Does the data suggest that true average maximum lean angle for older females (OF) is more than 10 degrees smaller than it is for younger females (YF)?

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As the population ages, there is increasing concern about accident-related injuries to the elderly. An article reported on an experiment in which the maximum lean angle—the farthest a subject is able to lean and still recover in one step—was determined for both a sample of younger females (21–29 years) and a sample of older females (67–81 years). The following observations are consistent with summary data given in the article:

YF: 28, 34, 32, 27, 28, 32, 31, 35, 32, 28
OF: 19, 14, 21, 13, 12
 

Does the data suggest that true average maximum lean angle for older females (OF) is more than 10 degrees smaller than it is for younger females (YF)? State and test the relevant hypotheses at significance level 0.10. (Use ?1 for younger females and ?2 for older females.)

H0: ?1 − ?2 = 10
Ha: ?1 − ?2 > 10H0: ?1 − ?2 = 10
Ha: ?1 − ?2 < 10    H0: ?1 − ?2 = 0
Ha: ?1 − ?2 > 0H0: ?1 − ?2 = 0
Ha: ?1 − ?2 < 0


Calculate the test statistic and determine the P-value. (Round your test statistic to one decimal place and your P-value to three decimal places.)

t  =   
P-value  =   



State the conclusion in the problem context.

Fail to reject H0. The data suggests that true average lean angle for older females is not more than 10 degrees smaller than that of younger females.Reject H0. The data suggests that true average lean angle for older females is not more than 10 degrees smaller than that of younger females.    Fail to reject H0. The data suggests that true average lean angle for older females is more than 10 degrees smaller than that of younger females.Reject H0. The data suggests that true average lean angle for older females is more than 10 degrees smaller than that of younger females.
Expert Solution
Step 1

Given data,

For Younger Female: 28, 34, 32, 27, 28, 32, 31, 35, 32, 28

Sample Size, N1 = 10

Sum, x = 307

Mean, x1¯ = xN1

Mean, x1¯ = 30710

Mean,  x1¯ = 30.7

Standard Deviation, s1 = 1N1-1i=1N(xi-x1¯)2

s1 = (xi-x1¯)2N1-1

s1 = (28-30.7)2+(34-30.7)2+(32-30.7)2+(27-30.7)2+(28-30.7)2+(32-30.7)2+(31-30.7)2+(35-30.7)2+(32-30.7)2+(28-30.7)210-1

s1 = 70.19

s1 = 7.7889

s1 = 2.791

 

 

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