1. If 15% of adults in a certain country work from home, what is the probability that fewer than 42 out of a random sample of 350 adults will work from home? (Round your answer to 3 decimal places)
2. Suppose we want to estimate the proportion of teenagers (aged 13-18) who are lactose intolerant. If we want to estimate this proportion to within 4% at the 95% confidence level, how many randomly selected teenagers must we survey?

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]

In this problem:

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]

The margin of error is:

[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In this problem:

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.

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