Chapter 10, Problem 12CRE

In a survey conducted by Yahoo Small Business, 1432 of 1813 adults surveyed said that they would alter their shopping habits if gas prices remain high (Associated Press, November 30, 2005). The article did not say how the sample was selected, but for purposes of this exercise, assume that it is reasonable to regard this sample as representative of adult Americans. Based on these survey data, is it reasonable to conclude that more than three-quarters of adult Americans plan to alter their shopping habits if gas prices remain high?

To determine

Check whether the sample data provide convincing evidence that more than three quarters of the proportion of adults of Country A plan to alter their shopping habits if gas prices remain high.

Answer to Problem 12CRE

Yes, the sample data provide convincing evidence that more than three quarters of the proportion of adults of Country A plan to alter their shopping habits if gas prices remain high.

Explanation of Solution

Calculation:

In a sample of 1,813 adults of Country A, 1,432 of adults were found to have planned to alter their shopping habits if gas prices remain high. It is given that the sample is representative of all the adults of Country A.

Step 1:

Population characteristic of interest:

Let p be the population characteristic of interest.

In this context, p denotes population proportion of adults of Country A who plan to alter their shopping habits if gas prices remain high.

Step 2:

Null hypothesis: $H_0:p=0.75$

That is, the proportion of adults of Country A who plan to alter their shopping habits if gas prices remain high is 0.75.

Step 3:

Alternative hypothesis: $H_a:p>0.75$

That is, the proportion of adults of Country A who plan to alter their shopping habits if gas prices remain high is greater than 0.75.

Step 4:

Significance level, $\alpha$:

It is assumed that the significance level, $\alpha =0.05$.

Step 5:

Test statistic, z:

$z=\frac{\widehat{p}-p}{\sqrt{\frac{p\left(1-p\right)}{n}}}$,

Where, $\widehat{p}$ be the sample proportion, p be the hypothesized proportion, and n be the sample size.

By substituting the hypothesized proportion, $p=0.75$ in the test statistic.

$z=\frac{\widehat{p}-0.75}{\sqrt{\frac{0.75\left(1-0.75\right)}{n}}}$

Here, the sample proportion $\widehat{p}$ is not known.

Step 6:

Assumptions:

• Let $\widehat{p}$ be the sample proportion from a random sample.
• The large sample z test can be used if the sample size n satisfies the conditions: $n\left(\text{hypothesized value}\right)\ge 10$ and $n\left(1-\text{hypothesized value}\right)\ge 10$.
• The sample size should not be greater than 10% of the population size.

Requirement check:

• It is assumed that the data are from a random sample.
• Check the conditions: $n\left(\text{hypothesized value}\right)\ge 10$ and $n\left(1-\text{hypothesized value}\right)\ge 10$.

$\begin{array}{c}n\left(\text{hypothesized value}\right)=np\\ =1,813\left(0.75\right)\\ =1,359.75\\ >10\end{array}$

$\begin{array}{c}n\left(1-\text{hypothesized value}\right)=n\left(1-p\right)\\ =1,813\left(1-0.75\right)\\ =1,813\left(0.25\right)\\ =453.25\\ >10\end{array}$

Since $n\left(\text{hypothesized value}\right)$ and $n\left(1-\text{hypothesized value}\right)$ are greater than 10, the sample size requirement is met.

Therefore, the large sample z test is appropriate.

• Although the population size is not known, it is reasonable to assume that the sample of size of 1,813 acts as a representative for all adults of Country A. It is also definite that the sample size is less than the 10% of the population of adults of Country A.

Step 7:

The value of the test statistic is obtained as follows:

$z=\frac{\widehat{p}-0.75}{\sqrt{\frac{0.75\left(1-0.75\right)}{n}}}$

The sample proportion, $\widehat{p}$ is obtained as follows:

$\widehat{p}=\frac{x}{n}$

Where, x be the number of observations of interest, and n be the sample size.

Substitute the corresponding values to get the sample proportion:

$\begin{array}{c}\widehat{p}=\frac{1,432}{1,813}\\ =0.79\end{array}$

By substituting the value of sample proportion, the test statistic is obtained as follows:

$\begin{array}{c}z=\frac{0.79-0.75}{\sqrt{\frac{0.75\left(1-0.75\right)}{1,813}}}\\ =\frac{0.04}{0.01017}\\ \approx 3.93\end{array}$

Thus, the value of test statistic is 3.93.

Step 8:

P-value:

In this context, the alternative hypothesis denotes that the test carried out is right-tailed test. Therefore, the P-value is the area under the z curve and to the right of the calculated z value.

The P-value for the test statistic value of 3 is obtained as follows:

$\begin{array}{c}P\text{-value}=\text{Area to the right of }3.93\\ =P\left(z>3.93\right)\\ =1-P\left(z<3.93\right)\end{array}$

Use Standard Normal Probabilities (Cumulative z curve areas) table to find the z-value.

However, the table has the value up to 3.8, such that $P\left(z<3.8\right)=1.0000$.

As $3.93>3.8$, the $P\left(z<3.93\right)=1.0000$.

The P-value is obtained as follows:

$\begin{array}{c}P\text{-value}=1-P\left(z<3.93\right)\\ =1-1\\ =0\end{array}$

Thus, the P-value for the test statistic of 3.93 is 0.

Step 9:

Decision rule:

If $P\text{-value}\le \text{significance level}$, then reject the null hypothesis $H_0$.

If $P\text{-value}>\text{significance level}$, then fail to reject the null hypothesis $H_0$.

Here, the $P\text{-value}$ of 0 is less than the significance level 0.05.

That is, $P\text{-value}\left(=0\right)<\text{significance level}\left(=0.05\right)$.

The decision is that the null hypothesis is rejected.

Conclusion:

Hence, the sample data provide the evidence that more than three quarters of the proportion of adults of Country A plan to alter their shopping habits if gas prices remain high.