Question

# A random sample of 324 medical doctors showed that 166 had a solo practice. (a) Let...

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

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

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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