Question

Our dataset has the following variables

Commitment- how committed the employee is to the organization [measured on a 5 point Likert scale:-1( strongly disagree= 5(strongly agree); higher number means more committed]

Satisfaction- How satisfied is the employee with his/her job?[ measured on a 5 point Likert scale:- 1 (strongly disagree) to 5 (strongly agree); higher number means more satisfied]

Performance- What was this employee’s rating on his/her last performance appraisal?[ measured on a Likert Scale:- 1 (poor) to 5 (excellent); higher number means higher performance]

Gender- What is the employee’s gender?[Male = 1 Female = 2]

Degree- What is the employee’s highest degree?[ Less than High School = 1 High School = 2 Some College = 3 Bachelor’s Degree = 4 Graduate Degree = 5]

Age- How many years old is the employee? (Years)

Absences- How many absences did this employee have this year? (Number of absent days)

Job- Is the employee management or a line worker? (Management = 1Line Worker = 2)

Self esteem- level of self-esteem [ measured on a 5 point Likert scale:-1 (low) to 5 (high)]

Use the dataset to test whether your employees’ commitment average is significantly different from the national average, which is 3.85 in companies like yours.

One-Sample Test
	Test Value = 3.85
	t	df	Sig. (2-tailed)	Mean Difference	95% Confidence Interval of the Difference
					Lower	Upper
COMMITMENT_B	2.006	84	.048	.185	.00	.37

i. Is there a statistically significant difference between your company and the national average? [report the t-value and p-value at the end of the sentence stating whether there is a difference: “…end of sentence (t = x.xx, p < 0.xx).”]

ii. How much error is there in your conclusions? [Be sure to write a complete statement using the variable names]

iii. Comment on the practical implication of your findings

Answer 1

(i)

(a)

t score = 2.006

=0.05

df = 84

Two Tail Test

From Table, criticaal values of t = 1.9886

Since calculated value of t = 2.006 is grater than critical value of t = 1.9886, Reject H0.

Conclusion:

There is a statistical difference between your company and national average.

(b) t score = 2.006

df = 84

Two Tail test

By Technology, p-value = 0.0481

Since p-value = 0.0481 less than =0.05, Reject H0.

Conclusion:

There is a difference between your company and the national average because t = 2.008 , p=0.0481<0.05.

(ii) Margin of Error = (0.37 - 0.00)/2 = 0.185. The conclusion that there is no difference between your company and national average is wrong at most by a margin of 18.50 %.

(iii) Practical Implication: Since it is concluded that there is statistically no significant difference between your company and the national average, we can confidently make use of any inferences about the national average for the use of the company.