A sample is selected from a population with a mean of 50. After a treatment is a
ID: 3023534 • Letter: A
Question
A sample is selected from a population with a mean of 50. After a treatment is administered to the individuals in the sample, the mean is found to be M=66, the variance is s^2=64 and n=16.
a. State the null hypothesis
b. Compute the becessary hypothesis test using a two-tailed test with alpha=.05. Show your calculations
c. Make a decision about youir null hypothesis. (Include the critical value you used to bade your decision.)
d. Write an APA format statement summarizing your findings (include a measure of effect size if necessary.)
Explanation / Answer
Set Up Hypothesis
Null, H0: U=66
Alternate, H1: U!=66
Test Statistic
Population Mean(U)=66
Given That X(Mean)=50
Standard Deviation(S.D)=8
Number (n)=16
we use Test Statistic (Z) = x-U/(s.d/Sqrt(n))
Zo=50-66/(8/Sqrt(16)
Zo =-8
| Zo | =8
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =8 & | Z | =1.96
Make Decision
Hence Value of | Zo | > | Z | and Here we Reject Ho
P-Value : Two Tailed ( double the one tail ) - Ha : ( P != -8 ) = 0
Hence Value of P0.05 > 0, Here we Reject Ho
[ANSWERS]
a. H0: U=66 , H1: U!=66
b. Zo =-8
c. | Z | =1.96, Reject Ho
d. We conclude that, treatment administered is n't founde be 66