Respuesta :
Answer:
(1) 0.058
(2) 601
Step-by-step explanation:
(1)
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]
As the sample size is large, i.e. n = 350 > 30, the Central limit theorem can be used to approximate the sampling distribution of sample proportion of adults in a certain country work from home.
Compute the probability that fewer than 42 out of a random sample of 350 adults will work from home:
Sample proportion: [tex]\hat p=\frac{42}{350}=0.12[/tex]
[tex]P(\hat p < 0.12)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}<\frac{0.12-0.15}{\sqrt{\frac{0.15(1-0.15)}{350}}})\\\\=P(Z<-1.57)\\\\=0.05821\\\\\approx 0.058[/tex]
Thus, the probability that fewer than 42 out of a random sample of 350 adults will work from home is 0.058.
(2)
The (1 - α)% confidence interval for population proportion is:
[tex]CI=\hat p\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
The margin of error for this interval is:
[tex]MOE= z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
Given:
MOE = 0.04
Confidence level = 95%
Assume that the sample proportion is 50%.
The critical z-value for 95% confidence level is 1.96.
Compute the required sample size as follows:
[tex]MOE= z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}[/tex]
[tex]n=[\frac{z_{\alpha/2}\cdot\sqrt{\hat p(1-\hat p)}}{MOE}]^{2}\\\\=[\frac{1.96\times \sqrt{0.50(1-0.50)}}{0.04}]^{2}\\\\=600.25\\\\\approx 601[/tex]
Thus, the required sample size is 601.
1. Using the normal approximation to the binomial, it is found that there is a 0.05 = 5% probability that fewer than 42 out of a random sample of 350 adults will work from home.
2. Solving the equation for the margin of error of a confidence interval of proportions, it is found that we must survey 601 randomly selected teenagers.
Question 1:
Normal Probability Distribution
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.
- The binomial distribution is the probability of x successes on n trials, with p probability of a success on each trial. It can be approximated to the normal distribution with [tex]\mu = np, \sigma = \sqrt{np(1-p)}[/tex].
In this problem:
- 15% work from home, thus [tex]p = 0.15[/tex].
- Sample of 350 adults, thus [tex]n = 350[/tex]
Then, for the approximation:
[tex]\mu = np = 350(0.15) = 52.5[/tex]
[tex]\sigma = \sqrt{np(1-p)} = \sqrt{350(0.15)(0.85)} = 6.68[/tex]
Using continuity correction, the probability is [tex]P(X < 42 - 0.5) = P(X < 41.5)[/tex], which is the p-value of Z when X = 41.5. So:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{41.5 - 52.5}{6.68}[/tex]
[tex]Z = -1.65[/tex]
[tex]Z = -1.65[/tex] has a p-value of 0.05.
0.05 = 5% probability that fewer than 42 out of a random sample of 350 adults will work from home.
Question 2:
In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
- z is the z-score that has a p-value of [tex]\frac{1+\alpha}{2}[/tex].
The margin of error is:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In this problem:
- Within 4%, thus [tex]M = 0.04[/tex].
- We do not have an estimate for the true proportion, thus [tex]\pi = 0.5[/tex].
- 95% confidence level, thus z has a p-value of [tex]\frac{1 + 0.95}{2} = 0.975[/tex], thus z = 1.96.
We have to solve for n, then:
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.04 = 1.96\sqrt{\frac{0.5(0.5)}{n}}[/tex]
[tex]0.04\sqrt{n} = 1.96(0.5)[/tex]
[tex]\sqrt{n} = \frac{1.96(0.5)}{0.04}[/tex]
[tex](\sqrt{n})^2 = (\frac{1.96(0.5)}{0.04})^2[/tex]
[tex]n = 600.25[/tex]
Rounding up:
We must survey 601 randomly selected teenagers.
A similar problem is given at https://brainly.com/question/24261244
