Using the t-distribution, as we have the standard deviation for the sample, it is found that it cannot be concluded that the diet has a long-term effect.
What are the hypotheses tested?
At the null hypotheses, it is tested if there is no difference in the weights, that is:
[tex]H_0: \mu_A - \mu_B = 0[/tex]
At the alternative hypotheses, it is tested if the weights decrease, that is:
[tex]H_1: \mu_A - \mu_B < 0[/tex]
What is the mean and the standard deviation of the distribution of the difference?
Using a calculator, for each sample, we have that:
[tex]\mu_B = 210.8, s_B = 33.85, sE_B = \frac{33.85}{\sqrt{15}} = 8.74[/tex]
[tex]\mu_A = 207.2, s_A = 33.47, sE_A = \frac{33.47}{\sqrt{15}} = 8.64[/tex]
Hence, for the distribution of differences, we have that:
[tex]\overline{x} = \mu_A - \mu_B = 207.2 - 210.8 = -3.6[/tex]
[tex]s = \sqrt{sE_B^2 + sE_A^2} = \sqrt{8.74^2 + 8.64^2} = 12.3[/tex]
What is the test statistic?
It is given by:
[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{\overline{x} - \mu}{s}[/tex]
[tex]t = \frac{-3.6 - 0}{12.3}[/tex]
[tex]t = -0.29[/tex]
What is the decision?
Considering a left-tailed test, as we are testing if the mean is less than a value, with a standard significance level of 0.05 and 15 + 15 - 2 = 28 df, the critical value is of [tex]t^{\ast} = -1.7[/tex].
Since the test statistic is greater than the critical value for the left tailed test, it cannot be concluded that the diet has a long-term effect, as the null hypothesis is not rejected.
To learn more about the t-distribution, you can check https://brainly.com/question/13873630
