9-64. Reconsider the data from Medicine and Science in Sports and Exercise descr

Question

9-64. Reconsider the data from Medicine and Science in Sports and Exercise described in Exercise 8-30. The sample size was seven and the sample mean and sample standard devi- ation were 315 W and 16 W, respectively (a) Is there evidence that leg strength exceeds 300 W at signifi- cance level 0.05? Find the P-value. (b) Compute the power of the test if the true strength is (c) What sample size would be required to detect a true (d) Explain how the question in part (a) could be answered 304 W mean of 304 W if the power of the test should be at least 0.90? with a confidence interval

Explanation / Answer

(a) Here

sample mean x = 315 W

sample standard deviation s= 16 W

standard error of sample mean = s/ sqrt(n) = 16/ sqrt(7) = 6.047

Test statistic

t = (x -H)/ (s/ sqrt(n)) = (315 - 300)/ (16/ sqrt(7) = 15/ 6.0474 = 2.48

p - value = Pr(t > 3.48; dF = 7-1 = 6) = 0..0239 < 0.05

so we can reject the null and can conclude that leg strength exceeds 300 W.

(b) If true strenth is 304 W.

then we have to calculate the power of the test that we would be able to reject the null hypothesis when it is wrong.

so we have to calculate the sample mean above which we will be able to reject the null hypothesis.

x = 300 + t6,0.05 se0 = 300 + 1.9432 * 6.047 = 311.75

now true mean = 304

so we will be able to reject the null if x > 311,75

Pr (x > 311.75) = 1 - Pr(x< 311.75; 304 ; 6.047)

Z = (311.75 - 304)/ 6.047 = 1.282

so Pr (x > 311.75) = 1 - Pr(x< 311.75; 304 ; 6.047) = 1 - Pr(Z < 1.282) = 1 - 0.9 = 0.1

so Power of the test = 1 - 0.9 = 0.1

(c) Here now we require power to be 0.90

so, let say sample size would be n

so we will reject the null if x > 300 + t0.05,dF (16/ n)

so as power of the test is 0.90

Pr(x > 300 + t0.05,dF (16/ n) ; 304; (16/ n) ) = 0.90

Pr(x > 300 + t0.05,dF (16/ n) ; 304; (16/ n) ) = 1 - Pr(x < 300 + t0.05,dF (16/ n) ; 304; (16/ n) )

Pr(x < 300 + t0.05,dF (16/ n) ; 304; (16/ n) ) = 0.1

so Z - value here is Z = -1.282

-1.282 =  (300 + t0.05,dF (16/ n) - 304) / (16/ n)

-20.512/ n = t0.05,dF (16/ n) - 4

if we take the Z - value as an approximation of t value than Z = 1.645

-20.512/ n = 26.32/n - 4

4 = 46.832/ n

n = 137.0777 or 138

so for 138 the t value for alpha = 0.05 is 1.656 so that would not make a great difference. so 138 is correct answer.

(d) Here the part (a) can be solved by the confidence interval. A confidence inteval of 90% can be build around the sample mean and than if the value encompasses the population mean than we will accept the null hypothesis , otherwise reject it.

90 % confidence intervla = x +- t6,0.05 se0 = 315 + 1.9432 * 6.47 = (302.43, 327.57)

the confidence interval doesn't contain the value of 300 W so we can say that the population mean is greater than 300W.

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