Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

Read the brief newspaper article on using a depression pill to help smokers quit

ID: 3054587 • Letter: R

Question

Read the brief newspaper article on using a depression pill to help smokers quit.

Depression Pill Seems To Help Smokers Quit:

Boston –Taking an anti-depression medicine appears to double smokers’ chances of kicking the habit, a study found. The Food and Drug Administration approved the marketing of this medicine called Zyban or bupropion, to help smokers in May. The results of several studies with the drug, including one publishing in today’s issue of the New England Journal of Medicine, were then made public. The newly published study was conducted on 615 volunteers who wanted to give up smoking and were not outwardly depressed. They took either Zyban or a placebo pill for 6 weeks. A year later, 23 percent of those getting Zyban were still off cigarettes, compared with 12 percent in the comparison group.

The results of this experiment were significant at the alpha=0.05 significance level. In your opinion, are the results practically significant? Justify your position.

Explanation / Answer

Solution:-

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: P1> P2
Alternative hypothesis: P1 < P2

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method is a two-proportion z-test.

Analyze sample data. Using sample data, we calculate the pooled sample proportion (p) and the standard error (SE). Using those measures, we compute the z-score test statistic (z).

p = (p1 * n1 + p2 * n2) / (n1 + n2)

p = 0.175

SE = sqrt{ p * ( 1 - p ) * [ (1/n1) + (1/n2) ] }
SE = 0.03447
z = (p1 - p 2) / SE

z = 3.19

where p1 is the sample proportion in sample 1, where p2 is the sample proportion in sample 2, n1 is the size of sample 1, and n2 is the size of sample 2.

Since we have a one-tailed test, the P-value is the probability that the z-score is greater than 3.19.

Thus, the P-value = 0.0007

Interpret results. Since the P-value (0.0007) is less than the significance level (0.05), we have to reject the null hypothesis.

The results of this experiment were significant at the alpha=0.05 significance level.

Yes the result are practically significant.

Hire Me For All Your Tutoring Needs
Integrity-first tutoring: clear explanations, guidance, and feedback.
Drop an Email at
drjack9650@gmail.com
Chat Now And Get Quote