Answer:
a. There is_<u><em>sufficient</em></u> 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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