Question

The following table shows the weights of nine subjects before and after following a particular diet for two months. Test the claim that the diet is effective in helping people lose weight Subject A B DEFGHI Before 168 180 157 132 202 124 190 210 171 After 162 178 145 125 171 126 180 195 163

Answer 1

Using the t-distribution, it is found that there is not enough evidence that the diet is helping people lose weight.

What are the hypothesis tested?

At the null hypothesis, it is tested if the mean hasn't decreased, that is:

$H_0: \mu_B \leq \mu_A$

$H_0: \mu_B - \mu_A \leq 0[qtex]At the alternative hypothesis, it is tested if the mean has decreased, that is:[tex]H_1: \mu_B - \mu_A > 0[qtex]What are the mean and the standard error for the distribution of differences?For each sample, they are given as follows:[tex]\mu_B = 170.44, s_B = \frac{29.275}{\sqrt{9}} = 9.7583$.

• $\mu_A = 160.56, s_A = \frac{24.203}{\sqrt{9}} = 8.067$.

Hence, for the distribution of differences, they are:

• $\overline{x} = \mu_B - \mu_A = 170.44 - 160.56 = 9.88$.

• $s = \sqrt{s_B^2 + s_A^2} = \sqrt{9.7583^2 + 8.067^2} = 11.66$

What is the test statistic and the decision?

The test statistic is:

$t = \frac{\overline{x} - \mu}{s}$

In which $\mu = 0$ is the value tested at the null hypothesis.

Hence:

$t = \frac{\overline{x} - \mu}{s}$

$t = \frac{9.88 - 0}{11.66}$

t = 0.85.

Considering a right-tailed test with 9 + 9 - 2 = 16 df, with a standard significance level of 0.05, the critical value is t = 1.7459. Since t = 0.85 < 0.85, there is not enough evidence that the diet is helping people lose weight.

More can be learned about the t-distribution at brainly.com/question/13873630

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