Question

According to the Census Bureau, $13\%$of American adults (aged $18$and over) are Hispanic. An opinion poll plans to contact an SRS of 1200 adults.

(a) What is the mean number of Hispanics in such samples? What is the standard deviation?

(b) Should we be suspicious if the sample selected for the opinion poll contains 15% Hispanic people? Compute an appropriate probability to support your answer.

Short answer

a). Mean, $\mu =156$and standard deviation, $\sigma \approx 11.6499$.

b). An appropriate probability is$0.0235$.

Step-by-step solution

Part (a) Step 1: Given Information

$p=13\%=0.13$

$n=1200$

Part (a) Step 2: Explanation 

The mean of a binomial random variable is the product of the sample size $n$and the probability $p$:

localid="1650079570201" $$\begin{gathered} \mu =np \\ =1200(0.13) \\ =156 \end{gathered}$$

The square root of the product of the sample size $n$and the standard deviation of a binomial random variable, the probability $p$and the probability $1-p$:

localid="1650079582715" $$\begin{gathered} \sigma =\sqrt{np(1-p)} \\ =\sqrt{1200(0.13)(1-0.13)} \\ \approx 11.6499 \end{gathered}$$

Part (b) Step 1: Given Information

Given: the sample contains Hispanic.

$p=13\%=0.13$

Part (b) Step 2: Explanation

We want to know the probability of the number of successes in the sample. Therefore, the distribution is binomial is:

$P(X=k)=\left(\begin{array}{l}n\\ k\end{array}\right)·p^k·(1-p{)}^{n-k}$

We should determine the probability of obtaining $1200\times 15\%==180$Hispanics or more extreme:

localid="1650079631185" $$\begin{gathered} P(X\ge 180)=\sum _{k=180}^{1200}\left(\begin{array}{c}n\\ 1200\end{array}\right)·0.{13}^k·(1-0.13{)}^{1200-k} \\ =0.0235 \end{gathered}$$

Since this probability is less than $0.05$or $5\%$, it is unlikely to obtain $180$Hispanics by chance and thus we should be suspicious.