Question

A recent study examined hearing loss data for 1981 U.S. teenagers. In this sample, 369 were found to have some level of hearing loss. News of this study spread quickly, with many news articles blaming the prevalence of hearing loss on the higher use of ear buds by teens. At MSNBC (8/17/2010), Carla Johnson summarized the study with the headline: "1 in 5 U.S. teens has hearing loss, study says." To investigate whether this is an appropriate or a misleading headline, you will conduct a test of significance with the following hypotheses: Null: π = 0.20 Alternative: π ≠ 0.20

Answer 1

Answer:

There is no enough evidence to support the claim that the proportion of US teens that have some level of hearing loss differs from 20%.

P-value=0.12

Step-by-step explanation:

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

The claim is that the proportion of US teens that have some level of hearing loss differs from 20%.

Then, the null and alternative hypothesis are:

$H_0: \pi=0.20\\\\H_a:\pi\neq0.20$

The significance level is assumed to be 0.05.

The sample, of size n=1981, has 369 positive cases. Then, the proportion is:

$p=X/n=369/1981=0.186$

The standard error of the proportion is:

$\sigma_p=\sqrt{\dfrac{\pi(1-\pi)}{n}}=\sqrt{\dfrac{0.2*0.8}{1981}}=\sqrt{ 0.000081 }= 0.009$

Now, we can calculate the statistic z:

$z=\dfrac{p-\pi+0.5/n}{\sigma_p}=\dfrac{0.186-0.20+0.5/1981}{0.009}=\dfrac{-0.014}{0.009}=-1.556$

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

$P-value=2*P(z$

The P-value is below the significance level, so the effect is not significant. The null hypothesis failed to be rejected.

There is no enough evidence to support the claim that the proportion of US teens that have some level of hearing loss differs from 20%.