The data is missing in the question. The data is provided below :
Document : 1 2 3 4 5 6 7 8
Brand A 17 29 18 14 21 25 22 29
Brand B 21 38 15 19 22 30 31 37
Solution :
State of the hypothesis of the null hypothesis and alternate hypothesis.
Null hypothesis : ![$h_A = h_B$](https://tex.z-dn.net/?f=%24h_A%20%3D%20h_B%24)
Alternate hypothesis : ![$h_A > h_B$](https://tex.z-dn.net/?f=%24h_A%20%3E%20h_B%24)
These hypothesis is a one tailed test. The null hypothesis will get rejected when the mean difference between the sample means is very small.
Significance level = 0.05
Therefore the standard error is : ![$SE = \sqrt{(\frac{s^2_1}{n_1})+(\frac{s^2_2}{n_2})}$](https://tex.z-dn.net/?f=%24SE%20%3D%20%5Csqrt%7B%28%5Cfrac%7Bs%5E2_1%7D%7Bn_1%7D%29%2B%28%5Cfrac%7Bs%5E2_2%7D%7Bn_2%7D%29%7D%24)
= 3.602
And the degree of freedom, DF = 14
![$t=\frac{(x_1-x_2)-d}{SE}$](https://tex.z-dn.net/?f=%24t%3D%5Cfrac%7B%28x_1-x_2%29-d%7D%7BSE%7D%24)
= -1.319
Here,
= standard deviation of the sample 1
= standard deviation of the sample 2
= size of the sample 1
= size of the sample 2
= mean of the sample 1
= mean of the sample 2
d = the hypothesis difference between the population mean
The observed difference in a sample means t static of -1.32. From t distribution calculator to determine P(
) = 0.1042
Since the P value of 0.1042 is greater than significance level o 0.05, we therefore cannot reject the null hypothesis.
But from the test, we have no sufficient evidence that supports that Brand A is better than Brand B.