Respuesta :
Answer:
a) [tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{2 -0}{\frac{3.521}{\sqrt{6}}}=1.391[/tex]
b) [tex]p_v =P(t_{(6)}>1.391) =0.107[/tex]
The significance level would be [tex] \alpha= 1-0.95 =0.05[/tex]
So the p value is higher than the significance level given 0.05, so then we can conclude that we FAIL to reject the null hypothesis that the mean after minus the mean before is lower or equal than 0.
So we don't have enoug evidence to conclude that the course actually increased scores on the driving exam
Step-by-step explanation:
Previous concepts
A paired t-test is used to compare two population means where you have two samples in which observations in one sample can be paired with observations in the other sample. For example if we have Before-and-after observations (This problem) we can use it.
Let put some notation :
x=test value before , y = test value after
x: 83 89 93 77 86 79
y: 87 88 91 77 93 83
Part a
The system of hypothesis for this case are:
Null hypothesis: [tex]\mu_y- \mu_x \leq 0[/tex]
Alternative hypothesis: [tex]\mu_y -\mu_x >0[/tex]
The first step is calculate the difference [tex]d_i=y_i-x_i[/tex] and we obtain this:
d: 4, -1, -2, 0, 7, 4
The second step is calculate the mean difference
[tex]\bar d= \frac{\sum_{i=1}^n d_i}{n}= \frac{12}{6}=2[/tex]
The third step would be calculate the standard deviation for the differences, and we got:
[tex]s_d =\frac{\sum_{i=1}^n (d_i -\bar d)^2}{n-1} =3.521[/tex]
The 4 step is calculate the statistic given by :
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{2 -0}{\frac{3.521}{\sqrt{6}}}=1.391[/tex]
Part b
The next step is calculate the degrees of freedom given by:
[tex]df=n-1=6-1=5[/tex]
Now we can calculate the p value, since we have a left tailed test the p value is given by:
[tex]p_v =P(t_{(6)}>1.391) =0.107[/tex]
The significance level would be [tex] \alpha= 1-0.95 =0.05[/tex]
So the p value is higher than the significance level given 0.05, so then we can conclude that we FAIL to reject the null hypothesis that the mean after minus the mean before is lower or equal than 0.
So we don't have enoug evidence to conclude that the course actually increased scores on the driving exam
