Question

Consider a sampling distribution with p=0.14 and samples of size n each. Using the appropriate formulas, find the mean and the standard deviation of the sampling distribution of the sample proportiona. For a random sample of size n = 4000b. For a random sample of size n = 1000c. For a random sample of size n=250.a. The mean is

Answer 1

Given:

p = 0.15

The mean and standard deviation(S.D) can be calculated using the formula:

(a) A random sample of size n = 4000

mean = 0.14

SD:

`\begin{gathered} =\text{ }\sqrt[]{(0.14(1-0.14))/(4000)} \\ =\text{ }\sqrt[]{(0.1204)/(4000)} \\ \approx\text{ 0.005486} \end{gathered}`

SD = 0.005486

(b) A random sample of size n = 1000

mean = 0.14

SD:

`\begin{gathered} =\text{ }\sqrt[]{(0.14(1-0.14))/(1000)} \\ =\text{ }\sqrt[]{(0.1204)/(1000)} \\ \approx\text{ }0.01097 \end{gathered}`

SD = 0.01097

(C) A random sample of size n = 250

mean = 0.14

SD:

`\begin{gathered} =\text{ }\sqrt[]{(0.14(1-0.14))/(250)} \\ =\sqrt[]{(0.1204)/(250)} \\ \approx\text{ 0.02195} \end{gathered}`

SD = 0.02195