Given data suggest that the paired "t" test should be used.
The paired "t" test
When "D=X-Y, \\mu_D=\\mu_1-\\mu_2" and null hypothesis
the test statistic value for testing the hypothesis is
where "\\bar{d}" and "s_D" are the sample mean and the sample standard deviation of differences "d_i," respectively.
In order to use this test assume that the differences "D_i" are from a normal distribution.
The sample mean "\\bar{d}" for the difference
"\\bar{d}={7.62+8.00+9.09+6.06+1.39+16.07+8.40+8.89+2.88\\over 9}=7.60"
The sample variance is
"s_D^2="
"={1 \\over9-1}((7.62-7.60)^2+(8.00-7.60)^2+(9.09-7.60)^2+"
"+(6.06-7.60)^2+(1.39-7.60)^2+(16.07-7.60)^2+"
"+(8.40-7.60)^2+(8.89-7.60)^2+(2.88-7.60)^2)="
and the sample standard deviation
The hypotheses are:
"H_1: \\mu_D>5"
The test statistic value is
The degrees of the freedom are: "df=n-1=9-1=8."
The P value
"P=0.049431<0.05<0.1"
Reject the null hypothesis at a significance level 0.1 and 0.05.
Do not reject the null hypothesis at a significance level 0.01.
Comments
Leave a comment