Answer:
a) The null and alternative hypotesis are:
![H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu_1-%5Cmu_2%3D0%5C%5C%5C%5CH_a%3A%20%5Cmu_1-%5Cmu_2%3C0)
1: productivity when boss absent, 2: productivity when boss present
b) t=-3.492
c) df=60
d) tc=-1.6706
e) Yes, it is significant.
Step-by-step explanation:
Hypothesis test for the difference between means.
The claim is that productivity decreases when the boss leaves the office.
Then, the null and alternative hypotesis are:
![H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2](https://tex.z-dn.net/?f=H_0%3A%20%5Cmu_1-%5Cmu_2%3D0%5C%5C%5C%5CH_a%3A%20%5Cmu_1-%5Cmu_2%3C0)
The significance level is α=0.05.
The degrees of freedom are calculated as:
![df=(n_1-1)+(n_2-1)=(26-1)+(36-1)=60](https://tex.z-dn.net/?f=df%3D%28n_1-1%29%2B%28n_2-1%29%3D%2826-1%29%2B%2836-1%29%3D60)
The difference between sample means Md is:
![M_d=M_1-M_2=5-8=-3](https://tex.z-dn.net/?f=M_d%3DM_1-M_2%3D5-8%3D-3)
Now, we have to calculate the standard error for the difference of the means.
As the sample sizes are not equal, we have to calculate the harmonic mean of the sample size:
The standard deviation of sample 1 (boss absent) is:
The standard deviation of sample 2 (boss present) is:
We calculate the mean square error (MSE) as:
![MSE=\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{(n_1-1)+(n_2-1)}=\dfrac{25\cdot 3.16^2+35\cdot 3.46^2}{60}=\dfrac{249.64+419.006}{60}\\\\\\MSE=\dfrac{668.646}{60}\\\\\\MSE=11.14](https://tex.z-dn.net/?f=MSE%3D%5Cdfrac%7B%28n_1-1%29s_1%5E2%2B%28n_2-1%29s_2%5E2%7D%7B%28n_1-1%29%2B%28n_2-1%29%7D%3D%5Cdfrac%7B25%5Ccdot%203.16%5E2%2B35%5Ccdot%203.46%5E2%7D%7B60%7D%3D%5Cdfrac%7B249.64%2B419.006%7D%7B60%7D%5C%5C%5C%5C%5C%5CMSE%3D%5Cdfrac%7B668.646%7D%7B60%7D%5C%5C%5C%5C%5C%5CMSE%3D11.14)
Then, the standard error can be calculated as:
![s_{M_d}=\sqrt{\dfrac{2MSE}{n}}=\sqrt{\dfrac{2\cdot 11.14}{30.194}}=0.86](https://tex.z-dn.net/?f=s_%7BM_d%7D%3D%5Csqrt%7B%5Cdfrac%7B2MSE%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cdfrac%7B2%5Ccdot%2011.14%7D%7B30.194%7D%7D%3D0.86)
Now, we can calculate the test statistic t:
![t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-3-(0)}{0.86}=-3.492](https://tex.z-dn.net/?f=t%3D%5Cdfrac%7BM_d-%28%5Cmu_1-%5Cmu_2%29%7D%7Bs_%7BM_d%7D%7D%3D%5Cdfrac%7B-3-%280%29%7D%7B0.86%7D%3D-3.492)
If we apply the critical value approach, the critical value of t for a significance level α=0.05 and 60 degrees of freedom is:
![t_c=-1.6706](https://tex.z-dn.net/?f=t_c%3D-1.6706)
As this is a left-tail test, the decision rule is to reject the null hypothesis if the test statistic is smaller than the critical value.
In this case, the test statistic t=-3.492 is smaller than the critical value t_c=-1.6706, the effect is significant.