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 $10\%$or less Hispanic people? Calculate an appropriate probability to support your answer.

Short answer

1. The obtained values are $\mu =156$and $\sigma =11.6499$.
2. Regardless of the fact that this likelihood is less than $0.05$or$5\%$, 180 Hispanics are unlikely to appear by chance, and hence suspicion should be raised.

Step-by-step solution

Part (a) Step 1: Given information

Given:

$\begin{array}{rcl}p& =& 13\%\\ & =& 0.13n\\ & =& 1200\end{array}$

Formula used:

$$\begin{gathered} \mu =np \\ \sigma =\sqrt{np(1-p)} \end{gathered}$$

Part (a) Step 2: Calculation

The average is

$\begin{array}{rcl}\mu & =& np\\ & =& 1200(0.13)\\ & =& 156\end{array}$

A binomial random variable's standard deviation is

$\begin{array}{rcl}\sigma & =& \sqrt{np(1-p)}\\ & =& \sqrt{1200(0.13)(1-0.13)}\\ & =& 11.6499\end{array}$

Part (b) Step 1: Given information

Given:

$15\%$ of the sample contains Hispanic.

$\begin{array}{rcl}p& =& 13\%\\ & =& 0.13\end{array}$

Formula used:

$P(X=k)=\left({}^{n}C_k\right)\times p^k\times (1-p)n-k$

Part(b) Step 2: Calculation

Because the likelihood of the number of successes in the sample is important, the distribution is binomial.

$P(X=k)=\left({}^{n}C_k\right)\times p^k\times (1-p{)}^{n-k}$

Determine if you should be suspicious or not by calculating the likelihood of receiving anything.

$1200\times 15\%=180$Hispanics or more extreme:

$P(X\ge 180)=\sum _{k=180}^{1200}\left({}^{n}C_{1200}\right)\times 0.{13}^k\times (1-0.13{)}^{1200-k}=0.0235$

Despite the fact that this likelihood is less than $0.05$ or $5\%$, 180 Hispanics are unlikely to appear by chance, and hence suspicion should be raised.