Question

3) [4] Below are the observed and expected (in parentheses) frequencies for athletic injuries for those athletes who properly stretched versus those who did not. Assuming a ? 2 Test for Independence test is being conducted, at ? = 0.05, are athletic injuries independent of stretching? Only list steps 3-7 of the hypothesis test.

	Stretched	Not stretched	Totals
Injury	18 (20.8)	22 (19.2)	40
No Injury	211 (208.2)	189 (191.8)	400
Totals	229	211	440

Answer 1

The hypothesis for testing is:

The null hypothesis: H₀: Athletic injuries are independent of stretching

The alternative hypothesis: H_a: Athletic injuries are not independent of stretching

Load the data into Excel.

Go to Data>Megastat.

Select the option Chi-square/Crosstab and go to Contingency Table.

Select the Input Range as the data set.

Click OK.

The output obtained will be as follows:

		Stretched	Not stretched	Total
Injury	Observed	18	22	40
	Expected	20.82	19.18	40.00
	O - E	-2.82	2.82	0.00
	(O - E)² / E	0.38	0.41	0.80
No Injury	Observed	211	189	400
	Expected	208.18	191.82	400.00
	O - E	2.82	-2.82	0.00
	(O - E)² / E	0.04	0.04	0.08
Total	Observed	229	211	440
	Expected	229.00	211.00	440.00
	O - E	0.00	0.00	0.00
	(O - E)² / E	0.42	0.46	0.88
		.88	chi-square
		1	df
		.3495	p-value

The chi-square test statistic, X² from the output is 0.88.

The chi-square critical value, X²_criticalfor the degree of freedom, df = 1 and significance level = 0.05 from the chi-square critical value table is 3.841.

The p-value from the output is 0.3495.

Decision:

Since the p-value (0.3495) is greater than the significance level (0.05), we fail to reject the null hypothesis.

Decision:

Since the chi-square test statistic, X² (0.88) is less than the chi-square critical value, X²_critical (3.841), we fail to reject the null hypothesis.

Conclusion:

There is not sufficient evidence to warrant rejection of the claim that athletic injuries are independent of stretching.