Question

Bag check Thousands of travelers pass through the airport in Guadalajara, Mexico, each day. Before leaving the airport, each passenger must pass through the customs inspection area. Customs agents want to be sure that passengers do not bring illegal items into the country. But they do not have time to search every traveler's luggage. Instead, they require each person to press a button. Either a red or a green bulb lights up. If the red light flashes, the passenger will be searched by customs agents. A green light means "Go ahead." Customs agents claim that $30\%$of all travelers will be stopped (red light), because the light has probability of showing red on any push of the button. To test this claim, a concerned citizen watches a random sample of 100 travelers push the button. Only 20 get a red light.

a. Assume that the customs agents' claim is true. Find the probability that the proportion of travelers who get a red light in a random sample of 100 travelers is less than or equal to the result in this sample.

b. Based on your results in part (a), is there convincing evidence that less than $30\%$of all travelers will be stopped? Explain your reasoning.

Short answer

(a)the corresponding probability using table A:

$P(\hat{p}\le 0.20)=P(z<-2.18)=0.0146$

(b)there is convincing evidence that less than $30\%$of all travellers will be stopped.

Step-by-step solution

Part (a) Step 1: Given Information

$$\begin{gathered} p=0.30 \\ x=20 \\ n=100 \end{gathered}$$

Formula used:

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

Part (a) Step 2: Simplification

The mean of the sampling distribution of $\hat{p}$is equal to the population proportion:

${\mu }_{\hat{p}}=p=0.30$

The standard deviation of the sampling distribution of $\hat{p}$is

$$\begin{gathered} {\sigma }_{\hat{p}}=\sqrt{\frac{p(1-p)}{n}} \\ =\sqrt{\frac{0.30(1-0.30)}{100}} \\ =0.0458 \end{gathered}$$

The sampling distribution of $\hat{p}$is about normal if n p and $n(1-p)$are both at least 10 .

$$\begin{gathered} np=100\times 0.30=30 \\ n(1-p)=100\times (1-0.30)=70 \\ \hat{p}=\frac{x}{n}=\frac{20}{100}=0.2 \end{gathered}$$

The z-score is

$$\begin{gathered} z=\frac{x-\mu }{\sigma } \\ =\frac{0.2-0.30}{0.0458} \\ =-2.18 \end{gathered}$$

Determine the corresponding probability using table A:

$P(\hat{p}\le 0.20)=P(z<-2.18)=0.0146$