Here is the full question.
Calculate the mean and the standard deviation of the sampling distribution of possible sample proportions for each combination of sample size (n) and population proportion (p).
a.) n = 64, p = 0.8
b.) n = 256, p = 0.8
Answer:
a.) Mean = 0.8
Standard deviation = 0.05
b.) Mean = 0.8
Standard deviation = 0.025
Step-by-step explanation:
From the given information:
a.)
The population parameter (p) is the mean of the sampling distribution for the sample proportion.
Thus; The Mean = 0.8
The standard deviation of the distribution can be calculated as follows:
Let's first represent the standard deviation with Sd([tex]\hat p[/tex])
Then:
[tex]sd(\hat p) = \sqrt{\dfrac{p*(1-p)}{n} }[/tex]
where;
p = 0.8
n = 64
Then:
[tex]sd(\hat p) = \sqrt{\dfrac{0.8*(1-0.8)}{64} }[/tex]
[tex]sd(\hat p) = \sqrt{\dfrac{0.8*(0.2)}{64} }[/tex]
[tex]sd(\hat p) = \sqrt{\dfrac{0.16}{64} }[/tex]
[tex]sd(\hat p) = \sqrt{0.0025}[/tex]
[tex]\mathbf{sd(\hat p) =0.05}[/tex]
b.)
The population parameter (p) is the mean of the sampling distribution for the sample proportion.
Thus; The Mean = 0.8
The standard deviation of the distribution can be calculated as follows:
Let's first represent the standard deviation with Sd([tex]\hat p[/tex])
Then:
[tex]sd(\hat p) = \sqrt{\dfrac{p*(1-p)}{n} }[/tex]
where;
p = 0.8
n = 256
Then:
[tex]sd(\hat p) = \sqrt{\dfrac{0.8*(1-0.8)}{256} }[/tex]
[tex]sd(\hat p) = \sqrt{\dfrac{0.8*(0.2)}{256} }[/tex]
[tex]sd(\hat p) = \sqrt{\dfrac{0.16}{256} }[/tex]
[tex]sd(\hat p) = \sqrt{6.25 \times 10^{-4}}[/tex]
[tex]\mathbf{sd(\hat p) =0.025}[/tex]
