Problem 2: The Centers for Disease Control and Prevention reported a survey of randomly selected Americans age 65 and older, which found that 535 of 106 women and 411 of 1012 men suffered from some form of arthritis. a) Find the sample proportion of arthritis for both genders. b) Do we have some evidence to suggest that there is one gender that suffers some form of arthritis more than the other gender? Explain your answer.

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Can I get some help on problem 2 A-C

Problem 2: The Centers for Disease Control and Prevention reported a survey of randomly
selected Americans age 65 and older, which found that 535 of 106 women and 411 of 1012 men
suffered from some form of arthritis.
a) Find the sample proportion of arthritis for both genders.
b) Do we have some evidence to suggest that there is one gender that suffers some form of
arthritis more than the other gender? Explain your answer.
c) Construct a 95% confidence interval for the true difference in proportions of arthritis between
these two genders. Use Confidence Interval for the Difference in Proportions Calculator -
Statology to do the calculations.
Transcribed Image Text:Problem 2: The Centers for Disease Control and Prevention reported a survey of randomly selected Americans age 65 and older, which found that 535 of 106 women and 411 of 1012 men suffered from some form of arthritis. a) Find the sample proportion of arthritis for both genders. b) Do we have some evidence to suggest that there is one gender that suffers some form of arthritis more than the other gender? Explain your answer. c) Construct a 95% confidence interval for the true difference in proportions of arthritis between these two genders. Use Confidence Interval for the Difference in Proportions Calculator - Statology to do the calculations.
Expert Solution
Part a)

For sample 1 (Women), we have that the sample size is , the number of favorable cases are , so then the sample proportion is \hat p_1 = \frac{X_1}{N_1} = \frac{ 535}{ 1062} = 0.5038p^1 = 53510620.5038.

For sample 2 (Men), we have that the sample size is , the number of favorable cases are , so then the sample proportion is \hat p_1 = \frac{X_1}{N_1} = \frac{ 535}{ 1062} = 0.5038p^2 = 41110120.4061.

 

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