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2 years ago

If you solve this i give you 50 points

Mathematics

1 answer:

PIT_PIT [208]2 years ago

7 0

Step-by-step explanation:

2) 8,6,5,3 ,57 7,4,2/8

=92/8

mean =11.5

Arrange the terms in ascending order

2,3, 4, 5, 6, 7, 8, 57

the median of the ascending set= 5.5

3) 72 75 85 82 37 108/ 6

459/6

mean =76.5

Arrange the terms in ascending order

37, 72, 75, 82, 85, 108

the median of the ascending set= 78.5

4) 10 students were asked because there were 10 data values in the list.

5) it should be 55/10 =5.5

so its false

6) Sum first and find the mean number

5+4+3+2+1/5

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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:

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

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

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.

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

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2 years ago

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egoroff_w [7]

Answer:

6.3 ?

Step-by-step explanation:

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2 years ago

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xz_007 [3.2K]

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3 years ago

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guapka [62]

Answer:

Step-by-step explanation:

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2 years ago

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