Chapter PV, Problem 5RE

(a)

To determine

To find out what are the null and alternative hypotheses.

Answer to Problem 5RE

  $\begin{array}{l}{H}_0:p=0.40\\ H_a:p<0.40\end{array}$

Explanation of Solution

It is given in the question that nationwide, it is estimated that $40\%$ of service stations have gas tanks that leak to some extent. Thus, the null hypothesis will be the proportion value is equal to $40\%$ and the alternative hypothesis will be the opposite of the alternative hypothesis that it will be less than $40\%$ . So, we have,

  $\begin{array}{l}{H}_0:p=0.40\\ H_a:p<0.40\end{array}$

(b)

To determine

To check the assumptions necessary for inference.

Answer to Problem 5RE

All are met.

Explanation of Solution

It is given in the question that nationwide, it is estimated that $40\%$ of service stations have gas tanks that leak to some extent. So, we have,

  $\begin{array}{l}{H}_0:p=0.40\\ H_a:p<0.40\end{array}$

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

Random condition: It is satisfied as the sample was randomly selected from the population.

Independent condition: It is satisfied as the sample is independent of each other.

  $10\%$ condition: It is satisfied as the sample is less than $10\%$ of all service stations.

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

  $\begin{array}{l}np=0.4(27)=10.8>10\\ n(1-p)=0.6(27)=16.2>10\end{array}$

Thus, all the conditions are satisfied.

(c)

To determine

To test the null hypothesis.

Explanation of Solution

It is given in the question that nationwide, it is estimated that $40\%$ of service stations have gas tanks that leak to some extent. So, we have,

  $\begin{array}{l}{H}_0:p=0.40\\ H_a:p<0.40\end{array}$

Thus, to calculate an appropriate hypothesis we will use $\text{TI}-89$ calculator. Thus, for this first we have to select $5:1-\text{PropZInt}$ from the STAT TESTS menu. Then, enter the observed counts and sample sizes for the sample. Then we will indicate what kind of test we want: one-tail lower tail, two-tail, or one-tail upper tail. Then, we will specify whether to calculate the result or draw the result. Hence we have,

  $\begin{array}{l}x=10.8\\ n=27\\ c=0.95=95\%\\ H_0:p=0.40\\ H_a:p<0.40\end{array}$

So, by putting the values above by using the calculator we will get the result as,

The value of z-statistics and the P-value will be then,

  $\begin{array}{l}z=-1.49\\ P=0.0677\end{array}$

And as we know if the P-value is smaller than the significance level, then reject the null hypothesis, so, we have,

  $P>0.05\Rightarrow \text{ Fail to Reject }{H}_0$

(d)

To determine

To explain what do you conclude.

Explanation of Solution

It is given in the question that nationwide, it is estimated that $40\%$ of service stations have gas tanks that leak to some extent. So, we have,

  $\begin{array}{l}{H}_0:p=0.40\\ H_a:p<0.40\end{array}$

The value of z-statistics and the P-value will be then,

  $\begin{array}{l}z=-1.49\\ P=0.0677\end{array}$

And as we know if the P-value is smaller than the significance level, then reject the null hypothesis, so, we have,

  $P>0.05\Rightarrow \text{ Fail to Reject }{H}_0$

Thus, we conclude that there is no evidence that the proportion of leaking gas tanks is less than $40\%$ .

(e)

To determine

To explain have you made an error and what kind.

Answer to Problem 5RE

It is type II error.

Explanation of Solution

It is given in the question that nationwide, it is estimated that $40\%$ of service stations have gas tanks that leak to some extent. So, we have,

  $\begin{array}{l}{H}_0:p=0.40\\ H_a:p<0.40\end{array}$

The value of z-statistics and the P-value will be then,

  $\begin{array}{l}z=-1.49\\ P=0.0677\end{array}$

And as we know if the P-value is smaller than the significance level, then reject the null hypothesis, so, we have,

  $P>0.05\Rightarrow \text{ Fail to Reject }{H}_0$

Thus, we conclude that there is no evidence that the proportion of leaking gas tanks is less than $40\%$ . Thus, yes, it has made an error and it is Type II error.

(f)

To determine

To explain what two things could you do to decrease the probability of making this kind of error.

Answer to Problem 5RE

We will increase $\alpha$ and increase the sample size.

Explanation of Solution

It is given in the question that nationwide, it is estimated that $40\%$ of service stations have gas tanks that leak to some extent. So, we have,

  $\begin{array}{l}{H}_0:p=0.40\\ H_a:p<0.40\end{array}$

The value of z-statistics and the P-value will be then,

  $\begin{array}{l}z=-1.49\\ P=0.0677\end{array}$

And as we know if the P-value is smaller than the significance level, then reject the null hypothesis, so, we have,

  $P>0.05\Rightarrow \text{ Fail to Reject }{H}_0$

Thus, we conclude that there is no evidence that the proportion of leaking gas tanks is less than $40\%$ . Thus, two things we could do to decrease the probability of making this kind of error is increase $\alpha$ and increase the sample size.

(g)

To determine

To explain what are the advantages and disadvantages of taking those two courses of action.

Explanation of Solution

It is given in the question that nationwide, it is estimated that $40\%$ of service stations have gas tanks that leak to some extent. So, we have,

  $\begin{array}{l}{H}_0:p=0.40\\ H_a:p<0.40\end{array}$

The value of z-statistics and the P-value will be then,

  $\begin{array}{l}z=-1.49\\ P=0.0677\end{array}$

And as we know if the P-value is smaller than the significance level, then reject the null hypothesis, so, we have,

  $P>0.05\Rightarrow \text{ Fail to Reject }{H}_0$

Thus, we conclude that there is no evidence that the proportion of leaking gas tanks is less than $40\%$ . Thus, two things we could do to decrease the probability of making this kind of error is increase $\alpha$ and increase the sample size. Thus, if we increase $\alpha$ , then there is increase in power of test and lowers the chance of Type II error but increases the chance of Type I error. And if we increase the sample size, then we will increase the power of test and it will costs more time and money.