple choice tests(1) than on essay tests (2))

(step 2) critical t(0.01/2,10)=3.17

statistic t=|(mean1-mean2)|/((s_p*(1/n1 +1/n2)^1/2)=0.4060 with df is n=n1+n2-2 =10

and s_p²=((n1-1)s1²+(n2-1)s2²)/n=4.55

(step 3) Accept null hypothesis H0

(step4) since two tailed p-value is more than alpha=0.01, so we accept H0 and conclude that science students perform same (or not different)  on multiple choice tests(sample1) than on essay tests (sample2))and

	sample	mean	s	s2	n	(n-1)s2
	multiple choice	6.5000	1.8708	3.5000	6	17.5000
	essay	6.0000	2.3664	5.6000	6	28.0000
	difference=	0.5000	sum=	9.1000	12	45.5000
	sp2=	4.5500
	sp=	2.1331
	SE=	1.2315
	t=	0.4060
one tailed	p-value=	0.3466
two tailed	p-value=	0.6933
two tialed critical	t(0.05)	2.2281
one tailed critical	t(0.05)	1.8125