In a trial of 400 patients who received 10 mg doses of a drug daily, 48 reported

Q: In a trial of 400 patients who received 10 mg doses of a drug daily, 48 reported

a)Point Estimate=48/400=0.12 mean=0.12, stddev=sqrt(0.12*0.88/400)=0.01625,n=400 b)np=400*0.12=48>10 n(1-p)=400*0.88=352>10 Hence yes both the requirements are satisfied---------------------->Option A) c)Z(0.995)=2.576 CI=[0.12-2.576*0.01625,0.12+2.576*0.01625]=[0.078,0.162] d)We are 99% Confident that the interval contains the true value of p------------Option B)

ID: 3133737 • Letter: I

Question

In a trial of 400 patients who received 10 mg doses of a drug daily, 48 reported headaches as a side effect. Use this information to complete parts a through d below.

A- Obtain a point estimate for the population proportion of patients who received 10-mg doses of a drug daily and reported headache as a side effect P^=

B- Verify that the requirements for constructing a confidence interval about p are satisfied. Are the requires for constructing a confidence satisfied?

a- Yes, the requirements for constructing a confidence interval are satisfied

b- No, the requirement that each trial be independent is not satisfied.

c- No, the requirement that np (1-p) is greater than 10 is not satisfied.

d- No, the requirement that the sample size is no more than 5% of the population is not satisfied.

C- Construct a 99% confidence interval for the population proportion of patients who receive the drug and report headache as a side effect. The 99% confidence interval is (___,____). Round to three decimal places as needed.

D- Interpret the confidence interval. Which statement below best interprets the interval?

a- WE are 99% confident that the interval does not contain the true value of p.

b- We are 99% confident taht the interval contains the true value of p.

c- There is a 99% chance that the true value of p will not fall in the interval.

d- There is a 99% chance that the true value of p will fall in the interval.

Explanation / Answer

a)Point Estimate=48/400=0.12

mean=0.12, stddev=sqrt(0.12*0.88/400)=0.01625,n=400

b)np=400*0.12=48>10

n(1-p)=400*0.88=352>10

Hence yes both the requirements are satisfied---------------------->Option A)

c)Z(0.995)=2.576

CI=[0.12-2.576*0.01625,0.12+2.576*0.01625]=[0.078,0.162]

d)We are 99% Confident that the interval contains the true value of p------------Option B)

Navigate

In a trial of 350 patients who received 10-mg doses of a drug daily, 42 reported

In a trial of 400 patients who received 10-mg doses of a drug daily, 48 reported

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.

In a trial of 400 patients who received 10 mg doses of a drug daily, 48 reported

Question

Explanation / Answer

Related Questions

Navigate