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# Summary of 5 Hypotheses

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I have posted this two times and paid two times (\$50) and still have not received a summary.

I need help summarizing the hypotheses of the attached Unit3 file. I am also attaching the Excel database the question references.

Here is the question:
In Unit 3 you tested 5 different hypotheses using the database from Unit 2. Provide a summary of EACH of the 5 hypotheses tested. This should be a summary, not more statistical data....it should be a couple of pages.

I am looking for help with the summary of each hypotheses.

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1. Test a hypothesis to see whether the average overall job satisfaction (in the population of all workers in the USA) is equal to 4.5 with a = .05.
a. State the null hypothesis, the alternative hypothesis, and the significance level.
Let M = average overall job satisfaction = SUM(OVERALL) / N-1
where N = sample size = 288
Ho: M = 4.5
Ha: 4.5

b. Using the data in our database, calculate the test statistic.
The sample average overall job satisfaction is M = SUM(OVERALL) / N-1 = 4.33
(refer to EXCEL for calculation)
The standard deviation of overall job satisfaction is
SD = SUM(OVERALL - M)2 / N = 1.364
Then the standard error is
SE = SD / SQRT(N) = 1.364 / SQRT(288) = 0.08
Then compute t = (4.5 - M) / SD = (4.5 - 4.33) / 0.08 = 2.11

c. What is the critical level for the significance level?
This is a two-tailed test, so the critical value for the significance level = 0.05 is 97.5% t value with degree of freedom = n-1 = 287
From a t-table, the critical value t* = 1.97

d. What is your conclusion? Do we accept or reject the null hypothesis?
Since the computed t = 2.11, which is greater than the critical value t* = 1.97
We have to reject the null hypothesis at the 0.05 significance level.

You may use Excel for the calculations, but you need to answer all four parts of this question.

2. Propose a hypothesis test for the mean intrinsic job satisfaction, similar to the test from problem 1, and answer parts a, b, c, and d of problem 1 for this hypothesis test.
a. State the null hypothesis, the alternative hypothesis, and the significance level.
Let Mi = average overall job satisfaction = SUM(INTRINSIC) / N-1
Ho: Mi = 4.5
Ha: Mi = 4.5

b. Using the data in our database, calculate the test statistic.
The sample average overall job satisfaction is Mi =SUM(INTRINSIC)/N-1 = 5.22
(refer to EXCEL for calculation)
The standard deviation of overall job satisfaction is
SDi = SUM(INTRINSIC - Mi)2 / N = 1.00
Then the standard error is
SEi = SDi / SQRT(N) = 1.00 / SQRT(288) = 0.059
Then compute t = (Mi - 4.5) / SDi = (5.22 - 4.5) / 0.059 =12.21

c. What is the critical level for the significance level?
This is also a two-tailed test, so the critical value for the significance level = 0.05 is 97.5% t value with degree of freedom = n-1 = 287
From a t-table, the critical value t* = 1.97

d. What is your conclusion? Do we accept or reject the null hypothesis?
Since the computed t = 12.21, which is greater than the critical value t* = 1.97
We have to reject the null hypothesis at the 0.05 significance level.

3. We believe that half of the population would have an extrinsic job satisfaction of 5.0 or greater. Answer parts a, b, c, and d of problem 1 for this hypothesis test of a proportion.
a. State the null hypothesis, the alternative hypothesis, and the significance level.
Let the proportion of extrinsic job satisfaction of 5.0 or greater be
P = (Number of extrinsic 5.0) / N
Ho: P = 0.5
Ha: P=0.5

b. Using the data in our database, calculate the test statistic.
The sample proportion is P = (Number of extrinsic &#8805; 5.0) / N = 185/288 = 0.642
(refer to EXCEL for calculation)
The standard error of the proportion is
SE = SQRT[(1-P)*P /N] = SQRT[(1-0.642)*0.642 /288] = 0.028
Then compute t = (P - 0.5) / SE = (0.642 - 0.5) / 0.028 = 5.04

c. What is the critical level for the significance level?
This is also a two-tailed test, so the critical value for the significance level = 0.05 is 97.5% t value with degree of freedom = n-1 = 287
From a t-table, the critical value t* = 1.97

d. What is your conclusion? Do we accept or reject the null hypothesis?
Since the computed t = 5.04, which is greater than the critical value t* = 1.97
We have to reject the null hypothesis at the 0.05 significance level.

4. We believe that the variance of the overall job satisfaction is equal to 1.0 Answer parts a, b, c, and d of problem 1 for this hypothesis test of a variance.
a. State the null hypothesis, the alternative hypothesis, and the significance level.
Let the variance of the overall job satisfaction be:
V = SUM(OVERALL0-M)2 / N-1 = SD2
Ho: V = 1.0
Ha: V=1.0

b. Using the data in our database, calculate the test statistic.
The sample proportion is V = = SUM(OVERALL0-M)2 / N-1 = SD2 = 1.861
(refer to EXCEL for calculation)
Then compute F = V / 1.0 = 1.861 / 1.0 = 1.861

c. What is the critical level for the significance level?
This is also an F test, the critical value for the significance level = 0.05 with degree of freedom 1 = n-1 = 287 and degree of freedom 2 = n-1 = 287 is
F* = 1.215

d. What is your conclusion? Do we accept or reject the null hypothesis?
Since the computed F = 1.861, which is greater than the critical value t* = 1.97
We have to reject the null hypothesis at the 0.05 significance level.

5. We will call a "deskbody" a person whose intrinsic job satisfaction level is higher than their extrinsic job satisfaction level (i.e. happy with their job more than their office). We will call a "socialbody" a person whose extrinsic job satisfaction level is higher than their intrinsic job satisfaction level (i.e. happy with the office more than their job). We believe that there are equal deskbodies and socialbodies in the work force.
a. State an appropriate null hypothesis and its alternative hypothesis.
Let the proportion of deskbody be Pd = Number of deskbody / N
the the proportion of socialbody is Ps= Number of socialbody / N
Ho: Pd = Ps or Pd - Ps=0
Ha: Pd = Ps or Pd - Ps = 0

b. In our database, what percent of the employees are deskbodies? Are socialbodies?
I created a new variable D= intrinsic- extrinsic
then for the samples D>0, he is a deskbody, and a socialbody when D<0
and we use the EXCEL command =COUNTIF(J2:J289,">0")
then the percent of the employees are deskbodies becomes Pd = 0.493
And the percent of the employees are socialbodies is Ps = 0.472

c. What did you do with the employees who had equal intrinsics and extrinsics?
By definition, we cannot count these employees who had equal intrinsics and extrinsics into either category. So these samples have to be left alone. And this is the reason that Pd + Ps&#8800; 1

d. What do you suggest as a good test statistic (i.e. a way to calculate part b of problem 1)?

We can use a t-test for the difference in proportions.
D=Pd - Ps = 0.493 - 0.472 = 0.021
Note: this time, the standard error is:
SE = SQRT( [ (1-Pd)*Pd + Ps(1-Ps)] /N)
= SQRT(((1-0.493)*0.493+0.472*(1-0.472))/288) = 0.042
And compute t = D / SE = 0.021 / 0.042 = 0.50
Since t < critical t value = 1.96, we fail to reject Ho and say that the two kind people are equal.

6. Determine the required sample size if you need to estimate the number of workers in the United States who are highly satisfied with their job and you want the estimate to be within 2 percentage points with a 96% confidence interval.

Since the sample size is generally large, we can use z value for calculation.
The two-tailed 96% z value is z(P=0.98) = 2.054
Let N = sample size
The confidence interval is + - 0.08
That is: z*SE = 0.02 (1)
where SE is the standard error = SQRT[P*(1-P)/N]
Substitute z and SE into (1):
2.054 * SQRT(P*(1-P)/N) = 0.08
we can compute P by P=Number of Overall>5.0 / N = 101/288 = 35%
therefore, we are able to substitute P into the equation
2.054 * SQRT(.35*(1-.35)/N) = 0.08
solve for N=2895, which is the required sample size.
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Thank you.

https://brainmass.com/statistics/hypothesis-testing/summary-of-5-hypotheses-54643

#### Solution Preview

Please see response attached. I hope this helps and take care.

FROM ATTACHED RESPONSE:

1.
The variable for job satisfaction is qualitative, since its level of overall job satisfaction is not numerical, but a descriptive label. The general purpose of the study is to determine the overall level of job satisfaction in the population of all workers in the USA. Specifically, we set up two hypotheses to test whether the average overall job satisfaction in the population of all workers in the USA is equal to 4.5 (null hypothesis) or not equal to 4.5 (alternative hypothesis) with the confidence level set at 5%. This means then we can be 95/100 confident (accept or reject the NULL hypothesis) that are results will be correct. In other words, there is a 5% margin of error, meaning that 5 times out of 100 our conclusions may not be correct.

We used the t-test statistic to test our hypothesis (which takes into consideration the standards deviation and the sample size), and rejected the null hypothesis at the 0.05 significance level (t = 2.11, d.f. = 287, &#945; = .05). In other words, this means that we can be 95% confident that the overall job satisfaction of the all workers of America was not equal to 4.5. However, we do not in what direction.

2.
The purpose of the study is to determine the overall intrinsic level of job satisfaction in the population of all workers in the USA. Specifically, we set up two hypotheses to test for the mean intrinsic level of job satisfaction and whether the ...

#### Solution Summary

The solution provides a summary of the five hypotheses from the file attached. Specifically, it explains the purpose of the study, explains the null and alternative hypotheses and what variables are being tested, and interprets the findings in summary form.

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