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anygoal [31]
3 years ago
12

Quit Smoking: The New England Journal of Medicine published the results of a double-blind, placebo-controlled experiment to stud

y the effect of nicotine patches and the antidepressant bupropion on quitting smoking. With the data from the experiment we calculate the sample difference in the "quit smoking" rates for the nicotine treatment group and the placebo group ("treatment" minus "placebo"). We get 0.8% = 0.008. Which of the following is an appropriate conclusion based on this finding? Group of answer choices n this experiment the nicotine treatment had a higher success rate than the placebo group, but the improvement was less than 1%. In this experiment, the placebo group had a higher success rate than the nicotine group by 0.8%. Nicotine patches will produce a slightly higher success rate when compared to a placebo, but the difference is not statistically significant. Nicotine patches and the antidepressant bupropion work equally well on "quit smoking" rates.
Mathematics
1 answer:
Juliette [100K]3 years ago
5 0

Answer:

Option A is correct.

In this experiment the nicotine treatment had a higher success rate than the placebo group, but the improvement was less than 1%. It was exactly 0.8%

Step-by-step explanation:

In a drug treatment experiment, there are usually two groups.

- The treatment group; they get the full dosage and content of the drug to be tested.

- The placebo group; here, they sort of serve as the control group. They get the dosage of the drug to be tested with the active ingredient of the drug removed.

Double blind means that both the participants and the conductors do not know which members belong to which treatment group. It is used to increase the randomness of the sampling technique.

For this experiment, nicotine patches, antidepressants and bupropion are used to help people quit smoking.

It is explained that the the rate at which the participants quit smoking is recorded and at the end, a sample mean is obtained.

The difference in sample means of both groups is then obtained by solving

(Rate of quit smoking for the treatment group) - (Rate of quit smoking for the placebo group)

This difference was obtained to be positive 0.8%. It is evident that this figure represents a higher success rate for the treatment group than the placebo group, but the improvement was less than 1%. (0.8%).

All the other options are either incorrect or not clearly stated in the description ofthe experiment.

- In this experiment, the placebo group had a higher success rate than the nicotine group by 0.8%.

This is obviously not true, it is the reverse of the true statement.

- Nicotine patches will produce a slightly higher success rate when compared to a placebo, but the difference is not statistically significant.

This started with a true statement bit it ended with a statement we cannot confirm because we don't know the sample size for either of the groups involved in the experiment. So, it isn't totally true.

- Nicotine patches and the antidepressant bupropion work equally well on "quit smoking" rates.

They don't have an equal rate of working. The nicotine slightly works better as stated in the question.

Hope this Helps!!!

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3. The Food Marketing Institute shows that 17% of households spend more than $100 per week on groceries. Assume the population p
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Answer:

A)sample proportion = 0.17,  the sampling distribution of p can be calculated/approximated with normal distribution of sample proportion = 0.17 and standard error/deviation = 0.013281

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Step-by-step explanation:

A) p ( proportion of population that spends more than $100 per week) = 0.17

sample size (n)= 800

the sample proportion of p = 0.17

standard error of p = \sqrt{\frac{p(1-p)}{n} } = 0.013281

the sampling distribution of p can be calculated/approximated with

normal distribution of sample proportion = 0.17 and standard error/deviation = 0.013281

B) probability that the sample proportion will be +-0.02 of the population proportion

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z value corresponding to P

Z = \frac{P - p}{standard deviation}

at P = 0.15

Z =  (0.15 - 0.17) / 0.013281 = = -1.51

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z = ( 0.19 - 0.17) / 0.013281 = 1.51

therefore the required probability will be

p( -1.5 ≤ z ≤ 1.5 ) = p(z ≤ 1.51 ) - p(z ≤ -1.51 )

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standard deviation/ error = 0.009391 (applying the equation for calculating standard error as seen in part A above)

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