Respuesta :
Using the t-distribution, it is found that since the absolute value of the test statistic is less than the critical value for the two-tailed test, you cannot conclude that the mean sales differ between the two programs.
At the null hypothesis, it is tested if the mean sales do not differ between the two programs, that is, the subtraction of the means is 0, hence:
[tex]H_0: \mu_1 - \mu_2 = 0[/tex]
At the alternative hypothesis, it is tested if they differ, that is:
[tex]H_1: \mu_1 - \mu_2 \neq 0[/tex]
The mean and the standard errors for each sample are:
[tex]\mu_1 = 42, s_1 = 5.9675[/tex]
[tex]\mu_2 = 44.56, s_2 = 5.6177[/tex]
The distribution of the difference has mean and standard deviation given by:
[tex]\overline{x} = \mu_2 - \mu_1 = 44.56 - 42 = 2.56[/tex]
[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{5.9675^2 + 5.6177^2} = 8.1957[/tex]
The test statistic is:
[tex]t = \frac{\overline{x} - \mu}{s}[/tex]
In which [tex]\mu = 0[/tex] is the value tested at the null hypothesis.
Then:
[tex]t = \frac{2.56 - 0}{8.1957}[/tex]
[tex]t = 0.31[/tex]
The critical value for a two-tailed test, as we are testing if the mean is different of a value, with 9 + 9 - 2 = 16 df and a significance level of 0.05 is [tex]|t^{\ast}| = 2.12[/tex]
Since the absolute value of the test statistic is less than the critical value for the two-tailed test, you cannot conclude that the mean sales differ between the two programs.
A similar problem is given at https://brainly.com/question/13873630
