Question

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

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:

$\mu_{\hat p}=p$

The standard deviation of this sampling distribution of sample proportion is:

$\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}$

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: $\hat p=\frac{42}{350}=0.12$

$P(\hat p < 0.12)=P(\frac{\hat p-\mu_{\hat p}}{\sigma_{\hat p}}$

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:

$CI=\hat p\pm z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}$

The margin of error for this interval is:

$MOE= z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}$

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:

$MOE= z_{\alpha/2}\cdot\sqrt{\frac{\hat p(1-\hat p)}{n}}$

      $n=[\frac{z_{\alpha/2}\cdot\sqrt{\hat p(1-\hat p)}}{MOE}]^{2}\\\\=[\frac{1.96\times \sqrt{0.50(1-0.50)}}{0.04}]^{2}\\\\=600.25\\\\\approx 601$

Thus, the required sample size is 601.