Respuesta :
Answer:
The correct answer to the following question will be "0.0367".
Step-by-step explanation:
The given values are:
[tex]p1=p2=0.06[/tex]
[tex]q1=q2=1-p1=0.94[/tex]
[tex]n1=n2=400[/tex]
As we know,
[tex]E(p1-p2)=p1-p2=0\\[/tex]
[tex]SE(p1-p2)=\sqrt{\frac{p1q1}{n1}+\frac{p2q2}{n2}}[/tex]
On putting the given values in the above expression, we get
[tex]= \sqrt{p1q1(\frac{1}{400}+\frac{1}{400})}[/tex]
[tex]=0.0168[/tex]
Now, consider
[tex]P(p1-p2>0.03)=P[\frac{(p1-p2)-E(p1-p2)}{SE(p1-p2)}>\frac{0.03-0}{0.0168}][/tex]
[tex]=P(Z>1.7857)[/tex]
[tex]=P(Z>1-79)[/tex]
[tex]=0.036727[/tex]
Therefore, "0.0367" is the right answer.
Using the normal distribution and the central limit theorem, it is found that:
The probability that the difference between the first sample proportion which possess the given characteristic and the second sample proportion which possess the given characteristic being more than .03 is of 0.0375 = 3.75%.
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
- By the Central Limit Theorem, for a proportion p in a sample of size n, the mean is [tex]\mu = p[/tex], while the standard deviation is [tex]s = \sqrt{\frac{p(1 - p)}{n}}[/tex]
- When two normal variables are subtracted, the mean is the subtraction of the means, while the standard deviation is the square root of the sum of the variances.
In this problem, two samples of 400 from a population of proportion 0.06, hence:
[tex]p_1 = p_2 = 0.06[/tex]
[tex]s_1 = s_2 = \sqrt{\frac{0.06(0.94)}{400}} = 0.0119[/tex]
For the distribution of the difference, we have that:
[tex]\mu = p_1 - p_2 = 0.06 - 0.06 = 0[/tex]
[tex]s = \sqrt{s_1^2 + s_2^2} = \sqrt{0.0119^2 + 0.0119^2} = 0.0168[/tex]
The probability of the difference being more than 0.03 is 1 subtracted by the p-value of Z when X = 0.03, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
By the Central Limit Theorem
[tex]Z = \frac{X - \mu}{s}[/tex]
[tex]Z = \frac{0.03 - 0}{0.0168}[/tex]
[tex]Z = 1.78[/tex]
[tex]Z = 1.78[/tex] has a p-value of 0.9625.
1 - 0.9625 = 0.0375.
The probability that the difference between the first sample proportion which possess the given characteristic and the second sample proportion which possess the given characteristic being more than .03 is of 0.0375 = 3.75%.
A similar problem is given at https://brainly.com/question/15352354
