Employee's productivity | ||
Group 1 (Home) | Group 2 (OnSide Work) | |
Week 1 | 138 | 104 |
Week 2 | 102 | 94 |
Week 3 | 142 | 107 |
Week 4 | 120 | 110 |
Week 5 | 101 | 83 |
Are employees more productive working from home or working onsite? | ||
Sample size (n) | Sample size (n) | |
Mean | Mean | |
Meadian | Meadian | |
Mode | Mode | |
Max | Max | |
Range | Range | |
Standard Deviation | Standard Deviation | |
Null Hypothesis | ||
Alternative Hypothesis | ||
Test Statistic | ||
Degrees of freedom | ||
P value |
Group 1 (Home) | Group 2 (OnSide Work) | |
Sample size | 5 | 5 |
Mean | 120.60 | 99.60 |
Standard deviation | 19.31 | 11.06 |
Mode | #N/A | #N/A |
Median | 120.00 | 104.00 |
Range | 41 | 27 |
The hypothesis being tested is:
H_{0}: µ_{1} = µ_{2}
H_{1}: µ_{1} ≠ µ_{2}
Group 1 (Home) | Group 2 (OnSide Work) | |
120.60 | 99.60 | mean |
19.31 | 11.06 | std. dev. |
5 | 5 | n |
8 | df | |
21.000 | difference (Group 1 (Home) - Group 2 (OnSide Work)) | |
247.550 | pooled variance | |
15.734 | pooled std. dev. | |
9.951 | standard error of difference | |
0 | hypothesized difference | |
2.110 | t | |
.0678 | p-value (two-tailed) |
The test statistic is 2.110.
The df is 8.
The p-value is 0.0678.
Since the p-value (0.0678) is greater than the significance level (0.05), we fail to reject the null hypothesis.
Therefore, we cannot conclude that µ_{1} ≠ µ_{2}.
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