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 5400 physicians in Colorado showed that 3305
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 all confidence intervals 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 the confidence intervals created using this method 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

n= 5400, x= 3305

a)

c= 99%

**Point estimate =
0.6120**

b)

formula for conffidence inteval is

Where Zc is the z critical value for c=99%

Zc= 2.58

0.595 < P < 0.629

(0.595 , 0.629)

**lower
limit= 0.595
upper limit= 0.629**

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*p = 5400* 0.6120 > 5

n*q = 5400* ( 1 - 0.6120 ) > 5

**Yes; np > 5 and nq >
5**

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