Question

A recent survey in the N.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?

Answer 1

Answer:

There is enough statistical evidence to support the claim that the survey is not accurate.

Step-by-step explanation:

We have  to perform a test of hypothesis on the proportion.

The claim is that the proportion of families that own stock differs from 48.8%.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.488\\\\H_a:\pi\neq0.488$

The significance level is 0.05.

The sample. of size n=250, has a proportion of p=0.568.

$p=X/n=142/250=0.568$

The standard error of the proportion is

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.488*0.512}{250}}=\sqrt{0.00099}=0.032$

The z-statistic can now be calculated:

$z=\dfrac{p-\pi-0.5/n}{\sigma_p}=\dfrac{0.568-0.488-0.5/250}{0.032}=\dfrac{0.078}{0.032}=2.4375$

The P-value for this two-tailed test is then:

$P-value=2*P(z>2.4375)=0.015$

As the P-value is smaller than the significance level, the effect is significant. The null hypothesis is rejected.

There is enough evidence to support the claim that the survey is not accurate.