Chapter 10.1, Problem 22E

(a)

To determine

To explain do these data give convincing evidence that the proportion of all teens with hearing loss has increased at the $\alpha =0.01$ significance level or not.

Answer to Problem 22E

Yes, there is convincing evidence that the proportion of all teens with hearing loss has increased over the years.

Explanation of Solution

Given:

  $\begin{array}{l}{x}_1=450\\ x_2=351\\ n_1=3000\\ n_2=1800\end{array}$

  $\alpha =0.01$

Define Hypotheses:

So, the given claim that: proportion of all teens with hearing loss in $2010$ is higher.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

Where we have,

  $p_1=$ the proportion of teens in $1990$ with hearing loss.

  $p_2=$ the proportion of teens in $2010$ with hearing loss.

Conditions to be satisfied:

There are three conditions to be satisfied:

Random: It is satisfied because the teenagers are independent random samples.

Independent: It is satisfied because the $3000$ teens in $1990$ are less than $10\%$ of all teens in $1990$ and $1800$ teens in $2010$ are less than $10\%$ of all teens in $2010$ .

Normal: It is satisfied because there are $450$ successes and $3000-450=2550$ failures in the first sample and there are $351$ successes and $1800-351=1449$ failures in the second sample, which are all at least ten.

Thus, all the conditions are satisfied.

Calculation:

The sample proportion is the number of successes divided by the sample size. Then, we have,

  $\begin{array}{l}{\widehat{p}}_1=\frac{x_1}{n_1}\\ =\frac{450}{3000}\\ =0.15\\ {\widehat{p}}_2=\frac{x_2}{n_2}\\ =\frac{351}{1800}\\ =0.195\\ {\widehat{p}}_p=\frac{x_1+x_2}{n_1+n_2}\\ =\frac{450+351}{3000+1800}\\ =\frac{800}{4800}\\ =0.1667\end{array}$

Now, we will calculate the value of test statistics as:

  $\begin{array}{l}z=\frac{{\widehat{p}}_1-{\widehat{p}}_2-(p_1-p_2)}{\sqrt{{\widehat{p}}_p(1-{\widehat{p}}_p)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{0.15-0.195-0}{\sqrt{0.1667(1-0.1667)}\sqrt{\frac{1}{3000}+\frac{1}{1800}}}\\ \approx -4.05\end{array}$

The P-value is the probability of obtaining the value of the test statistics or a value more extreme assuming that the null hypothesis is true. Thus, we have,

  $\begin{array}{l}P=P(Z<-4.05)\\ \approx 0\end{array}$

Thus, if the P-value is smaller than the significance level, then we will reject the null hypothesis, thus, we have,

  $P<0.01\Rightarrow \text{Reject }{H}_0$

Thus, we conclude that there is convincing evidence that the proportion of all teens with hearing loss has increased over the years.

(b)

To determine

To interpret the P value from part (a) in the context of this study.

Answer to Problem 22E

There is a very small chance of obtaining similar test results or more extreme when the proportion of all teens with hearing loss is greater in $2010$ compared to $1990$ .

Explanation of Solution

From part (a), we have that,

  $\begin{array}{l}{H}_0:p_1=p_2\\ H_a:p_1<p_2\end{array}$

Where we have,

  $p_1=$ the proportion of teens in $1990$ with hearing loss.

  $p_2=$ the proportion of teens in $2010$ with hearing loss.

And, the value of test statistics is as:

  $\begin{array}{l}z=\frac{{\widehat{p}}_1-{\widehat{p}}_2-(p_1-p_2)}{\sqrt{{\widehat{p}}_p(1-{\widehat{p}}_p)}\sqrt{\frac{1}{n_1}+\frac{1}{n_2}}}\\ =\frac{0.15-0.195-0}{\sqrt{0.1667(1-0.1667)}\sqrt{\frac{1}{3000}+\frac{1}{1800}}}\\ \approx -4.05\end{array}$

The P-value is the probability of obtaining the value of the test statistics or a value more extreme assuming that the null hypothesis is true. Thus, we have,

  $\begin{array}{l}P=P(Z<-4.05)\\ \approx 0\end{array}$

From this we can conclude that there is a very small chance of obtaining similar test results or more extreme when the proportion of all teens with hearing loss is greater in $2010$ compared to $1990$ .