Question

The Centers for Disease Control and Prevention reported a survey of randomly selected Americans age 65 and older, which found that 411 of 1012 men and 535 of 1062 women suffered from some form of arthritis.
(a) Are the assumptions and conditions necessary for inference satisfied? Explain.
(b) Create a 95% confidence interval for the difference in the proportions of senior men and women who have this disease.
(c) Interpret your interval in this context.
(d) Does this confidence interval suggest that arthritis is more likely to afflict women than men? Explain

Answer 1

Answer:

a) We need to satisfy three conditions in order to apply inference:

1) Random: We assume that the data comes from a random sample

2) Normal: We assume that the normal distribution for the estimated proportion can be used

np >10 , n(1-p) >10

2) Independent: We asume independence between the trials and the sample is <10% of the population size

b) $(0.406-0.504) - 1.96 \sqrt{\frac{0.406(1-0.406)}{1012} +\frac{0.504(1-0.504)}{1062}}=-0.141$  

$(0.406-0.504) + 1.96 \sqrt{\frac{0.406(1-0.406)}{1012} +\frac{0.504(1-0.504)}{1062}}=-0.055$  

And the 95% confidence interval would be given (-0.141;-0.055).

c) We are confident at 95% that the difference between the two proportions is between $-0.141 \leq p_A -p_B \leq -0.055$  

d) For this case sine the confidence interval contain just negative numbers we have enough evidence to conclude that the proportion of women suffered from some form of arthritis is significantly higher than the proportion for mean at 5% of significance.

Step-by-step explanation:

Previous concepts

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

$p_A$ represent the real population proportion of men

$\hat p_A =\frac{411}{1012}=0.406$ represent the estimated proportion for men

$n_A=1012$ is the sample size for men

$p_B$ represent the real population proportion for female

$\hat p_B =\frac{535}{1062}=0.504$ represent the estimated proportion for women

$n_B=1062$ is the sample size required for women

$z$ represent the critical value for the margin of error  

The population proportion have the following distribution  

$p \sim N(p,\sqrt{\frac{p(1-p)}{n}})$  

Part a

We need to satisfy three conditions in order to apply inference:

1) Random: We assume that the data comes from a random sample

2) Normal: We assume that the normal distribution for the estimated proportion can be used

np >10 , n(1-p) >10

2) Independent: We asume independence between the trials and the sample is <10% of the population size

Part b

The confidence interval for the difference of two proportions would be given by this formula  

$(\hat p_A -\hat p_B) \pm z_{\alpha/2} \sqrt{\frac{\hat p_A(1-\hat p_A)}{n_A} +\frac{\hat p_B (1-\hat p_B)}{n_B}}$  

For the 95% confidence interval the value of $\alpha=1-0.95=0.05$ and $\alpha/2=0.025$, with that value we can find the quantile required for the interval in the normal standard distribution.  

$z_{\alpha/2}=1.96$  

And replacing into the confidence interval formula we got:  

$(0.406-0.504) - 1.96 \sqrt{\frac{0.406(1-0.406)}{1012} +\frac{0.504(1-0.504)}{1062}}=-0.141$  

$(0.406-0.504) + 1.96 \sqrt{\frac{0.406(1-0.406)}{1012} +\frac{0.504(1-0.504)}{1062}}=-0.055$  

And the 95% confidence interval would be given (-0.141;-0.055).

Part c  

We are confident at 95% that the difference between the two proportions is between $-0.141 \leq p_A -p_B \leq -0.055$  

Part d

For this case sine the confidence interval contain just negative numbers we have enough evidence to conclude that the proportion of women suffered from some form of arthritis is significantly higher than the proportion for mean at 5% of significance.