Question

Question 2 (1 point) (Quesitons 1-5 are in one hypothesis problem using Teaching Methodology data) A teacher of statistics wants to know if a new teaching methodology that includes IT is efficient in terms of increased average score. He took a class with old methodology and a class with new methodology for samples and gave a same test. Once you opened the file in question 1 and ran Excel, you need not open it again. Just use the Excel output to answer this question For the new methodology to be efficient, the difference in average scores between two population means (old - new, 11->) should be negative. Consider that the class with old methodology is the first group. We do not Inow the population variances, but assume they are NOT equal. We do hypothesis testing to see if the new methodology is efficient. What is the t-test statistic? Find it from Excel output that you generate a) 0.62 Ob) 0.5 OC) 0.31 Od) -0.5 Question 3 (1 point) (Quesitons 1-5 are in one hypothesis problem using Teaching Methodology cata). A teacher of statistics wants to know if a new teaching methodology that includes IT is efficient in terms of increased average score. He took a class with old methodology and a class with new methodology for samples and gave a same test. Once you opened the file in question 1 and ran Excel, you need not open it again Just use the Excel output to answer this question For the new methodology to be efficient, the difference in average scores between two population means (old - new, 11-42) should be negative. Consider that the class with old methodology is the first group. We do not know the population variances, but assume they NOT are equal. We do hypothesis testing to see if the new methodology is efficient. What is the p-value? Find it from Excel output that you generate. a) 0.62 Ob) 0.31 Oc) -0.5 d) 0.5

Answer 1

Solution:

Null and alternative hypotheses:

$H0:\mu_{1}-\mu_{1}\geq 0$

$Ha:\mu_{1}-\mu_{1}< 0$

Following is the excel output:

Hypothesis Test: Independent Groups (t-test, unequal variance)
	Old method	New method
	73.56	75.71	mean
	12.57	12.06	std. dev.
	16	17	n
		30	df
		-2.143	difference (Old method - New method)
		4.293	standard error of difference
		0	hypothesized difference
		-0.5	t
		.31	p-value (one-tailed, lower)

Here, t-statistic = -0.5

P-value = 0.31

Decision: P-value >0.05, fail to reject H0.

Conclusion: There is not sufficient evidence to support the claim that the difference in average scores between two population means should be negative.

Done