Question

An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than 10 pounds. A consumer activist decides to test the authenticity of the claim. She follows the progress of 18 women who recently joined the weight-reduction program. She calculates the mean weight loss of these participants as 10.8 pounds with a standard deviation of 2.4 pounds. It may be assumed that the weight loss of participants to the weight-reduction program is normally distributed.
a) set up the competing hypotheses to test the advertisement's claim
b) at the 5% significance level, what is the critical value(s)? specify the decision rule
c) what does the consumer activist conclude?

Answer 1

Testing the hypothesis, it is found that:

a)

The null hypothesis is: $H_0: \mu \leq 10$

The alternative hypothesis is: $H_1: \mu > 10$

b)

The critical value is: $t_c = 1.74$

The decision rule is:

• If t < 1.74, we do not reject the null hypothesis.
• If t > 1.74, we reject the null hypothesis.

c)

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

Item a:

At the null hypothesis, it is tested if the mean loss is of at most 10 pounds, that is:

$H_0: \mu \leq 10$

At the alternative hypothesis, it is tested if the mean loss is of more than 10 pounds, that is:

$H_1: \mu > 10$

Item b:

We are having a right-tailed test, as we are testing if the mean is more than a value, with a significance level of 0.05 and 18 - 1 = 17 df.

Hence, using a calculator for the t-distribution, the critical value is: $t_c = 1.74$.

Hence, the decision rule is:

• If t < 1.74, we do not reject the null hypothesis.
• If t > 1.74, we reject the null hypothesis.

Item c:

We have the standard deviation for the sample, hence the t-distribution is used. The test statistic is given by:

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

The parameters are:

• $\overline{x}$ is the sample mean.
• $\mu$ is the value tested at the null hypothesis.
• s is the standard deviation of the sample.
• n is the sample size.

For this problem, we have that:

$\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18$

Thus, the value of the test statistic is:

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

$t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}$

$t = 1.41$

Since t = 1.41 < 1.74, we do not reject the null hypothesis, that is, it cannot be concluded that the mean weight loss is of more than 10 pounds.

A similar problem is given at brainly.com/question/25147864