Can Training Reduce Workplace Offensiveness?

[MWR]

How do I go about analyzing this scenario? Consider the following: You’ve been hired by a company who is interested in finding a way to decrease incidents of offensive interpersonal behavior at work. The company has identified one department that seems to have a big problem with offensive interpersonal behavior. The company hopes to resolve this problem by putting all of the employees in this department through a training course. The company measures their offensiveness (on a scale of 1 indicating no offensiveness to 20 indicating highly offensive) both before (pre) and after (post) the training course. The company also provided the rank of the employees, with 1 = manager and 2 = employee.

Family Name Pre-Training Post-Training Rank
Taub 6 3 2
Hadley 5 5 2
Cameron 10 9 1
House 17 15 1
Wilson 2 3 1
Cuddy 12 10 1
Foreman 15 13 1
Volakis 6 4 2
Chase 9 4 2

1. Before they begin their analysis, the company wants to know whether the sample of employees in this study had different levels of offensiveness before the study compared to other individuals in the organization. For the purposes of this study, the overall organization should be considered the population of interest. The company knows that in this organization, employees typically engage in an average of 7 incidents of offensive interpersonal behavior at work with a standard deviation of 3 incidents. Does this sample of employees engage in a significantly different number of incidents of offensive interpersonal behavior before the study (Pre-training) than employees in the organization?

How to calculate the appropriate test:
The company is interested in testing your intervention on a department who has significantly more incidents of interpersonal offensiveness than in the overall organization. Should they proceed with their analysis using this department? Why or why not?
2. The company wants to know whether the training was able to reduce offensiveness for its employees. Based on the data provided above, determine whether offensiveness after the training (Post-Training) was lower than offensiveness before the training (Pre-Training).

How to calculate the appropriate test:

If appropriate, how to calculate a confidence interval:
Report a formal conclusion:
3. The company wants to know whether the training was more effective for managers than it was for employees. Based on the data provided above, determine whether there were lower incidents of offensiveness after training (Post-Training) for managers than there were for employees (Rank).

How to calculate the appropriate test:
State a formal test of the hypothesis:4. One of the researchers has expressed some concern that the results from #3 might have been influenced by individual differences in this sample. Is this a reasonable concern? Why or why not?

 

[Jameson]

Hello there. This is going to be a problem with lots of steps so you might not get help with every part until you show us where you are stuck.

Assuming that the 1-20 scale is actually counting "incidents of offensive interpersonal behavior at work" then you can do a t-test to determine if the averages of the sample is statistically different than the whole company.

 

[MWR]

Hello there. This is going to be a problem with lots of steps so you might not get help with every part until you show us where you are stuck.

Assuming that the 1-20 scale is actually counting "incidents of offensive interpersonal behavior at work" then you can do a t-test to determine if the averages of the sample is statistically different than the whole company.

So the appropriate test is the t-test, or z-test.

Here is what I have:

x numbers represent pre-training.
y numbers represent post-training.

Sx = 82
Mx = 9.11

Sx^2 = 24.11

Sy = 66
My = 7.33

Sy^2 = 20.75

After calculating the square root of Sx^2 and Sy^2, I obtain t = 0.16, which is statistically significant.

Am I on the right track?

 

[Jameson]

(1) is asking if the group they chose to look at pre-training differ from the company's numbers in general. The only issue I see is that we don't know how many people are in the company. When calculating a 2 sample t-test or z-test, we need the sample size of both groups. I think something is missing here or they want you to do another test...

Or perhaps they want you to do a one-sample t-test to see if the sample average is statistically different than 7. If so you could do it like this:

\(\displaystyle T_0=\frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}}=\frac{\bar{x}-7}{\frac{s}{\sqrt{n}}}\)

As for comparing before and after, you need to use the formula for a 2 sample t-test with unequal variances:

\(\displaystyle T_0=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\)

 

[MWR]

(1) is asking if the group they chose to look at pre-training differ from the company's numbers in general. The only issue I see is that we don't know how many people are in the company. When calculating a 2 sample t-test or z-test, we need the sample size of both groups. I think something is missing here or they want you to do another test...

Or perhaps they want you to do a one-sample t-test to see if the sample average is statistically different than 7. If so you could do it like this:

\(\displaystyle T_0=\frac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}}=\frac{\bar{x}-7}{\frac{s}{\sqrt{n}}}\)

As for comparing before and after, you need to use the formula for a 2 sample t-test with unequal variances:

\(\displaystyle T_0=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\)

I believe it is asking if the pre-training sample average is different from the rest of the organization, which is 7.

(9.11-7)/(1.74 x 2.65) = (2.11)/(4.61) = 0.46.

I am a little confused as to how to use this information to determine the answer.

 

[Jameson]

Ok so here is our data set for the pre-training sample:

$\{6,5,10,17,2,12,15,6,9\}$

I get an average of $9.11$ and a standard deviation of $4.1$. So this would be the test statistic:

\(\displaystyle T_0=\frac{9.11-7}{\frac{4.1}{\sqrt{9}}}\)

Once you simplify that you go to a t-table or online calculator with 8 degrees of freedom and find the two sided probability.

 

[MWR]

Ok so here is our data set for the pre-training sample:

$\{6,5,10,17,2,12,15,6,9\}$

I get an average of $9.11$ and a standard deviation of $4.1$. So this would be the test statistic:

\(\displaystyle T_0=\frac{9.11-7}{\frac{4.1}{\sqrt{9}}}\)

Once you simplify that you go to a t-table or online calculator with 8 degrees of freedom and find the two sided probability.

So with that I get 1.54. And a two-sided probability of 0.2?

 

[Jameson]

So with that I get 1.54. And a two-sided probability of 0.2?

I also got 1.54 but how did you get .2 for the probability? I would use a website like this one to find it.

Also, to do this part and the rest of the problem you need to be familiar with setting up a null hypothesis. Does this sound familiar? In this part we have:

$H_0: \mu=7$, $H_1: \mu \ne 7$.

When can reject the null hypothesis and when can't we?

 

[MWR]

So with that I get 1.54. And a two-sided probability of 0.2?

Also, I keep getting 4.91 as the SD with a variance of 24.11.

- - - Updated - - -

I also got 1.54 but how did you get .2 for the probability? I would use a website like this one to find it.

Also, to do this part and the rest of the problem you need to be familiar with setting up a null hypothesis. Does this sound familiar? In this part we have:

$H_0: \mu=7$, $H_1: \mu \ne 7$.

When can reject the null hypothesis and when can't we?

Now I get 0.16. Thus not making it significant. In that case, we reject the null hypothesis, correct?

 

[Jameson]

The opposite. When we have a p-value greater than 0.05 (sometimes we set this threshold at other values, but standardly it's 0.05) then we fail to reject the null hypothesis. This doesn't mean we prove the null hypothesis, we just says it's possible or feasible the means are actually equivalent.

I didn't calculate the variance by hand to be honest because no one does in practice. I used this site and confirmed it on Wolfram Alpha. Are you diving by 9 or by 8? Remember you divide by $n-1$ when calculating the sample S.D..

 

[MWR]

The opposite. When we have a p-value greater than 0.05 (sometimes we set this threshold at other values, but standardly it's 0.05) then we fail to reject the null hypothesis. This doesn't mean we prove the null hypothesis, we just says it's possible or feasible the means are actually equivalent.

I didn't calculate the variance by hand to be honest because no one does in practice. I used this site and confirmed it on Wolfram Alpha. Are you diving by 9 or by 8? Remember you divide by $n-1$ when calculating the sample S.D..

Makes perfect sense, yes. So how do we tie this into answering number 1?

Do we provide our answer relative to the p-value being greater than 0.05 indicating that, yes, this sample, or department, engages in more interpersonal behavior than the rest of the organization?

 

[Jameson]

Makes perfect sense, yes. So how do we tie this into answering number 1?

Do we provide our answer relative to the p-value being greater than 0.05 indicating that, yes, this sample, or department, engages in more interpersonal behavior than the rest of the organization?

No, we completed a hypothesis test with $H_0: \mu=7$ and from our results we can conclude that it's plausible the sample average is the same as the company average, meaning the sample is probably ok to use for the rest of the experiment. If we had a statistically different average then one could argue that the sample wouldn't lead to useful results in the whole company.

 

[MWR]

The opposite. When we have a p-value greater than 0.05 (sometimes we set this threshold at other values, but standardly it's 0.05) then we fail to reject the null hypothesis. This doesn't mean we prove the null hypothesis, we just says it's possible or feasible the means are actually equivalent.

I didn't calculate the variance by hand to be honest because no one does in practice. I used this site and confirmed it on Wolfram Alpha. Are you diving by 9 or by 8? Remember you divide by $n-1$ when calculating the sample S.D..

So this would be my answer to number 1 (I did it by hand so the numbers are slightly different).

Pre-training scores: 6, 5, 10, 17, 2, 12, 15, 6, 9 = 82/9 = 9.11 mean

(6 – 9.11)^2 = 9.67
(5-9.11)^2 = 16.89
(10-9.11)^2 = 0.79
(17-9.11)^2 = 62.25
(2-9.11)^2 = 50.55
(12-9.11)^2 = 8.35
(15-9.11)^2 = 34.69
(6-9.11)^2 = 9.67
(9-9.11)^2 = 0.01
(192.87/9-1) = 192.87/8 = 24.11. The square root of 24.11 is 4.91, the sample standard deviation.

2.11/1.63 = 1.29. A T score of 1.29 and DF of 8 gives a p value of 0.12.
When we have a p-value greater than 0.05, then we fail to reject the null hypothesis. This doesn't mean we prove the null hypothesis, we just says it's possible or feasible that the means are actually equivalent.

Regarding the second part of question 1, the answer is no because the department represents the average of the organization.

 

[Jameson]

I made a mistake when copying over the S.D. - sorry. You are right, it is 4.91. I missed the middle "9".

Anyway, that looks good to me and it how I would answer it. It's always good to have the context of what you are covering in your class but I feel this is an appropriate approach.

 

[MWR]

I made a mistake when copying over the S.D. - sorry. You are right, it is 4.91. I missed the middle "9".

Anyway, that looks good to me and it how I would answer it. It's always good to have the context of what you are covering in your class but I feel this is an appropriate approach.

Thank you, Jameson, for your help. I greatly appreciate it. Regarding the second part of the question, would we do another t-test to show the differences between pre-training and post-training?

 

[Jameson]

Hi again.

So now you'll be comparing two sample groups, thus be conducting a two-sample t-test. Since the variances are not equal you will use this formula.

\(\displaystyle \displaystyle T_0=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\)

Calculate the mean and sample standard deviation for each group then plug that data in. The tricky part of doing this is the degrees of freedom is a horribly messy formula, but that will come later. See what you get when you do this.

Also, what do you think $H_0$ will be for this test? Previously we had $H_0: \mu=7$ because we were testing one sample against a set value, but now we are testing two samples against each other.