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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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.

Jenny Bolton · Accepted Answer

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^(1−p^)n  =0.506−(1.645×0.506(1−0.506)324)  =0.506−0.046  =0.460  upper limit=p^+zcriticalp^(1−p^)n  =0.506+(1.645×0.506(1−0.506)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.

Navreaiw · Accepted Answer

Step 1 a) Sample proportion =0.494 b) Sample size, n=324 Standard error, SE=p∩×(1−p∩)n SE=0.494×(1−0.494}{324}=0.0278 Given Cl level is 90%, hence α=1−0.9=0.1 α2=0.12=0.05 Zc=Zα2=1.64 Cl=(p∩−z×SE, p∩+z×SE) Cl=(0.494−1.64×0.0278, 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

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

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