In a clinical trial of 2165 subjects treated with a certain drug, 21 reported headaches. In a control group of 1718 subjects gi
ven a placebo, 23 reported headaches. Denoting the proportion of headaches in the treatment group by pt and denoting the proportion of headaches in the control (placebo) group by pc, the relative risk is pt/pc. The relative risk is a measure of the strength of the effect of the drug treatment. Another such measure is the odds ratio, which is the ratio of the odds in favor of a headache for the treatment group to the odds in favor of a headache for the control (placebo) group, found by evaluating StartFraction p Subscript t Baseline divided by (1 minus p Subscript t Baseline )Over p Subscript c Baseline divided by (1 minus p Subscript c Baseline )EndFraction . The relative risk and odds ratios are commonly used in medicine and epidemiological studies. Find the relative risk and odds ratio for the headache data. What do the results suggest about the risk of a headache from the drug treatment?
1 answer:
You might be interested in
Answer:
We conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.
Step-by-step explanation:
We are given that a particular brand of tires claims that its deluxe tire averages at least 50,000 miles before it needs to be replaced. From past studies of this tire, the standard deviation is known to be 8000.
From the 28 tires surveyed, the mean lifespan was 46,500 miles with a standard deviation of 9800 miles.
<u><em>Let </em></u><u><em> = average miles for deluxe tires</em></u>
So, Null Hypothesis, :
50,000 miles {means that deluxe tire averages at least 50,000 miles before it needs to be replaced}
Alternate Hypothesis, :
< 50,000 miles {means that deluxe tire averages less than 50,000 miles before it needs to be replaced}
The test statistics that will be used here is <u>One-sample z test statistics</u> as we know about population standard deviation;
T.S. = ~ N(0,1)
where, = sample mean lifespan = 46,500 miles
= population standard deviation = 8000 miles
n = sample of tires = 28
So, <u><em>test statistics</em></u> =
= -2.315
The value of the test statistics is -2.315.
Now at 5% significance level, the z table gives critical value of -1.6449 for left-tailed test. Since our test statistics is less than the critical value of z as -2.315 < -1.6449, so we have sufficient evidence to reject our null hypothesis as it will fall in the rejection region due to which we reject our null hypothesis.
Therefore, we conclude that deluxe tire averages less than 50,000 miles before it needs to be replaced which means that the claim is not supported.
Integrate the force field along the given path (call it <em>C</em>):