A survey was taken of randomy selected Americans, age 65 and older, which found

Question

A survey was taken of randomy selected Americans, age 65 and older, which found that 418 of 1006 men and 539 of 1057 women suffered from some form of arthrits. a) Are the assumptions and condišons necessary for inference satisfied? Explain. b) Create a 95% oorfidence interval for the dfference in the proportions of serior men and women who have this disease. c) Interpret your interval in this context d) Does this suggest that arthritis is more lely to affict women than men? Explain a) Are the assumptions and condiions necessary for inference satisfied? Explain. 0 A No, more than 10% of the population was sampled B. Yes, it was a random sample, less than 10% of the population was sampled, te goups were independent, and there were more than 10 successes and fal res in each goup C. No, the groups were not independent. OD. No, it was not a random sample Create @95confidence interval fr b) Let P1 be the sample proportion of senior wo en sufferigfrom somo form of artviss, and let p2 be the sample proportion of senior men s Mering from som e form of arthritis. the diferenco in the proportions of senior men and women who have this diseas Pr-P2. The confidence interval is Round to three decimal places as needed) % ad % than the proportion of American men of the c) There 95% oonddenoe, based on these samples, that the proportion of American women, age 68 and older, who suffer from arthritis is between 1 same age who suffer from arthris (Round to one decimal place as needed.) d) Does this suggest that arthritis is more liely to afflict women than men? Select the comect answer below and, if necessary, fil in the answer boxes within your choice O A No, the interval is too close to 0 B. Yes, entire interval lies above 0. O C. No, a conclusion cannot be made based on the confidence interval OD. Yes, there 95% corfidenoe, based on tese samples that about %ofsenior women suffer from artits, while only %ofsonor men suffer from arthna

Explanation / Answer

Part a

Answer: B.

Yes, it was a random sample, less than 10% of the population was sampled, the groups were independent, and there are more than 10 successes and failures in each group.

(Adults are randomly selected from the entire population of Americans, and we know that population size is very large. Selected sample is very small as compared to population and less than 10% of the population was sampled.)

Part b

Here, we have to find the 95% confidence interval for difference between two proportions.

Confidence interval = (P1 – P2) -/+ Z*sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)]

Where, P1 and P2 are sample proportions for first and second groups respectively.

We are given

X1 = 539, N1 = 1057, X2 = 418, N2 = 1006,

P1 = X1/N1 = 539/1057 = 0.509933775

P2 = X2/N2 = 418/1006 = 0.415506958

Confidence level = 95%

Critical Z value = 1.96

Estimate for difference in two proportions = (0.509933775 - 0.415506958) = 0.094426817

sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)] = sqrt[(0.509933775*(1 - 0.50993377) / 1057) + (0.415506958*(1 - 0.415506958) / 1006)]

sqrt[(P1*(1 – P1)/N1) + (P2*(1 – P2)/N2)] = 0.0219

Confidence interval = 0.094426817 -/+ 1.96*0.0219

Confidence interval = 0.094426817 -/+ 0.0428

Lower limit = 0.094426817 - 0.0428 = 0.051626817

Upper limit = 0.094426817 + 0.0428 = 0.137226817

The confidence interval is (0.052, 0.137).

Part c

There is 95% confidence, based on these samples,that the proportion of American women, age 65and older, who suffer from arthivitis is between 5.2% and 13.7% is greater than the proportion of American men same age who suffer from arthiritis.

Part d

Answer: D

Explanation:

For the given scenario, the confidence interval only includes positive values for differences, which indicate that proportion for women is more than men.

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