A sample is selected from a population with µ = 50. After a treatment is adminis

Question

A sample is selected from a population with µ = 50. After a treatment is administered to the individuals in the sample, the mean is found to be M = 55 and the variance is s 2 = 64. a. For a sample of n = 4 scores, conduct a hypothesis test to evaluate the significance of the treatment effect and calculate Cohen’s d to measure the size of the treatment effect. Use a two-tailed test with = .05. b. For a sample of n = 16 scores, conduct a hypothesis test to evaluate the significance of the treatment effect and calculate Cohen’s d to measure the size of the treatment effect. Use a two-tailed test with = .05. c. Describe how increasing the size of the sample affects the likelihood of rejecting the null hypothesis and the measure of effect size.
A sample is selected from a population with µ = 50. After a treatment is administered to the individuals in the sample, the mean is found to be M = 55 and the variance is s 2 = 64. a. For a sample of n = 4 scores, conduct a hypothesis test to evaluate the significance of the treatment effect and calculate Cohen’s d to measure the size of the treatment effect. Use a two-tailed test with = .05. b. For a sample of n = 16 scores, conduct a hypothesis test to evaluate the significance of the treatment effect and calculate Cohen’s d to measure the size of the treatment effect. Use a two-tailed test with = .05. c. Describe how increasing the size of the sample affects the likelihood of rejecting the null hypothesis and the measure of effect size.
A sample is selected from a population with µ = 50. After a treatment is administered to the individuals in the sample, the mean is found to be M = 55 and the variance is s 2 = 64. a. For a sample of n = 4 scores, conduct a hypothesis test to evaluate the significance of the treatment effect and calculate Cohen’s d to measure the size of the treatment effect. Use a two-tailed test with = .05. b. For a sample of n = 16 scores, conduct a hypothesis test to evaluate the significance of the treatment effect and calculate Cohen’s d to measure the size of the treatment effect. Use a two-tailed test with = .05. c. Describe how increasing the size of the sample affects the likelihood of rejecting the null hypothesis and the measure of effect size.

Explanation / Answer

Set Up Hypothesis
Null, H0: U=50
Alternate, H1: U!=50
Test Statistic
Population Mean(U)=50
Sample X(Mean)=55
Standard Deviation(S.D)=8
Number (n)=4
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =55-50/(8/Sqrt(4))
to =1.25
| to | =1.25
Critical Value
The Value of |t | with n-1 = 3 d.f is 3.182
We got |to| =1.25 & | t | =3.182
Make Decision
Hence Value of |to | < | t | and Here we Do not Reject Ho
P-Value :Two Tailed ( double the one tail ) -Ha : ( P != 1.25 ) = 0.2999
Hence Value of P0.05 < 0.2999,Here We Do not Reject Ho
b.
Number (n)=16
we use Test Statistic (t) = x-U/(s.d/Sqrt(n))
to =55-50/(8/Sqrt(16))
to =2.5
| to | =2.5
Critical Value
The Value of |t | with n-1 = 15 d.f is 2.131
We got |to| =2.5 & | 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 != 2.5 ) = 0.0245
Hence Value of P0.05 > 0.0245,Here we Reject Ho

c)

increasing the size of the sample increases the likelihood of rejecting the

null hypothesis and the measure of effect size.

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