Respuesta :
Answer:
The sampling distribution of [tex]\hat p[/tex] is N (0.25, 0.0354²).
Step-by-step explanation:
According to the Central limit theorem, if from an unknown population large samples of sizes n > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.
The mean of this sampling distribution of sample proportion is:
[tex]\mu_{\hat p}=p[/tex]
The standard deviation of this sampling distribution of sample proportion is:
[tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}[/tex]
Let p = proportion of adults in the city having credit card debts of more than $2000.
It is provided that the proportion of adults in the city having credit card debts of more than $2000 is, p = 0.25.
A random sample of size n = 150 is selected from the city.
Since n = 150 > 30 the Central limit theorem can be used to approximate the distribution of p by the Normal distribution.
The mean is:
[tex]\mu_{\hat p}=p=0.25[/tex]
The standard deviation is:
[tex]\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.25(1-0.25)}{150}}=0.0354[/tex]
Thus, the sampling distribution of [tex]\hat p[/tex] is N (0.25, 0.0354²).
Following are the calculation to the given value:
Given:
Based on one town's study, the population percentage of adults in the city had credit card bills totaling and over [tex]\$2000\ (p) = 0.25[/tex]
A city-wide sample of adults was recruited [tex](n) = 150[/tex]
Solution:
For step 1:
To get the sampling distribution of probability value [tex]\hat{p}[/tex], we should first determine whether the sample is sufficient using the following two conditions:
[tex]\to np \geq 10\\\\\to n(1-P) \geq 10\\\\\to np = (150 \times 0.25)= 37.5 \geq 10\\\\\to n(1-P) = 150(1 - 0.25) = 112.5\geq 10\\\\[/tex]
The two conditions are fulfilled. We understand from the "Central limit theorem" that the as being [tex]\hat{p}[/tex] is roughly normally distributed with the mean [tex]\mu_{\hat{p}}=P[/tex] and the standard deviation [tex]\sigma_{\hat{p}} = \sqrt{\frac{(p(1-p))}{n}}[/tex] along with [tex]np \geq 10 \ \ and \ \ n(1-P) \geq 10[/tex]
Therefore the 2 conditions which satisfy the proportion of sample [tex]\hat{p}[/tex] that were normally distributed by the mean [tex]\mu_{\hat{p}}=P[/tex] and the standard deviation[tex]\sigma_{\hat{p}} = \sqrt{\frac{(p(1-p))}{n}}[/tex]
For step 2:
Sample proportion mean [tex](\hat{p})\ \ is,\ \ \mu _{\hat{p}}=p=0.25[/tex]
For step 3:
The sample proportion [tex]\hat{p}[/tex] of the standard deviation:
[tex]\to \sigma_{\hat{p}} = \sqrt{\frac{(p(1-p))}{n}}[/tex]
[tex]= \sqrt{\frac{(0.25(1-0.25))}{150}} \\\\= \sqrt{\frac{(0.25-0.0625)}{150}} \\\\= \sqrt{\frac{0.1875}{150}} \\\\= \sqrt{0.00125}\\\\=0.035 \\\\[/tex]
Since the distribution of the sample proportion that is [tex]\hat{p}[/tex] is adults which has to credit card debts of more than [tex]\$2000[/tex] is approximately normal; [tex]\bold{\mu _{\hat{p}}=0.25 ,\ \sigma_{\hat{p}}=0.035}[/tex] .
Learn more:
brainly.com/question/10664750
