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skelet666 [1.2K]
3 years ago
12

A therapist wanted to determine if yoga or meditation is better for relieving stress. The therapist recruited 100 of her high-st

ress patients. Fifty of them were randomly assigned to take weekly yoga classes, and the other 50 were assigned weekly meditation classes. After one month, 30 of the 50 patients in the yoga group reported less stress, and 35 of the 50 patients in the meditation group reported less stress.
Based on the 95% confidence interval, (–0.29, 0.09), is there convincing evidence of a difference in the true proportions of patients experiencing stress relief in the yoga and meditation groups?

There is convincing evidence because the two sample proportions are different.

There is not convincing evidence because the interval contains 0.

There is not convincing evidence because the interval contains both negative and positive values for the true difference.
Mathematics
1 answer:
schepotkina [342]3 years ago
5 0

Answer:  Choice B

There is not convincing evidence because the interval contains 0.

========================================================

Explanation:

The confidence interval is (-0.29, 0.09)

This is the same as writing -0.29 < p1-p1 < 0.09

The thing we're trying to estimate (p1-p2) is between -0.29 and 0.09

Because 0 is in this interval, it is possible that p1-p1 = 0 which leads to p1 = p2.

Therefore, it is possible that the population proportions are the same.

The question asks " is there convincing evidence of a difference in the true proportions", so the answer to this is "no, there isn't convincing evidence". We would need both endpoints of the confidence interval to either be positive together, or be negative together, for us to have convincing evidence that the population proportions are different.

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We need to sample at least 1069 parents.

Step-by-step explanation:

We have that to find our \alpha level, that is the subtraction of 1 by the confidence interval divided by 2. So:

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Now, we have to find z in the Ztable as such z has a pvalue of 1-\alpha.

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Now, find the margin of error M as such

M = z*\frac{\sigma}{\sqrt{n}}

In which \sigma is the standard deviation of the population and n is the size of the sample.

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We need to sample at least n parents.

n is found when M = 4, \sigma = 79.5. So

M = z*\frac{\sigma}{\sqrt{n}}

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4\sqrt{n} = 1.645*79.5

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