Statistical Analysis

Statistical Anlaysis

Unit 4 IP

Test 1

Perform the following two tailed hypothesis test using a .05 level of significance

Intrinsic by gender

State the null and alternate hypothesis for the test

Use Microsoft excel to process your data and run the appropriate test. Copy and paste the result of the output to your report in Microsoft Word

Identify the significance level, test static, and the critical value

State whether you are rejecting or failing to reject the null hypothesis statement.

Explain how the results could be used y the manager of the company.

Solution

Gender selected = 1 (Male)

Null hypothesis, Ho: µ= 5.0 and Alternate hypothesis, Ha: µ= 5.0

Level of significance = 0.05

Test used is z-test for population mean

Critical scores are z=1.96 and z=-1.96

Rejection rule is: reject Ho if z calculated is >1.96 or <-1.96

Test static is z = (x-bar- µ)/(s/vn) = (5.06-5.0)/(1.0309/v52) = 0.419

Decision: since 0.419<1.96, we fail to reject Ho

Conclusion: there is not sufficient evidence that mean intrinsic score for males is different from 5.0

Z Test of Hypothesis for the Mean

Data

Null Hypothesis m=

5

Level of Significance

0.05

Population Standard Deviation

1.03

Sample Size

52

Sample Mean

5.06

Intermediate Calculations

Standard Error of the Mean

0.1429601

Z Test Statistic

0.419

Two-Tail Test

Lower Critical Value

-1.959963985

Upper Critical Value

1.959963985

p-Value

0.674709

Do not reject the null hypothesis

(Data collected for the analysis can be seen from the appendix)

The fact that mean intrinsic satisfaction rating for men is not different from 5 is encouraging. A right tailed test would have been more appropriate but two tailed test also serves the purpose. The manager should try to retain this level of satisfaction among the male employees.

Test 2

Perform the following two-tailed hypothesis test, using a .05 significance level:

Extrinsic variable by Position Type

State the null and an alternate statement for the test

Use Microsoft Excel (Data Analysis Tools) to process your data and run the appropriate test.

Copy and paste the results of the output to your report in Microsoft Word.

Identify the significance level, the test statistic, and the critical value.

State whether you are rejecting or failing to reject the null hypothesis statement.

Solution

Position type selected = 1 (Hourly employees)

Null hypothesis, Ho: µ=5.0 and alternate hypothesis, Ha: µ= 5.0

Level of significance = 0.05

Test used is z-test for population mean

Critical scores are z=1.96 and z=-1.96

Rejection rule is: reject Ho if z calculated is >1.96 or <-1.96

Test static is z = (x-bar- µ)/(s/vn) = (5.15-5.0)/(.91/v56) = 1.239

Decision rule: sine z calculated value of 1.239 is less than 1.96 so we fail to reject null hypothesis.

Conclusion: there is not sufficient evidence that mean extrinsic score for hourly employees in no different from 5

Z Test of Hypothesis for the Mean

Data

Null Hypothesis m=

5

Level of Significance

0.05

Population Standard Deviation

0.91

Sample Size

56

Sample Mean

5..15



Intermediate Calculations

Standard Error of the Mean

0.1210603

Z Test Statistic

1.239



Two-Tail Test

 

Lower Critical Value

-1.959963985

Upper Critical Value

1.959963985

p-Value

0.21534549

Do not reject the null hypothesis

 

(Data collected for the analysis can be seen from the appendix)

The fact that mean extrinsic satisfaction rating for men is not different from 5 and most approximately greater than 5 is encouraging. The manager should make sure that he or she maintains this level of satisfaction.

References Unit 4IP (Test1 and Test2)

Boyd, ...