Q7 a human resource manager was of the view that employees...

Q7. A human resource manager was of the view that employees are more likely to stay longer in an organization where they enjoy better job satisfaction than in organization where job satisfaction is poorer. Data was collected to test this assertion. The results are presented below. Independent t-Test result, mean and standard deviation of perceived turnover scores classified by high and low job satisfaction of employees. Job Satisfaction N Mean St Deviation df t p Poorer job satisfaction 23 79.6 4.1 48 4.451 .009 Better job satisfaction 27 87.3 7.3 Based on the claim in the case, state the null and the alternative hypotheses Analyze the data and interpret the results The dependent variable was measured on an interval scale, state and explain the characteristics of an interval scale

Transcript

00:01 We have provided an independent details results, mean and standard deviation of a perceived turn -vers course classified by high and low job satisfaction of employees.

00:11 And we have here on the basis of the claim, the case, we have been first asked to state the null hypothesis and the alternative hypothesis.

00:22 So here, let us denote that the better job satisfaction is the first group is better job satisfaction and second is the poor job satisfaction now here the claim which is made by the hr manager is that employees are more likely to stay longer in an organization where they enjoy better job satisfaction right so here the null hypothesis will be equal will be that new 1 is equal to new 2 and the alternative hypothesis will be that new 1 is greater than new 2 here it means that the employees are more likely to to stay longer in an organization where they enjoy better job satisfaction as compared to the organization where the job satisfaction is poor right now in the second part we have been asked to analyze the given data and interpret the results all right so what information we have here we have the mean one is 87 .3 mean 2 is 79 .6 standard deviation 1 is 7 .3 and standard deviation 1 is 7 .3 and deviation 2 is 4 .1 right so here the group 1 n1 is 27 and 2 is 23 right now we have to conduct the analysis so here are the level let us assume that the two samples are independent and our for normal population have the same variance let's take the level of significance of 0 .05 so t will be the test statistics will be 1 minus 2 divided by standard deviation of 1 divided by n 1 plus 1 divided by n 2 and this follows the t distribution with n1 plus n 2 minus 2 with the degrees of freedom right so here the variance will be equal to n1 minus 1 to variance 1 square plus n 2 minus 1 variance 2 squared divided by n 1 plus n 2 minus 2 right so it will be 27 minus 1 to 7 .3 square plus 23 minus 1 into 4 .1 squared divided by 27 plus 23 minus 2 that would be equal to 1755 .3, 6 by 48, and this is equal to 36 .57...

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