Question

For this assignment, you will use the six-step hypothesis testing process (noted below) to run and interpret a correlation analysis using SPSS. The following vignette will inform you of the context for this assignment. A data file is provided in the week’s resources for use in this assignment.

A manager is interested in studying the associations between a number of variables. These variables are age, years of experience, level of education, engagement, job satisfaction, and performance level. She thinks that employees with more years of experience are more engaged and satisfied. She thinks that younger employees will perform at a higher level, on average.

1. State the null and alternative hypotheses.
2. Select the significance level.
3. Select the test statistics and calculate its value.
4. Identify critical values for the test statistics and state the decision rule concerning when to reject or fail to reject the null hypothesis.
5. Compare the calculated and critical values to reach a conclusion for the null hypothesis.
6. Explain the related business decision.

After conducting the above analysis please structure your paper as follows:

• Introduction to the assignment
• Hypotheses
• Results of the analysis (Hint. You may use an APA-style table to display these data. SPSS output images are not in APA style).
• Did the analysis support each of the hypotheses?
• Explain what decisions the manager might make using these findings.

Length: 4 to 6 pages not including title and reference page

References: Include a minimum of 3 scholarly resources.

Your paper should demonstrate thoughtful consideration of the ideas and concepts presented in the course and provide new thoughts and insights relating directly to this topic. Your response should reflect scholarly writing and current APA standards. Be sure to adhere to Northcentral University's Academic Integrity Policy.

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Answer 1

1. The hypothesis being tested is:

Null Hypothesis:

There is no significant relationship between employees with more years of experience are more engaged and satisfied.

Alternative Hypothesis:

There is significant relationship between employees with more years of experience are more engaged and satisfied.

To test the above null and alternative hypothesis we may use Chi-square test.

SPSS Output

Chi-Square Test

Frequencies

                             Experiance

	Observed N	Expected N	Residual
1	1	2.0	-1.0
2	1	2.0	-1.0
3	1	2.0	-1.0
4	2	2.0	.0
5	1	2.0	-1.0
6	2	2.0	.0
7	1	2.0	-1.0
8	4	2.0	2.0
9	3	2.0	1.0
10	2	2.0	.0
11	2	2.0	.0
12	3	2.0	1.0
13	1	2.0	-1.0
14	2	2.0	.0
15	4	2.0	2.0
Total	30

NPar Tests

                                      Engagement

	Observed N	Expected N	Residual
very low engagement	3	6.0	-3.0
low engagement	8	6.0	2.0
neutral	3	6.0	-3.0
engaged	9	6.0	3.0
highly engaged	7	6.0	1.0
Total	30

                                JobSatisfaction

	Observed N	Expected N	Residual
very dissatisfied	2	6.0	-4.0
dissatisfied	7	6.0	1.0
neutral	6	6.0	.0
satisfied	9	6.0	3.0
very satisfied	6	6.0	.0
Total	30

                                  Test Statistics

	Experiance	Engagement	JobSatisfaction
Chi-Square(a,b)	8.000	5.333	4.333
df	14	4	4
Asymp. Sig.	.889	.255	.363

a 15 cells (100.0%) have expected frequencies less than 5. The minimum expected cell frequency is 2.0.

b 0 cells (.0%) have expected frequencies less than 5. The minimum expected cell frequency is 6.0.

Conclusion:

From the above test statistics the significant value (0.889) is greater than the probability value 0.05 so we don’t have significant evidence to reject null hypothesis.

2.

Null Hypothesis:

There is no significant relationship between employees age and performance.

Alternative Hypothesis:

There is significant relationship between employees age and performance.

To test the above null and alternative hypothesis we may use Chi-square test.

Chi-Square Test

Frequencies

                          EmployeeAge

	Observed N	Expected N	Residual
25	2	1.6	.4
26	1	1.6	-.6
27	1	1.6	-.6
28	1	1.6	-.6
29	5	1.6	3.4
31	2	1.6	.4
32	1	1.6	-.6
34	1	1.6	-.6
35	2	1.6	.4
36	2	1.6	.4
38	1	1.6	-.6
39	1	1.6	-.6
40	2	1.6	.4
41	1	1.6	-.6
42	1	1.6	-.6
43	1	1.6	-.6
47	3	1.6	1.4
49	1	1.6	-.6
50	1	1.6	-.6
Total	30

                                 JobPerformance

	Observed N	Expected N	Residual
Low performer	3	6.0	-3.0
Avergae performer	6	6.0	.0
High performer	4	6.0	-2.0
4	8	6.0	2.0
5	9	6.0	3.0
Total	30

                                             

Test Statistics

	EmployeeAge	JobPerformance
Chi-Square(a,b)	11.800	4.333
df	18	4
Asymp. Sig.	.857	.363

a 19 cells (100.0%) have expected frequencies less than 5. The minimum expected cell frequency is 1.6.

b 0 cells (.0%) have expected frequencies less than 5. The minimum expected cell frequency is 6.0.

Conclusion:

From the above test statistics the significant value (0.857) is greater than the probability value 0.05 so we don’t have significant evidence to reject null hypothesis.