A random sample of 336 medical doctors showed that 160 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 What is the margin of error based on a 95% confidence interval? (Use 3 decimal places.)

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Asked Mar 22, 2019
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A random sample of 336 medical doctors showed that 160 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  

What is the margin of error based on a 95% confidence interval? (Use 3 decimal places.)
 

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Step 1

(a) Obtain a point estimate for p

Here, a randm sample of 336 (n) medical doctors showed that 160 (x) had a solo practice.

The point estimate is obtained as 0.476 from the calculation given below:

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Step 2

(b) Obtain the critical value

From, the standard normal table, the critical value at 5% level is ±1.96.

Step 3

Find a 95% confidence interval for p.

The 95% confidence interval is (0.423, 0.529), the lower limit is 0.423 an...

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Hypothesis Testing

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