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Alex_Xolod [135]
2 years ago
13

Forty-two percent of primary care doctors think their patients receive unnecessary medical care (reader's digest, december 2011/

january 2012). suppose a sample of 300 primary care doctors were taken. what is the standard error of the sampling distribution of the sample proportion?
Mathematics
1 answer:
igomit [66]2 years ago
5 0

Answer:

The standard error is 0.02849

Step-by-step explanation:

<u>Explanation:</u>-

given data is 42% of primary case doctors think their patients receive un-necessary medical care.

That is The proportion 'p' = 42% = 0.42

Given sample size is n =300

The standard error of the sampling distribution of the sample proportion is

se (p) = \frac{\sqrt{p(1-p)} }{\sqrt{n} }

se (p) = \frac{\sqrt{0.42(1-0.42))} }{\sqrt{300} }

use calculator on simplification , we get

standard error = 0.02849

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