Answered

Welcome to Westonci.ca, the place where your questions are answered by a community of knowledgeable contributors. Get immediate and reliable answers to your questions from a community of experienced professionals on our platform. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

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?

Sagot :

Testing the hypothesis, it is found that:

a)

The null hypothesis is: [tex]H_0: \mu \leq 10[/tex]

The alternative hypothesis is: [tex]H_1: \mu > 10[/tex]

b)

The critical value is: [tex]t_c = 1.74[/tex]

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:

[tex]H_0: \mu \leq 10[/tex]

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

[tex]H_1: \mu > 10[/tex]

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: [tex]t_c = 1.74[/tex].

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:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

The parameters are:

  • [tex]\overline{x}[/tex] is the sample mean.
  • [tex]\mu[/tex] 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:

[tex]\overline{x} = 10.8, \mu = 10, s = 2.4, n = 18[/tex]

Thus, the value of the test statistic is:

[tex]t = \frac{\overline{x} - \mu}{\frac{s}{\sqrt{n}}}[/tex]

[tex]t = \frac{10.8 - 10}{\frac{2.4}{\sqrt{18}}}[/tex]

[tex]t = 1.41[/tex]

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 https://brainly.com/question/25147864

Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.