Question

A recent survey in theN.Y. Times Almanac indicated that $48.8\%$of families own stock. A broker wanted to determine if this survey could be valid. He surveyed a random sample of $250$ families and found that$142$owned some type of stock. At the $0.05$ significance level, can the survey be considered to be accurate?

Short answer

There is insufficient evidence to conclude that$48.8\%$of families own stock.

Step-by-step solution

Given information

$x=142,n=250,p_0=0.488,a=0.05$

Explanation

State the hypothesis:

The null hypothesis states that $48.8\%$of families own stock. In symbols:

$H_0:p=0.488$

The alternative hypothesis states that not $48.8\%$of families own stock. In symbols:

$H_a:p\ne 0.488$

The sample proportion is:

$$\begin{gathered} \stackrel{⏜}{p}=\frac{x}{n} \\ =\frac{142}{250} \\ =0.568 \end{gathered}$$

The test statistic is:

$$\begin{gathered} z=\frac{\stackrel{⏜}{p}-p_0}{\sqrt{{\displaystyle \frac{p_0\left(1-p_0\right)}{n}}}} \\ =\frac{0.568-0.488}{\sqrt{{\displaystyle \frac{0.488\left(1-0.488\right)}{250}}}} \\ =\frac{0.080}{0.0316} \\ =2.53 \end{gathered}$$

The p-value is:

$$\begin{gathered} p-value=P\left(z<-2.53\right)+P\left(z>2.53\right) \\ =0.0057+0.0057 \\ =0.0114 \end{gathered}$$

Since the null hypothesis is rejected, there is insufficient evidence to conclude that $48.8\%$of families own stock.