Respuesta :
Answer:
[tex] d = 0.3[/tex] represent the sample difference
[tex] s^2_d = 0.15[/tex] represent the sample variance
[tex] s_d = \sqrt{0.15} =0.387[/tex] represent the sample deviation for the difference
[tex]n_1= n_2 =10[/tex] represent the sample size
The statistic on this case is given by:
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.3 -0}{\frac{0.387}{\sqrt{10}}}=2.449[/tex]
Step-by-step explanation:
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
1=test value before , 2 = test value after
The system of hypothesis for this case are:
Null hypothesis: [tex]\mu_1- \mu_2 = 0[/tex]
Alternative hypothesis: [tex]\mu_1 -\mu_2 \neq 0[/tex]
For this case we know the following info:
[tex] d = 0.3[/tex] represent the sample difference
[tex] s^2_d = 0.15[/tex] represent the sample variance
[tex] s_d = \sqrt{0.15} =0.387[/tex] represent the sample deviation for the difference
[tex]n_1= n_2 =10[/tex] represent the sample size
The statistic on this case is given by:
[tex]t=\frac{\bar d -0}{\frac{s_d}{\sqrt{n}}}=\frac{0.3 -0}{\frac{0.387}{\sqrt{10}}}=2.449[/tex]
