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Testing the Difference Between Two Sample Means

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Please describe the process and paste the output. List the description to enable me generate mine output. Please remember the assignment is using SPSS not excel. You may refer also to Posting 622444 Library Download on Fundamental Statistics One Sample Mean in which the author answered same questions using excel.

I have added additional information for question 9-10 . see attached document.
And data for question 9;10

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Solution Summary

Testing the Difference Between Two Sample Means are given. The fundamental statistics for one sample mean is determined.

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Please see attached and let me know if you have any question.
I did the most important ones, as you suggested.

Testing the Difference Between Two Sample Means
From
Tokunaga, H, T. (2015). Fundamental Statistics for the Social and Behavioral Sciences. SAGE Publications, Inc.


• Locate the 7.13 exercises on page 256 of the Tokunaga (2016) text.
• Complete Exercise 2, 6, 10 and 32 using SPSS.
• Locate the 8.12 exercises on page 301 of the Tokunaga (2016) text.
• Complete Exercise 8 and 12 using SPSS.
• Locate the 10.10 exercises on page 411 of the Tokunaga (2016) text.
• Complete Exercise 4 and 16 using SPSS.
The questions are bellow:
2. For each of the following situations, calculate the population standard error of the mean
a. σ = 18; N = 36
Standard error of the mean is:
In SPSS: Go to Transform/ Compute variable

b. σ = 9.42; N = 49
c. σ = 1.87; N = 60
d. d. σ = .91; N = 22
e. e. σ = 21.43; N = 106

6. For each of the following situations, calculate the z-statistic (z), make a decision about the null hypothesis (reject, do not reject), and indicate the level of significance (p > .05, p < .05, p < .01).

In SPSS, same as before using Transformation variable
Transform/ Compute variable
Since , we reject the Null Hypothesis and p < .05

10. In the GRE test example (Exercise 9), what if it was believed that the only possible alternative to the null hypothesis is one in which the students' GRE scores increase (i.e., they cannot decrease).
a. State the null and alternative hypotheses (H0 and H1).
H0: μ1=1000
H1: μ1>1000

b. Make a decision about the null hypothesis.
We will compute variable as before

In SPSS: Transform/ Compute variable:
(MeanVar-1000)/ Standard_dev/sqrt(Size)

(1) Set alpha (α), identify the critical values, and state a decision rule. Why is the decision rule different in this situation?
Α, critical value is Zα=1.645
The decision rule is different when the alternative Hypothesis is with greater or equal sign instead of different than.

(2) Calculate a statistic: z-test for one mean. Is it a different value in this situation? Why or why not?
From the above calculation in SPSS we got Z=1.88
It is different because the alternative is one way instead of two-ways

(3) Make a decision whether to reject the null hypothesis.
Since Z> Zα we reject the Null ...

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