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 5751 physicians in Colorado showed that 2981
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.

1% of the confidence intervals created using this method would include the true proportion of Colorado physicians providing at least some charity care.99% 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.99% 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.

Yes; *np* < 5 and *nq* < 5.No; *np*
> 5 and *nq* < 5. No;
*np* < 5 and *nq* > 5.Yes; *np* > 5
and *nq* > 5.

Answer #1

(a)

p = 2981/5751 =0.5183

(b)

q = 1- p = 0.4817

SE =

= 0.01

From Table, critical values of Z = 2.576

99 % Confidence interval:

p Z SE

= 0.5183 (2.576 X 0.0066)

= 0.5183 0.0170

= (0.5013 , 0.5353)

Correct option:

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

(c)

n = 5751

p = 0.5183

q = 0.4817

np = 5751 X 0.5183 = 2980.7433

nq = 5751 X 0.4817 = 2770.2567

So,

correct option:

Yes; np > 5 and nq > 5.

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