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A random sample of 311 medical doctors showed that 135 had a solo practice. As a news writer, how would you report the survey results regarding the percentage of medical doctors in solo practice? What is the margin of error based on a 90% confidence interval?a. A recent study shows that about 57% of medical doctors have a solo practice with a margin of error of 4.6 percentage points.b. A recent study shows that about 43% of medical doctors have a solo practice with a margin of error of 2.3 percentage points.c. A recent study shows that about 57% of medical doctors have a solo practice with a margin of error of 2.3 percentage points.d. A recent study shows that about 57% of medical doctors have a solo practice with a margin of error of 9.2 percentage points.e. A recent study shows that about 43% of medical doctors have a solo practice with a margin of error of 4.6 percentage points.

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Answer:

A recent study shows that about 43% of medical doctors have a solo practice with a margin of error of 4.6 percentage points.

Explanation:

Given that a random sample of 311 medical doctors showed that 135 had a solo practice.

Sample proportion = 0.43

Std error of sample proportion =
\sqrt{(0.43*0.57)/(311) } \\=0.0281

Margin of error 90%
= 1.645 *0.0281\\=0.046

Margin of error in percent = 4.6%

Hence correction option is:

A recent study shows that about 43% of medical doctors have a solo practice with a margin of error of 4.6 percentage points.

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