Question

A random sample of 324 medical doctors showed that 166 had a solo practice.

(a) Let *p* represent the proportion of all medical
doctors who have a solo practice. Find a point estimate for
*p*. (Use 3 decimal places.)

(b) Find a 99% confidence interval for *p*. (Use 3 decimal
places.)

lower limit | |

upper limit |

Give a brief explanation of the meaning of the interval.

1% of the all confidence intervals would include the true proportion of physicians with solo practices.99% of the all confidence intervals would include the true proportion of physicians with solo practices. 99% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.1% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.

(c) As a news writer, how would you report the survey results
regarding the percentage of medical doctors in solo practice?

Report *p̂* along with the margin of error.Report the
confidence interval. Report the margin of
error.Report *p̂*.

What is the margin of error based on a 99% confidence interval?
(Use 3 decimal places.)

Answer #1

a)

point estimate of proportion,

pcap = 166/324 = 0.512

b)

The z value at 99% confidence interval is,

alpha = 1 - 0.99 = 0.01

alpha/2 = 0.01/2 = 0.005

Zalpha/2 = Z0.005 = 2.58

Margin of error = E =z *sqrt(p *(1-p)/n)

= 2.58 *sqrt(0.512 *(1-0.512)/324)

= 0.072

CI = p +/- E

= 0.512 +/-0.072

= (0.440 , 0.584)

lower limit = 0.440

upper limit = 0.584

99% of the confidence intervals created using this method would
include the true proportion of physicians with solo practices

c)

Pcap along with margin of error as:

pcap +/- E

0.512 +/.- 0.072

the confidence interval are as:

(0.440 , 0.584)

Margin of error = 0.072

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