Question

"Hit" Group	"Smashed Into" Group
25	50
25	45
34	55
30	40
36	45
37	41
31	50
35	35
30	37
35	45
30	55
25	50
20	45
25	43
30	42
24	40
34	36
33	45
37	50
38	41

A. Null Hypothesis?

Ha: Smashed into group estimated higher speed than hit group

Ho: Smashed into group estimated lower or same speed than hit group(directional hypothesis)

B. Alternative hypothesis?

Ha: Smashed into group estimated higher speed than hit group

Ha: Smashed into group estimated lower or same speed than hit group

C. Risk Level?

.01

.5

.05

.25

D. What is the best statistical test to use?

correlation test

t-Test for dependent samples

t-Test for independent samples

z-test

E. Compute the test statistic.

Answer 1

A. Null Hypothesis?

Ho: Smashed into group estimated lower or same speed than hit group(directional hypothesis)

B. Alternative hypothesis?

Ha: Smashed into group estimated higher speed than hit group

C. Risk Level?

.05

D:

t-Test for independent samples

E:

Let sample 1: Hit group

Sample 2: "Smashed Into" Group

Following table shows the calculations for mean and sd for sample 1:

	X1	(X1-mean)^2
	25	32.49
	25	32.49
	34	10.89
	30	0.49
	36	28.09
	37	39.69
	31	0.09
	35	18.49
	30	0.49
	35	18.49
	30	0.49
	25	32.49
	20	114.49
	25	32.49
	30	0.49
	24	44.89
	34	10.89
	33	5.29
	37	39.69
	38	53.29
Total	614	516.2

Sample size n-20 Sample sum: Σ614 Sample mean: Σ =614/20 30.7 Sample standard deviation: Σ(-3) -=sqrt(516.2/(20-1)) -1 =5.2123

Following table shows the calculations for mean and sd for sample 2:

	X2	(X2-mean)^2
	50	30.25
	45	0.25
	55	110.25
	40	20.25
	45	0.25
	41	12.25
	50	30.25
	35	90.25
	37	56.25
	45	0.25
	55	110.25
	50	30.25
	45	0.25
	43	2.25
	42	6.25
	40	20.25
	36	72.25
	45	0.25
	50	30.25
	41	12.25
Total	890	635

Sample size n220 Sample sum Σ-390 2 Sample mean: Σ3 --890/20 44.5 Sample standard deviation: 2 Σ(-3) -=sqrt(635/(20-1)) S2 n-1 =5.7811

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Hypotheses are: 0 1- 7:°н о > - 7:"н a 0.05 Test is left tailed. Level of significance: Since we can assume that variances are equal so degree of freedom of t test will be: df n n2 =38 The pooled standard deviation is: (т -1)s+ (n-1)s'. 2 -5.5041 п +п, -2 Test Statistics: (2)-(4-P2) = ({30.7-44.5)-0)/(5.5041* sqrt((1/20)+(1/20)) 1 1 _ -7.929 п n2 Critical value: = -1.686 Rejection Region: Ift-1.686, Reject HO reject the null hypothesis. Decision: Since test statistics lies in rejection region so we P-value: P-value 0 Decision: Since p-value is less than level of significance so we reject the null hypothesis. Excel function for critical value: Excel function for p-value: TINV(0.1,38) TDIST (7.929,38,1)