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Zielflug [23.3K]
4 years ago
14

An experiment will be conducted to test the effectiveness of a weight-loss supplement. Volunteers will be randomly assigned to t

ake either the supplement or a placebo for 90 days, with 12 volunteers in each group. The subjects will not know which treatment they receive. At the end of the experiment, researchers plan to calculate the mean weight loss for each of the two groups and to construct a two-sample t-confidence interval for the difference of the two treatment means. Which of the following assumptions is necessary for the confidence interval to be valid?
a. The sample size is greater than or equal to 10 percent of the population size.
b. Each of the two groups has at least 5 successes and at least 5 failures.
c. The distributions of weight loss of the two treatments are approximately normally distributed.
d. The volunteers in the supplement group are paired with volunteers in the placebo group.
e. The expected number of people who lose weight in each group is at least 5.
Mathematics
1 answer:
DerKrebs [107]4 years ago
3 0

Answer:

Option c. The distributions of weight loss of the two treatments are approximately normally distributed.

Step-by-step explanation:

To increase the confidence level, the results must be able to satisfy the following conditions:

  • The sample size is increased. This reduces the margin of the error in the sampling experiment.
  • Reduction of the variability. This means that the less the data varies, the more precise the data is.
  • Using the one-sided confidence level
  • Lowering the confidence level.

All these characteristics can only be satisfied by the normal distribution curve.

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The mean one-way commute to work in Chowchilla is 7 minutes. The standard deviation is 2.4 minutes, and the population is normal
adell [148]

Answer:

The answer is below

Step-by-step explanation:

Given that:

The mean (μ) one-way commute to work in Chowchilla is 7 minutes. The standard deviation (σ) is 2.4 minutes.

The z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

z=\frac{x-\mu}{\sigma}

a) For x < 2:

z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08

From normal distribution table,  P(x < 2) = P(z < -2.08) = 0.0188 = 1.88%

b) For x = 2:

z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08

For x = 11:

z=\frac{x-\mu}{\sigma}=\frac{11-7}{2.4} =1.67

From normal distribution table, P(2 < x < 11) = P(-2.08 < z < 1.67 ) = P(z < 1.67) - P(z < -2.08) = 0.9525 - 0.0188 = 0.9337  

c) For x = 11:

z=\frac{x-\mu}{\sigma}=\frac{11-7}{2.4} =1.67

From normal distribution table,  P(x < 11) = P(z < 1.67) = 0.9525

d) For x = 2:

z=\frac{x-\mu}{\sigma}=\frac{2-7}{2.4} =-2.08

For x = 5:

z=\frac{x-\mu}{\sigma}=\frac{5-7}{2.4} =-0.83

From normal distribution table, P(2 < x < 5) = P(-2.08 < z < -0.83 ) = P(z < -0.83) - P(z < -2.08) =  0.2033- 0.0188 = 0.1845  

e) For x = 5:

z=\frac{x-\mu}{\sigma}=\frac{5-7}{2.4} =-0.83

From normal distribution table,  P(x < 5) = P(z < -0.83) = 0.2033

8 0
4 years ago
Use the distributive property to expand the following expression:
KIM [24]

Answer:

This is guess of what the question is asking:

I had got 49x - 21y - 6.

Step-by-step explanation:

7 x 7 is 49 and 7 x 3 is 21.

4 0
3 years ago
Read 2 more answers
Find the center of the circle that can be circumscribed about triangle EFG with E(2, 2), F(2, -2), and G(6,-2).
nadezda [96]
Hello friend that would be the last option (3, 2)
it would be great if you gave me brainlest
6 0
3 years ago
Read 2 more answers
Line k has a slope of 2/3. If line m is parallel to line k, then it has a slope of
puteri [66]

Answer:

Line M would also have a slope of 2/3.

Step-by-step explanation:

When two line are parallel that means that they will never intersect.

In order for two lines to avoid intersection they would have to be at the same slope.

Since lines K and line M are parallel to each other they have to have the same slope.

6 0
3 years ago
The brain volumes ?( cm cubed cm3?) of 20 brains have a mean of 1083.9 1083.9 cm cubed cm3 and a standard deviation of 122.2 122
Colt1911 [192]

Answer:

Range: (844.9,1333.7)

A brain volume of 1348.3 cm cubed can be considered significantly high.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 1083.9 cm cubed

Standard Deviation, σ = 122.2 cm cubed

Range rule thumb:

  • This rule state that the the range of data is four times the standard deviation of the data.

\text{Range} = 4\times \sigma = 4\times 122.2 = 488.8

Upper Limit:

\mu + 2\sigma = 1089.3 + 2(122.2) = 1333.7

Lower limit:

\mu - 2\sigma = 1089.3 - 2(122.2) = 844.9

Since, 1348.3 does not lie in the range (844.9,1333.7),  a brain volume of 1348.3 cm cubed can be considered significantly high.

6 0
3 years ago
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