Respuesta :
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 : [tex]$h_A = h_B$[/tex]
Alternate hypothesis : [tex]$h_A > h_B$[/tex]
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 : [tex]$SE = \sqrt{(\frac{s^2_1}{n_1})+(\frac{s^2_2}{n_2})}$[/tex]
= 3.602
And the degree of freedom, DF = 14
[tex]$t=\frac{(x_1-x_2)-d}{SE}$[/tex]
= -1.319
Here, [tex]$s_1$[/tex] = standard deviation of the sample 1
[tex]$s_2$[/tex] = standard deviation of the sample 2
[tex]$n_1$[/tex] = size of the sample 1
[tex]$n_2$[/tex] = size of the sample 2
[tex]$x_1$[/tex] = mean of the sample 1
[tex]$x_2$[/tex] = 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([tex]$t \leq -1.32$[/tex]) = 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.
