A random sample of 332 medical doctors showed that 178 had a solo practice. USE SALT...

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A random sample of 332 medical doctors showed that 178 had a solo practice. USE SALT (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.) 0.536 Enter a number. (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 the margin of error. Report p along with the margin of error. Report p. Report the confidence interval. What is the margin of error based on a 90% confidence interval? (Use 3 decimal places.) A random sample of 332 medical doctors showed that 178 had a solo practice. USE SALT (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.) 0.536 Enter a number. (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 the margin of error. Report p along with the margin of error. Report p. Report the confidence interval. What is the margin of error based on a 90% confidence interval? (Use 3 decimal places.)

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a Find a point estimate for p The point estimate for p is p 178332 0536 b Find a 90 confidence inter... View the full answer