Respuesta :
Answer:
There is enough statistical evidence to say that the means are different.
Step-by-step explanation:
The question is incomplete:
"The two-sample t statistic for comparing the population means has value (±0.001)."
The null and alternative hypothesis are:
[tex]H_0: \mu_r-\mu_b=0\\\\H_1: \mu_r-\mu_b\neq0[/tex]
The level of significance is assumed to be α=0.01.
The difference of the sample means is:
[tex]M_d=M_r-M_b=9.64-15.84=-6.20[/tex]
The standard deviation for the difference of the means is:
[tex]s_d=\sqrt\frac{s_r^2+s_b^2}{n} } =\sqrt\frac{3.43^2+8.65^2}{21} }=\sqrt{\frac{85.59}{21} }=\sqrt{4.12} =2.03[/tex]
Then, the t-statistic is:
[tex]t=\frac{M_d-(\mu_r-\mu_b)}{s_d} =\frac{-6.20-0}{2.03}= -3.05[/tex]
The degrees of freedom are:
[tex]df=n_r+n_b-2=21+21-2=40[/tex]
With 40 degrees of freedom the t-critial for a significance of α=0.01 is t=±2.705.
As the t-statistic lies in the rejection region, the effect is significant and the null hypothesis is rejected.
There is enough statistical evidence to say that the means are different.
Using the t-distribution, it is found that the two-sample t statistic for comparing the population means has value t = -3.05.
The standard errors are given by:
[tex]s_1 = \frac{3.43}{\sqrt{21}} = 0.7485[/tex]
[tex]s_2 = \frac{8.65}{\sqrt{21}} = 1.8876[/tex]
The distribution of the differences has mean and standard deviation given by:
[tex]\overline{x} = 9.64 - 15.84 = -6.2[/tex]
[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.7485^2 + 1.8876^2} = 2.0306[/tex]
The test statistic is given by:
[tex]t = \frac{\overline{x}}{s}[/tex]
Hence:
[tex]t = \frac{-6.2}{2.0306}[/tex]
[tex]t = -3.05[/tex]
The two-sample t statistic for comparing the population means has value is t = -3.05.
A similar problem is given at https://brainly.com/question/13873630
