Question

A random sample of 322 medical doctors showed that 164 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 90% confidence interval for *p*. (Use 3 decimal
places.)

lower limit | |

upper limit |

Give a brief explanation of the meaning of the interval.

90% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.90% of the all confidence intervals would include the true proportion of physicians with solo practices. 10% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.10% of the all confidence intervals 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*.Report *p̂* along with the margin of
error. Report the confidence interval.Report
the margin of error.

What is the margin of error based on a 90% confidence interval?
(Use 3 decimal places.)

Answer #1

n = 322

x = 164

A) point estimate is

= x / n = 164 / 322 =0.509

#(b) Find a 90%
confidence interval for *p*. is

- ME < P < + ME

1 - = 1 - 0. 509 = 0 491

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_{/2} = Z_{0.05} =
1.645

Margin of error =M E

ME= Z_{ / 2} * (( * (1 - )) / n)

ME= 1.645 * (((0.509 * 0.491) / 322)

ME= 0.046

A 90 % confidence interval for population proportion p is ,

- ME < P < + ME

0.509 - 0.046 < p < 0.503 + 0.046

0.463< p < 0.555

lower limit=0.463 | |

upper limit=0.555 |

The 90% confidence interval for the population proportion p is : ( 0.463,0.555)

#c)What is the margin of error based on a 90% confidence interval?

Margin of error =M E

ME= Z_{ / 2} * (( * (1 - )) / n)

ME= 1.645 * (((0.509 * 0.491) / 322)

ME= 0.046

#Margin of error is 0.046

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