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Let the sample size of leg strengths to be 7 and the sample mean and sample standard deviation be 630 watts and 32 watts, respectively.

(a) Is there evidence that leg strength exceeds 600 watts at significance level 0.05? Find the P-value. There is_________ evidence that the leg strength exceeds 600 watts at ? = 0.05.

A. 0.001 < P-value < 0.005

B. 0.10 < P-value < 0.25

C. 0.010 < P-value < 0.025

D. 0.05 < P-value < 0.10

(b) Compute the power of the test if the true strength is 610 watts.

(c) What sample size would be required to detect a true mean of 610 watts if the power of the test should be at least 0.9? n=

Respuesta :

akiran007

Answer:

a. There is_sufficient evidence that the leg

C. 0.010 < P-value < 0.025

b. Power of test = 1- β=0.2066

c. So the sample size is 88

Step-by-step explanation:

We formulate the null and alternative hypotheses as

H0 : u1= u2 against Ha : u1 > u2 This is a right tailed test

Here n= 7 and significance level ∝= 0.005

Critical value for a right tailed test with 6 df is 1.9432

Sample Standard deviation = s= 32

Sample size= n= 7

Sample Mean =x`= 630

Degrees of freedom = df = n-1= 7-1= 6

The test statistic used here is

Z = x- x`/ s/√n

Z= 630-600 / 32 / √7

Z= 2.4797= 2.48

P- value = 0.0023890 > ∝ reject the null hypothesis.

so it lies between 0.010 < P-value < 0.025

b) Power of test if true strength is 610 watts.

For  a right tailed test value of z is = ± 1.645

P (type II error) β= P (Z< Z∝-x- x`/ s/√n)

Z = x- x`/ s/√n

Z= 610-630 / 32 / √7

Z=0.826

P (type II error) β= P (Z< 1.645-0.826)

= P (Z> 0.818)

= 0.7933

Power of test = 1- β=0.2066

(c)

true mean = 610

hypothesis mean = 600

standard deviation= 32

power = β=0.9

Z∝= 1.645

Zβ= 1.282

Sample size needed

n=( (Z∝ +Zβ )*s/ SE)²

n=  ((1.645+1.282) 32/ 10)²

Putting the values  and solving we get 87.69

So the sample size is 88

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