A random sample of 324 medical doctors showed that 164 had a

chezmarylou1i 2021-12-13 Answered
A random sample of 324 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
c) Give a brief explanation of the meaning of the interval.
10% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.
90% of all confidence intervals would include the true proportion of physicians with solo practices.
10% of all confidence intervals would include the true proportion of physicians with solo practices.
90% of the confidence intervals created using this method would include the true proportion of physicians with solo practices.
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Expert Answer

Jenny Bolton
Answered 2021-12-14 Author has 32 answers
Step 1
a) Let p represent the proportion of all medical doctors who have a solo practice.
The point estimate for p is,
p^=164324
=0.506
Step 2
b) 90% confidence interval for p:
From the given information, the confidence level is 90%.
Significance level is 10%.
From the standard normal table, the critical value corresponding with 10% significance level is 1.645.
lower limit=p^zcriticalp^(1p^)n
=0.506(1.645×0.506(10.506)324)
=0.5060.046
=0.460
upper limit=p^+zcriticalp^(1p^)n
=0.506+(1.645×0.506(10.506)324)
=0.506+0.046
=0.552
Step 3
c) Interpretation: 90% of all confidence intervals would include the true proportion of physicians with solo practices and 10% of all confidence intervals would not include the true proportion of physicians with solo practices. Correct option: 90% of all confidence intervals would include the true proportion of physicians with solo practices.
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Navreaiw
Answered 2021-12-15 Author has 34 answers

Step 1
a) Sample proportion =0.494
b) Sample size, n=324
Standard error, SE=p×(1p)n
SE=0.494×(10.494}{324}=0.0278
Given Cl level is 90%, hence α=10.9=0.1
α2=0.12=0.05
Zc=Zα2=1.64
Cl=(pz×SE, p+z×SE)
Cl=(0.4941.64×0.0278, 0.494+1.64×0.0278)
Cl=(0.448, 0.540)
c) Margin of Error, ME=zc×SE
ME=1.64×0.0278
ME=0.046

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