Respuesta :
Answer:
a) The sampling distribution for n=540 has a mean sample proportion of p=0.37 and a standard deviation of σs=0.0208.
b) probability = 0.99998
d) probability = 0.99874
e) You gain 0.12% in probability for an increase of 80% in sample size.
The increase in sample size is not justified by the increase in probability, for this margin of error (Δp=0.09).
Step-by-step explanation:
a) We have a known population proportion π=0.37 and we have to describe the sampling distribution when the sample size is n=540.
The mean sample proportion is expected to be the same as the population proportion:
[tex]\bar p = \pi = 0.37[/tex]
The standard deviation of the sampling will be the population standard deviation divided by the square root of the sample size:
[tex]\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.37*0.63}{540}}=\sqrt{0.000431667}=0.0208[/tex]
Then, we can say that the sampling distribution will have a p=0.37 and a standard deviation σs=0.0208.
b) We have to calculate the probability that the sample proportion will be within 0.09 of the population proportion.
We can calculate the z-value as:
[tex]z_1=\dfrac{p_1-\bar p}{\sigma_s}=\dfrac{0.09}{0.0208}=4.3269\\\\\\z_2=\dfrac{p_2-\bar p}{\sigma_s}=\dfrac{-0.09}{0.0208}=-4.3269[/tex]
As the distribution is symmetrical, we can calculate the probabilty that he sample proportion will be within 0.09 of the population proportion as:
[tex]P(|p-\bar p|<0.09)=P(|z|<4.3269)=0.99998[/tex]
probability = 0.99998
d. Now the sample is smaller (n=300), so the standard deviation of the samping distribution:
[tex]\sigma_s=\dfrac{\sigma}{\sqrt{n}}=\sqrt{\dfrac{p(1-p)}{n}}=\sqrt{\dfrac{0.37*0.63}{300}}=\sqrt{0.000777}=0.0279[/tex]
We have to recalculate the z-scores:
[tex]z_1=\dfrac{p_1-\bar p}{\sigma_s}=\dfrac{0.09}{0.0279}=3.2258\\\\\\z_2=\dfrac{p_2-\bar p}{\sigma_s}=\dfrac{-0.09}{0.0279}=-3.2258[/tex]
And the probability is:
[tex]P(|p-\bar p|<0.09)=P(|z|<3.2258)=0.99874[/tex]
probability = 0.99874
e. The increase in sample size is 80%
[tex]\Delta n\%=\dfrac{n_1}{n_2}-1=\dfrac{540}{300}-1=1.8-1=0.8=80\%[/tex]
and the increase in probability is 0.12%
[tex]\Delta P\%=\dfrac{P_1}{P_2}-1=\dfrac{0.99998}{0.99874}-1=1.0012-1=0.0012=0.12\%[/tex]
You gain 0.12% in probability for an increase of 80% in sample size.
The increase in sample size is not justified by the increase in probability, for this margin of error (Δp=0.09).
