Question

A study by Allstate Insurance Co. finds that 82% of teenagers have used cell phones while driving (The Wal/Street Journal, May 5, 2010, Suppose a random sample of 100 teen drivers is taken. [You may find it useful to reference the z table.] a-1. Calculate the standard error for the sampling distribution of the sample proportion. (Do not round intermediate calculations. Round your answer to 4 decimal places.) Standard error a-2. Is the sampling distribution of the sample proportion approximately normal? O Yes No b. What is the probability that the sample proportion is less than 0.80? (Round "" value to 2 decimal places, and final answer to 4 decimal places.) Probability

Answer 1

Solution:

a-1: Calculate the standard error for the sampling distribution of the sample proportion.

Answer: The standard error is:

$\dpi{100} \small \boldsymbol{\sqrt{\frac{p(1-p)}{n}}=\sqrt{\frac{0.82(1-0.82)}{100}}=0.0384}$

a-2: Is the sampling distribution of the sample proportion approximately normal?

Answer: Yes

b. What is the probability that the sample proportion is less than 0.80?

Answer: The probability that the sample proportion is less than 0.80 is:

$\dpi{100} \small P\left ( \hat{p}<0.80 \right )=P\left ( \frac{\hat{p}-p}{SE}<\frac{0.80-0.82}{0.0384} \right )$

  $\dpi{100} \small =P\left (z<-0.52 \right )$

Now using the standard normal table, we have:

$\dpi{100} \small P(\hat{p}<0.80)=P\left (z<-0.52 \right )=0.3015$

Therefore, the probability that the sample proportion is less than 0.80 is 0.3015   

c. What is the probability that the sample proportion is within $\dpi{100} \small \pm0.02$ of the population proportion?

Answer:

$\dpi{100} \small P\left (0.80 <\hat{p}<0.84 \right )=P\left ( \frac{0.80-0.82}{0.0384}<\frac{\hat{p}-p}{SE}<\frac{0.84-0.82}{0.0384} \right )$

$\dpi{100} \small =P\left (-0.52<\frac{\hat{p}-p}{SE}<0.52 \right )$

$\dpi{100} \small =P(z<0.52)-P(z<-0.52)$

Now using the standard normal table we have:

$\dpi{100} \small \boldsymbol{P(0.80<\hat{p}<0.84)=P(z<0.52)-P(z<-0.52)=0.6985-0.3015=0.3970}$

Therefore, the probability that the sample proportion is within $\dpi{100} \small \pm0.02$ of the population proportion is 0.3970