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

A group conducted a randomized experiment to see if hormone therapy was helpful

ID: 3157405 • Letter: A

Question

A group conducted a randomized experiment to see if hormone therapy was helpful for post-menopausal women. The women were randomly assigned to receive a hormone or a placebo. After 5 years. 106 of the 8524 on the hormone therapy developed cancer and 95 of the 8290 in the placebo group developed cancer. Is this a significant difference? a. State the assumptions and the hypotheses. What assumptions about the two samples are necessary for the significance test? A. The samples are independent, random, and selected from a population that is approximately normal. B. The samples are independent, random, and a point estimate of the population proportion is known. C. The response variable is qualitative with two outcomes, and the samples are independent, random, and large enough. Let p represent the probability that someone gets cancer. Let p_1 represent the probability that a person who received hormone therapy got cancer, and let P_2 represent the probability that a person who received a placebo got cancer. Write the claim as a hypothesis test. Choose the correct null and alternative hypotheses below. A. H_0: p_1 = p_2 versus H_a: p_1 notequalto p_2 B. H_0: p_1 p_2 C. H_0: p_1 = 0 versus H_a: p_1 notequalto 0 D. H_0: p_1 > p_2 versus H_a: p_1

Explanation / Answer

Solution:

the samples are independent,random and selected from population that is approximately normal

option(B)

solution2:Option C

let   

be sample proportions.

p1^ = 104/8563 =0.0121

p2^ =84/8075=0.0104

Null Hypothesis:

H0 : p1=p2

Alternate Hypothesis:

Ha:p1 p2

level of significance=0.1

test statistic:

z=(0.0121-0.0104)-0/sqrt({0.0121(1-.0121)/8563}+[0.0104(1-0.0104)/8075]}

= 0.0017/0.001634

z calcualted =1.0403

z tabulated at 10% level of significance for two tail test is (±)1.65

since

-1.65< 1.0403<1.65

since calculated z falls in the range of tabulated value

The P-Value is 0.298201.

as 0.298201>0.1

The result is not significant at p < 0.10.

Accept null hypothesis.

As there is not sufficient evidence to support the claim taht the population proportions are different.

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