Chapter PV, Problem 26RE

(a)

To determine

To write an appropriate hypotheses.

Answer to Problem 26RE

  $\begin{array}{l}{H}_0:p=17.7\%=0.177\\ H_a:p>0.177\end{array}$

Explanation of Solution

It is given in the question that in $1999$ the centres for disease control and prevention estimated that about $34.8\%$ of high school students smoke cigarettes. Thus, it is also given,

  $\begin{array}{l}\widehat{p}=19.5\%=0.195\\ n=5080\\ \alpha =0.05\end{array}$

Thus, the given claim is that the proportion is not reduced to $17.7\%$ . The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportion is equal to the value mentioned in the claim. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis. So, we have,

  $\begin{array}{l}{H}_0:p=17.7\%=0.177\\ H_a:p>0.177\end{array}$

(b)

To determine

To verify that the appropriate assumptions are satisfied.

Answer to Problem 26RE

All the conditions are satisfied.

Explanation of Solution

It is given in the question that in $1999$ the centres for disease control and prevention estimated that about $34.8\%$ of high school students smoke cigarettes. Thus, it is also given,

  $\begin{array}{l}\widehat{p}=19.5\%=0.195\\ n=5080\\ \alpha =0.05\end{array}$

  $\begin{array}{l}{H}_0:p=17.7\%=0.177\\ H_a:p>0.177\end{array}$

Thus, let us check the conditions and assumptions necessary for inference as:

Random condition: It is satisfied as the exercise prompt states that the sample was randomly selected from the population.

Independent condition: It is satisfied as we can assume that the sample is independent.

  $10\%$ condition: It is satisfied as $5080$ high school students are less than $10\%$ of all high school students.

Success/failure condition: It is satisfied because both are greater than ten as,

  $\begin{array}{l}np=5080(0.177)=899.16>10\\ n(1-p)=5080(1-0.177)=4180.84>10\end{array}$

Thus, all the conditions are satisfied.

(c)

To determine

To find the P-value of this test.

Answer to Problem 26RE

  $P=0.0004$ .

Explanation of Solution

It is given in the question that in $1999$ the centres for disease control and prevention estimated that about $34.8\%$ of high school students smoke cigarettes. Thus, it is also given,

  $\begin{array}{l}\widehat{p}=19.5\%=0.195\\ n=5080\\ \alpha =0.05\end{array}$

  $\begin{array}{l}{H}_0:p=17.7\%=0.177\\ H_a:p>0.177\end{array}$

Thus, to find out the value of test statistics we have,

  $\begin{array}{l}z=\frac{\widehat{p}-p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}}\\ =\frac{0.195-0.177}{\sqrt{\frac{0.177(1-0.177)}{5080}}}\\ =3.36\end{array}$

Thus, the P-value is the probability of obtaining the value of the test statistics or a value more extreme when the null hypothesis is true. So, we have using the normal probability table in the appendix that:

  $\begin{array}{l}P=P(Z<3.36)\\ =1-P(Z<3.36)\\ =1-0.9996\\ =0.0004\end{array}$

(d)

To determine

To explain what the P-value means in this context.

Explanation of Solution

It is given in the question that in $1999$ the centres for disease control and prevention estimated that about $34.8\%$ of high school students smoke cigarettes. Thus, it is also given,

  $\begin{array}{l}\widehat{p}=19.5\%=0.195\\ n=5080\\ \alpha =0.05\end{array}$

  $\begin{array}{l}{H}_0:p=17.7\%=0.177\\ H_a:p>0.177\end{array}$

Thus, the P-value is the probability of obtaining the value of the test statistics or a value more extreme when the null hypothesis is true. So, we have using the normal probability table in the appendix that:

  $\begin{array}{l}P=P(Z<3.36)\\ =1-P(Z<3.36)\\ =1-0.9996\\ =0.0004\end{array}$

Thus, there is a $0.0004$ chance of obtaining a sample of $5080$ high school students of which $19.5\%$ or more are current smokers when the proportion of current smokers is actually $17.7\%$ .

(e)

To determine

To state an appropriate conclusion.

Explanation of Solution

It is given in the question that in $1999$ the centres for disease control and prevention estimated that about $34.8\%$ of high school students smoke cigarettes. Thus, it is also given,

  $\begin{array}{l}\widehat{p}=19.5\%=0.195\\ n=5080\\ \alpha =0.05\end{array}$

  $\begin{array}{l}{H}_0:p=17.7\%=0.177\\ H_a:p>0.177\end{array}$

Thus, the P-value is the probability of obtaining the value of the test statistics or a value more extreme when the null hypothesis is true. So, we have using the normal probability table in the appendix that:

  $\begin{array}{l}P=P(Z<3.36)\\ =1-P(Z<3.36)\\ =1-0.9996\\ =0.0004\end{array}$

We know that if the P-value is smaller than the significance level than reject the null hypothesis, then.

  $P<0.05\Rightarrow \text{ Reject }{H}_0$

Thus there is sufficient evidence to support the claim that progress toward the goal of $16\%\text{ by 2010}$ is off track.

(f)

To determine

To find out which kind of error did you commit if your conclusion may be incorrect.

Answer to Problem 26RE

It is Type II error.

Explanation of Solution

It is given in the question that in $1999$ the centres for disease control and prevention estimated that about $34.8\%$ of high school students smoke cigarettes. Thus, it is also given,

  $\begin{array}{l}\widehat{p}=19.5\%=0.195\\ n=5080\\ \alpha =0.05\end{array}$

  $\begin{array}{l}{H}_0:p=17.7\%=0.177\\ H_a:p>0.177\end{array}$

As we know that Type I error is when we reject the null hypothesis if the null hypothesis is true. And Type II error is when we fail to reject the null hypothesis if the null hypothesis is false. So, in this case, since we failed to reject the null hypothesis it is possible that we make a Type II error.