A random sample of 328 medical doctors showed that 162 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 95% confidence interval for p. (Use 3 decimal
places.)
lower limit | |
upper limit |
Give a brief explanation of the meaning of the interval. (Choose one of the below)
5% of the all confidence intervals would include the true proportion of physicians with solo practices.
5% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.
95% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.
95% 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?
(Choose one of the below)
Report p̂ along with the margin of error.
Report p̂.
Report the margin of error.
Report the confidence interval.
What is the margin of error based on a 95% confidence interval?
(Use 3 decimal places.) ______
Solution :
Given that,
n = 328
x = 162
a)
Point estimate = sample proportion = = x / n = 328 / 162 = 0.494
1 - = 1 - 0.494 = 0.506
b)
At 95% confidence level the z is ,
= 1 - 95% = 1 - 0.95 = 0.05
/ 2 = 0.05 / 2 = 0.025
Z/2 = Z 0.025 = 1.96
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.96 * (((0.494*0.506) / 328)
= 0.054
A 95% confidence interval for population proportion p is ,
- E < p < + E
0.494 - 0.054 < p < 0.494 + 0.054
0.440 < p < 0.548
The 95% confidence interval for the population proportion p is : (0.440 , 0.548 )
Lower limit = 0.440
Upper limit = 0.548
95% of the all confidence intervals would include the true
proportion of physicians with solo practices.
c)
Report p̂.
Margin of error = E = Z / 2 * (( * (1 - )) / n)
= 1.96 * (((0.494*0.506) / 328)
= 0.054
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