In: Statistics and Probability
Once you have the dataset, please use knowledge gained in other business and/or economics classes to realize what topic and theory the data could relate and a research question that it could allow you to answer. More specifically, please put together an analysis by making sure your project report includes the following:
7.1. Specify the level of significance (Type I error associated with the null hypothesis),
7.2. Determine the test statistic (the appropriate statistical test as mentioned under point 5 above),
7.3. Determine the critical values (and region(s) if applicable),
ID | Nr. of units produced before new compensation system | Nr. of units produced after new compensation system |
1 | 20 | 23 |
2 | 6 | 8 |
3 | 12 | 11 |
4 | 34 | 35 |
5 | 55 | 57 |
6 | 43 | 76 |
7 | 54 | 54 |
8 | 24 | 26 |
9 | 33 | 35 |
10 | 21 | 26 |
11 | 34 | 29 |
12 | 33 | 31 |
13 | 54 | 56 |
14 | 23 | 22 |
15 | 33 | 35 |
16 | 44 | 41 |
17 | 65 | 56 |
18 | 43 | 34 |
19 | 53 | 51 |
20 | 22 | 21 |
21 | 34 | 31 |
22 | 32 | 33 |
23 | 44 | 38 |
24 | 17 | 15 |
25 | 28 | 27 |
26 | 47 | 50 |
27 | 35 | 78 |
28 | 29 | 29 |
29 | 36 | 44 |
30 | 25 | 23 |
Research question:
Does new compensation system affect the number of units produced?
Hypothesis:
That is new compensation system does not affect the number of units produced.
Que.7.1
Level of significance : It is the probability of rejecting null hypothesis when it is true.
Here we assume it is 5%.
Que.7.2
Test statistics:
Here we use paired t test.
Where
I used R software to solve this question.
R codes and output:
before=scan('clipboard');before
Read 30 items
[1] 20 6 12 34 55 43 54 24 33 21 34 33 54 23 33 44 65 43 53 22 34
32 44 17 28
[26] 47 35 29 36 25
> after=scan('clipboard');after
Read 30 items
[1] 23 8 11 35 57 76 54 26 35 26 29 31 56 22 35 41 56 34 51 21 31
33 38 15 27
[26] 50 78 29 44 23
> t.test(before,after,paired=TRUE)
Paired t-test
data: before and after
t = -1.0769, df = 29, p-value = 0.2904
alternative hypothesis: true difference in means is not equal to
0
95 percent confidence interval:
-5.991486 1.858153
sample estimates:
mean of the differences
-2.066667
Test statistic = t = -1.0769
DF = 29
Hence from t table critical value at 5% level of significance is 2.045
P-value = 0.2904
Since p-value is greater than 0.05, hence we accept null hypothesis and conclude that new compensation system does not affect the number of units produced.