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solmaris [256]

Answer:

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3 0

3 years ago

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An advertisement for a popular weight-loss clinic suggests that participants in its new diet program lose, on average, more than

Sedbober [7]

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

3 0

3 years ago

The following data points represent the number of classes that each teacher at Broxin High School teaches.

nika2105 [10]

Answer:

3

Step-by-step explanation:

4 0

3 years ago

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Find the area of the circle (radius = 6)

Rainbow [258]

Step-by-step explanation:

Area of the circle = pi×r²

= 22/7 × 6 × 6

= 113.14 unit²

4 0

3 years ago

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Simplify the expression x8 y-26/x14 y-5 X x-39 y-21.

KonstantinChe [14]

$\frac{x^8y^{-26}}{x^{14}y^{-5}}[x^{-36}y^{-21}]$

We will apply these following power rule

$x^m*x^n = x^{m+n}$
$x^m/x^n=x^{m-n}$

$[ \frac{x^8}{x^{14}}][ \frac{y^{-26}}{y^{-5}}][x^{-39}y^{-21}]$
$[x^{8-14}][y^{-26-5}][x^{-39}y^{-21}]$
$x^{-6}y^{-21}[x^{-39}y^{-21}]$
$[x^{-6-39}][y^{-21-21}]$
$x^{-45}y^{-42}$

8 0

3 years ago

Please help me.......​

Please help me.......