In: Statistics and Probability
A sample of students 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 s2 = 64.
If the sample has n = 16 scores, then conduct a hypothesis test to evaluate the significance of the treatment effect.
Use a two-tailed test with α = .05.
What is the est. standard error or est. s.e. value?
What is the tobt value?
What is the correct decison for this study?
Calculate Cohen's D
Calculate Cohen’s d to measure the size of the treatment effect.
(1) Given values:
The provided sample mean is X̅ = 55.0 and the known population standard deviation is σ = 8.0
and the sample size is n = 16.0. and with known population mean is μ0 = 50.0
(2) Our test hypothesis is :
The following null and alternative hypothesis need to be tested,
hypothesis =
H0: μ = 50.0 vs
H1: μ ≠ 50.0
This hypothesis corresponds to a two-tailed, for which a t-test for one population mean.
(3) Test statistics :
The calculation of the t-test proceeds as follows,
X̅ - μ0 55.0 - 50.0
t = -------- = --------------- = 2.5
σ / √n 8.0 / 4.0
(4) Rejection Criteria :
Based on the information provided, the significance level is α = 0.05, and the degrees of freedom are
df =15.0 and the critical value for a two-tailed test is t tabulated = 1.7531
And the rejection region for this two-tailed test is R =[t: |t| > 1.7531]
(5) Decision about the null hypothesis :
Since it is observed that t calculated means |t| = 2.5 > t tabulated = 1.7531
it is then concluded that the null hypothesis is Rejected.
Using the P-value approach:
The p-value is 0.0245, and since p = 0.0245 < α = 0.05
it is then concluded that the null hypothesis is Rejected.
(6) Conclusion :
It is concluded that the null hypothesis Ho is Rejected. Therefore, there is enough evidence to claim
that population mean μ is different than 50.0, at the 0.05 significance level.
(1) Given values:
The provided sample mean is X̅ = 55.0 and the known population standard deviation is σ = 8.0
and the sample size is n = 16.0. and with known population mean is μ0 = 50.0
(2) Our test hypothesis is :
The following null and alternative hypothesis need to be tested,
hypothesis =
H0: μ = 50.0 vs
H1: μ ≠ 50.0
This hypothesis corresponds to a two-tailed, for which a t-test for one population mean.
(3) Test statistics :
The calculation of the t-test proceeds as follows,
X̅ - μ0 55.0 - 50.0
t = -------- = --------------- = 2.5
σ / √n 8.0 / 4.0
(4) Rejection Criteria :
Based on the information provided, the significance level is α = 0.05, and the degrees of freedom are
df =15.0 and the critical value for a two-tailed test is t tabulated = 1.7531
And the rejection region for this two-tailed test is R =[t: |t| > 1.7531]
(5) Decision about the null hypothesis :
Since it is observed that t calculated means |t| = 2.5 > t tabulated = 1.7531
it is then concluded that the null hypothesis is Rejected.
Using the P-value approach:
The p-value is 0.0245, and since p = 0.0245 < α = 0.05
it is then concluded that the null hypothesis is Rejected.
(6) Conclusion :
It is concluded that the null hypothesis Ho is Rejected. Therefore, there is enough evidence to claim
that population mean μ is different than 50.0, at the 0.05 significance level.