Obtain the sampling distribution of p―

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mhalmantus · Accepted Answer

It is given that is the sample proportion of entrepreneurs whose first startup was at 30 years or more is $p = 0.45$ and the sample size $n = 200$
The sampling distribution of the proportion is approximately normal if $n p \Rightarrow 5 a n d n (l — p) \Rightarrow 5$ .
Verify the conditions:
$n p = 200 \times 0.45$
$= 90 \Rightarrow 5$ .
$n (1 — p) = 200 \times (1 — 0.45)$
And
$= 110 \Rightarrow 5$
The conditions are satisfied. Therefore, the sampling distribution of the proportion is normal.
The mean of the $\overset{―}{p}$ is $E (\overset{―}{p}) = p$ and standard deviation of $\overset{―}{p}$ is $σ_{\overset{―}{p}}, = \frac{\sqrt{p (1 - p)}}{n}$
Mean and standard deviation are calculated as below
$E (\overset{―}{p}) = p = 0.45$
$σ_{\overset{―}{p}} = \sqrt{p (1 - p)} = \frac{\sqrt{0.45 \times 0.55}}{200} = 0.0352$
$P y 0.45 \times 0.55$
Thus, the sampling distribution of the proportion J of entrepreneurs whose first startup was at 30 years or more is normal with mean $E (\overset{―}{p}) = 0.45$ and standard deviation $σ_{\overset{―}{p}}, = 0.0352$

Obtain the sampling distribution of overline{p}

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