A random sample of 332 medical doctors showed that 176 had a solo practice. (a) Let p...

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A random sample of 332 medical doctors showed that 176 had asolo practice.

(a) Let p represent the proportion of all medicaldoctors who have a solo practice. Find a point estimate forp. (Use 3 decimal places.)

(b) Find a 98% confidence interval for p. (Use 3 decimalplaces.)

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upper limit

What is the margin of error based on a 98% confidence interval?(Use 3 decimal places.)

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Solution Given thatn 332x 176Point estimate sample proportion x n 17633205301    See Answer
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