Question

For this problem, carry at least four digits after the decimal in your calculations. Answers may vary slightly due to rounding.

A random sample of 5967 physicians in Colorado showed that 3224 provided at least some charity care (i.e., treated poor people at no cost).

(a) Let p represent the proportion of all Colorado physicians who provide some charity care. Find a point estimate for p. (Round your answer to four decimal places.)


(b) Find a 99% confidence interval for p. (Round your answers to three decimal places.)

lower limit
upper limit

Give a brief explanation of the meaning of your answer in the context of this problem.

99% of the confidence intervals created using this method would include the true proportion of Colorado physicians providing at least some charity care.

99% of all confidence intervals would include the true proportion of Colorado physicians providing at least some charity care.    

1% of the confidence intervals created using this method would include the true proportion of Colorado physicians providing at least some charity care.

1% of all confidence intervals would include the true proportion of Colorado physicians providing at least some charity care.

(c) Is the normal approximation to the binomial justified in this problem? Explain.

No; np > 5 and nq < 5.

No; np < 5 and nq > 5.    

Yes; np < 5 and nq < 5.

Yes; np > 5 and nq > 5.

Answer 1

(a) Let p represent the proportion of all Colorado physicians who provide some charity care. Find a point estimate for p.

We are given X = 3224, n = 5967

Point estimate for p = P = X/n = 3224/5967 = 0.540305

Point estimate for p = 0.5403

(b) Find a 99% confidence interval for p.

Confidence interval for Population Proportion

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Where, P is the sample proportion, Z is critical value, and n is sample size.

We have

n = 5967

P = 0.540305011

Confidence level = 99%

Critical Z value = 2.5758

(by using z-table)

Confidence Interval = P ± Z* sqrt(P*(1 – P)/n)

Confidence Interval = 0.540305011 ± 2.5758* sqrt(0.540305011*(1 – 0.540305011)/ 5967)

Confidence Interval = 0.540305011 ± 2.5758*0.0065

Confidence Interval = 0.540305011 ± 0.0166

Lower limit = 0.540305011 - 0.0166 = 0.524           

Upper limit = 0.540305011 + 0.0166 = 0.557

Lower limit = 0.524

Upper limit = 0.557

Give a brief explanation of the meaning of your answer in the context of this problem.

99% of the confidence intervals created using this method would include the true proportion of Colorado physicians providing at least some charity care.

Explanation: For this confidence interval, we are 99% confident that the true proportion of Colorado physicians providing at least some charity care will lies between given limits. This means, about 99% of the confidence intervals of this method will include the true proportion of Colorado physicians providing at least some charity care.

(c) Is the normal approximation to the binomial justified in this problem? Explain.

We have

n = 5967

p = 0.5403

q = 1 – p = 1 – 0.5403 = 0.4597

np = 3223.97

nq = 2743.03

np and nq > 5

So, we can use normal approximation to the binomial because np and nq are greater than 5.

Yes; np > 5 and nq > 5.