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slega [8]
2 years ago
15

A large company that produces a "fat-burner" pill claims an average loss of 20 pounds in the first month. A consumer advocacy gr

oup believes that this claim is actually just "hype" intended to sell more of the compound. The advocacy group would like to obtain statistical evidence about this issue and takes a random sample of 100 consumers who responded that they had purchased the pill but didn't know what the survey was about. They find that these 100 people lost an average of 18 pounds with a standard deviation of 7.5 pounds. What are the null and alternative hypothesis in this situation
Mathematics
1 answer:
faust18 [17]2 years ago
4 0

Answer:

Based on the claim (the average loss is 20 pounds)  the system of hypothesis for this case are:

Null hypothesis \mu = 20

Alternative hypothesis: \mu \neq 20

Step-by-step explanation:

For thi case we want to test if the "fat-burner" pill present and average loss of 20 pounds in the first month.

And in order to test this hypothesis they have the following sample data:

n = 100 represent the sample size

\bar X = 18 the sample average fro the lost weight

s = 7.5

Based on the claim (the average loss is 20 pounds)  the system of hypothesis for this case are:

Null hypothesis \mu = 20

Alternative hypothesis: \mu \neq 20

And in order to test the hypothesis we can use the following statistic:

t = \frac{\bar X -\mu}{\frac{s}{\sqrt{n}}}

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What is the maximum number of pages the report can have so that it can be completely stored on a USB drive that holds 4 \cdot 10
lana [24]

Answer:

<em>2,300,000pages  report.</em>

Step-by-step explanation:

Given a USB drive with memory of 4.6*10⁶KB, in order to know the maximum number of pages a report can have so that it can be completely stored on the drive, the conversion factor must be used.

We must first understand that 1 page of a document uses up approximately 2kB of the storage.

Since 2kB = 1page

4.6*10⁶KB = x page

Cross multiply

2kB * x = 4.6*10⁶KB * 1

2x =  4.6*10⁶

x =  4.6*10⁶/2

x =  2.3*10⁶

x = 2,300,000

<em>Hence the maximum number of pages that the report can have so that it can be completely stored on a USB drive that holds 4.6*10⁶KB is approximately 2,300,000pages  report.</em>

4 0
2 years ago
Fabio and carlos play on a basketball team together. in the last game, fabio had 777 points less than 222 times as many points a
jasenka [17]

For this case, the first thing we must do is define variables.

We have then:

f: number of points that fabio scored.

c: number of points that Carlos scored.

We now write the equation that models the problem:

f = 2c - 7

Then, for f = 31 we have:

2c - 7 = 31

From here, we clear the number of points:

 2c = 31 + 7

2c = 38

c = \frac{38}{2}

c = 19

Answer:

The equation to find the number of carlos points is:

2c - 7 = 31

Then, Carlos scored:

c = 19

5 0
2 years ago
Read 2 more answers
A number cube has faces numbered 1 to 6. What is true about rolling the number cube one time? Select three options. S = {1, 2, 3
Marat540 [252]

Answer:

The correct answers are the options B. ; C. ; D.

Step-by-step explanation:

A numbered cube has face from 1 to 6.

The set S = {1, 2, 3, 4, 5, 6}.

A. If A is a subset of S, A could be {0, 1, 2}. This statement is false as A is a subset of B and A contains the element {0} which is not in S.

B. If A is a subset of S and A is given by {5, 6}. This statement is true as A elements of A are also in S.

C. If a subset A represents the complement of rolling a 5, then A = {1, 2, 3, 4, 6}. This statement is even true as A should contain all numbers from 1 to 6 except 5.

D. If a subset A represents the complement of rolling an even number, then A = {1, 3}. This statement is true as the numbers complement of an even number are 1, 3 and 5.

5 0
2 years ago
Read 2 more answers
there are 50 competitors in the mens ski jumping. 30 move on to the qualifying round. how many different ways can the qualifying
Andru [333]
They can take the 30 best abd remove the 20 worst, or remove the 20 best and send on the 39 worst.
8 0
2 years ago
A certain article indicates that in a sample of 1,000 dog owners, 610 said that they take more pictures of their dog than of the
lbvjy [14]

Answer:

(a) The 90% confidence interval is: (0.60, 0.63) Correct interpretation is (3).

(b) The 95% confidence interval is: (0.42, 0.46) Correct interpretation is (1).

(c) First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

Step-by-step explanation:

(a)

Let <em>X</em> = number of dog owners who take more pictures of their dog than of their significant others or friends.

Given:

<em>X</em> = 610

<em>n</em> = 1000

Confidence level = 90%

The (1 - <em>α</em>)% confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

The sample proportion is:

\hat p=\frac{X}{n}=\frac{610}{1000}=0.61

The critical value of <em>z</em> for a 90% confidence level is:

z_{\alpha/2}=z_{0.10/2}=z_[0.05}=1.645

Construct a 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.61\pm 1.645\sqrt{\frac{0.61(1-0.61}{1000}}\\=0.61\pm 0.015\\=(0.595, 0.625)\\\approx(0.60, 0.63)

Thus, the 90% confidence interval for the population proportion of dog owners who take more pictures of their dog than their significant others or friends is (0.60, 0.63).

<u>Interpretation</u>:

There is a 90% chance that the true proportion of dog owners who take more pictures of their dog than of their significant others or friends falls within the interval (0.60, 0.63).

Correct option is (3).

(b)

Let <em>X</em> = number of dog owners who are more likely to complain to their dog than to a friend.

Given:

<em>X</em> = 440

<em>n</em> = 1000

Confidence level = 95%

The (1 - <em>α</em>)% confidence interval for population proportion is:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}

The sample proportion is:

\hat p=\frac{X}{n}=\frac{440}{1000}=0.44

The critical value of <em>z</em> for a 95% confidence level is:

z_{\alpha/2}=z_{0.05/2}=z_{0.025}=1.96

Construct a 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend as follows:

CI=\hat p\pm z_{\alpha/2} \sqrt{\frac{\hat p(1-\hat p)}{n}}\\=0.44\pm 1.96\sqrt{\frac{0.44(1-0.44}{1000}}\\=0.44\pm 0.016\\=(0.424, 0.456)\\\approx(0.42, 0.46)

Thus, the 95% confidence interval for the population proportion of dog owners who are more likely to complain to their dog than to a friend  is (0.42, 0.46).

<u>Interpretation</u>:

There is a 95% chance that the true proportion of dog owners who are more likely to complain to their dog than to a friend falls within the interval (0.42, 0.46).

Correct option is (1).

(c)

The confidence interval in part (b) is wider than the confidence interval in part (a).

The width of the interval is affected by:

  1. The confidence level
  2. Sample size
  3. Standard deviation.

The confidence level in part (b) is more than that in part (a).

Because of this the critical value of <em>z</em> in part (b) is more than that in part (a).

Also the margin of error in part (b) is 0.016 which is more than the margin of error in part (a), 0.015.

First, the confidence level in part(b) is <u>more than</u> the confidence level in part(a) is, so the critical value of part(b) is <u>more than</u> the critical value of part(a), Second the <u>margin of error</u> in part(b) is <u>more than</u> than in part(a).

4 0
2 years ago
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