Answer:
a)
And the deviation would be just the square root of the variance:
And now we can calculate the statistic:
Now we can calculate the degrees of freedom given by:
And now we can calculate the p value using the altenative hypothesis:
If we compare the p value obtained and using the significance level assumed
we have
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
b) ![(\bar X_2 -\bar X_1) \pm t_{\alpha/2}* S_p \sqrt{\frac{1}{n_!} +\frac{1}{n_2}}](https://tex.z-dn.net/?f=%20%28%5Cbar%20X_2%20-%5Cbar%20X_1%29%20%5Cpm%20t_%7B%5Calpha%2F2%7D%2A%20S_p%20%5Csqrt%7B%5Cfrac%7B1%7D%7Bn_%21%7D%20%2B%5Cfrac%7B1%7D%7Bn_2%7D%7D)
And replacing we got:
![(15.8333-12.8333) -1.812 \sqrt{\frac{1}{6} +\frac{1}{6}}= 1.954](https://tex.z-dn.net/?f=%20%2815.8333-12.8333%29%20-1.812%20%5Csqrt%7B%5Cfrac%7B1%7D%7B6%7D%20%2B%5Cfrac%7B1%7D%7B6%7D%7D%3D%201.954)
![(15.8333-12.8333) +1.812 \sqrt{\frac{1}{6} +\frac{1}{6}}= 4.046](https://tex.z-dn.net/?f=%20%2815.8333-12.8333%29%20%2B1.812%20%5Csqrt%7B%5Cfrac%7B1%7D%7B6%7D%20%2B%5Cfrac%7B1%7D%7B6%7D%7D%3D%204.046)
Step-by-step explanation:
Part a
TYPE N MEAN STD. DEVIATION STD. ERROR MEAN
Type 1 6 12.8333 1.47196 0.60093
Type 2 6 15.3333 4.32049 1.7638
When we have two independent samples from two normal distributions with equal variances we are assuming that
And the statistic is given by this formula:
Where t follows a t distribution with
degrees of freedom and the pooled variance
is given by this formula:
This last one is an unbiased estimator of the common variance
The system of hypothesis on this case are:
Null hypothesis:
Alternative hypothesis:
Or equivalently:
Null hypothesis:
Alternative hypothesis:
Our notation on this case :
represent the sample size for group 1
represent the sample size for group 2
represent the sample mean for the group 1
represent the sample mean for the group 2
represent the sample standard deviation for group 1
represent the sample standard deviation for group 2
First we can begin finding the pooled variance:
And the deviation would be just the square root of the variance:
And now we can calculate the statistic:
Now we can calculate the degrees of freedom given by:
And now we can calculate the p value using the altenative hypothesis:
If we compare the p value obtained and using the significance level assumed
we have
so we can conclude that we have enough evidence to FAIL to reject the null hypothesis.
Part b
For the confidence interval we know that the confidence is 90% so then the value os
and
, the degrees of freedom are given by:
![df= n_1 + n_2 -2 = 6+6-2= 10](https://tex.z-dn.net/?f=%20df%3D%20n_1%20%2B%20n_2%20-2%20%3D%206%2B6-2%3D%2010)
And the crtitical value for this case would be ![t_{critc}= 1.812](https://tex.z-dn.net/?f=%20t_%7Bcritc%7D%3D%201.812)
The confidence interval for the difference of means is given by:
![(\bar X_2 -\bar X_1) \pm t_{\alpha/2}*S_p \sqrt{\frac{1}{n_!} +\frac{1}{n_2}}](https://tex.z-dn.net/?f=%20%28%5Cbar%20X_2%20-%5Cbar%20X_1%29%20%5Cpm%20t_%7B%5Calpha%2F2%7D%2AS_p%20%5Csqrt%7B%5Cfrac%7B1%7D%7Bn_%21%7D%20%2B%5Cfrac%7B1%7D%7Bn_2%7D%7D)
And replacing we got:
![(15.8333-12.8333) -1.812 \sqrt{\frac{1}{6} +\frac{1}{6}}= 1.954](https://tex.z-dn.net/?f=%20%2815.8333-12.8333%29%20-1.812%20%5Csqrt%7B%5Cfrac%7B1%7D%7B6%7D%20%2B%5Cfrac%7B1%7D%7B6%7D%7D%3D%201.954)
![(15.8333-12.8333) +1.812 \sqrt{\frac{1}{6} +\frac{1}{6}}= 4.046](https://tex.z-dn.net/?f=%20%2815.8333-12.8333%29%20%2B1.812%20%5Csqrt%7B%5Cfrac%7B1%7D%7B6%7D%20%2B%5Cfrac%7B1%7D%7B6%7D%7D%3D%204.046)