Respuesta :
Answer:
a) The null and alternative hypotesis are:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2<0[/tex]
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:
[tex]H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2<0[/tex]
The significance level is α=0.05.
The degrees of freedom are calculated as:
[tex]df=(n_1-1)+(n_2-1)=(26-1)+(36-1)=60[/tex]
The difference between sample means Md is:
[tex]M_d=M_1-M_2=5-8=-3[/tex]
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:
[tex]n=\dfrac{2}{1/n_1+1/n_2}=\dfrac{2}{1/26+1/36}=\dfrac{2}{0.066}=30.194[/tex]
The standard deviation of sample 1 (boss absent) is:
[tex]s_1=\sqrt{\dfrac{SS_1}{n_1-1}}=\sqrt{\dfrac{250}{26-1}}=\sqrt{10}=3.16[/tex]
The standard deviation of sample 2 (boss present) is:
[tex]s_2=\sqrt{\dfrac{SS_2}{n_2-1}}=\sqrt{\dfrac{420}{36-1}}=\sqrt{12}=3.46[/tex]
We calculate the mean square error (MSE) as:
[tex]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[/tex]
Then, the standard error can be calculated as:
[tex]s_{M_d}=\sqrt{\dfrac{2MSE}{n}}=\sqrt{\dfrac{2\cdot 11.14}{30.194}}=0.86[/tex]
Now, we can calculate the test statistic t:
[tex]t=\dfrac{M_d-(\mu_1-\mu_2)}{s_{M_d}}=\dfrac{-3-(0)}{0.86}=-3.492[/tex]
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:
[tex]t_c=-1.6706[/tex]
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.
