A sample of n 16 individuals is selected from a population

A sample of n = 16 individuals is selected from a population with µ = 30. After a treatment is administered to the individuals, the sample mean is found to be M = 33.

A. If the sample variance is s2 = 16, then conduct a hypothesis test to evaluate the significance of the treatment effect and calculate r2 to measure the size of the treatment effect. Use a two-tailed test with = .05.

B. If the sample variance is s2 =64, then conduct a hypothesis test to evaluate the significance of the treatment effect and calculate r2 to measure the size of the treatment effect. Use a two-tailed test with = .05.

C. Describe how increasing variance affects the likelihood of rejecting the null hypothesis and the measure of effect size

Solution

a)
Set Up Hypothesis
Null Hypothesis H0: U=30
Alternate Hypothesis H1: U!=30
Test Statistic
Population Mean(U)=30
Sample X(Mean)=33
Standard Deviation(S.D)=4
Number (n)=16
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =33-30/(4/Sqrt(15))
to =3
| to | =3
Critical Value
The Value of |t | with n-1 = 15 d.f is 2.131
We got |to| =3 & | t | =2.131
Make Decision
Hence Value of | to | > | t | and Here we Reject Ho
P-Value :Two Tailed ( double the one tail ) -Ha : ( P != 3 ) = 0.009
Hence Value of P0.05 > 0.009,Here we Reject Ho

b)
When sample variance = 64
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =33-30/(8/Sqrt(15))
to =1.5
| to | =1.5
Critical Value
The Value of |t | with n-1 = 15 d.f is 2.131
We got |to| =1.5 & | t | =2.131
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value :Two Tailed ( double the one tail ) -Ha : ( P != 1.5 ) = 0.1544
Hence Value of P0.05 < 0.1544,Here We Do not Reject Ho