Chapter 10.3, Problem 31E

The paper “Debt Literacy, Financial Experiences and Over-Indebtedness” (Social Science Research Network, Working paper W14808, 2008) included analysis of data from a national sample of 1000 Americans. One question on the survey was:

“You owe $3000 on your credit card. You pay a minimum payment of $30 each month. At an Annual Percentage Rate of 12% (or 1% per month), how many years would it take to eliminate your credit card debt if you made no additional charges?”

Answer options for this question were: (a) less than 5 years; (b) between 5 and 10 years; (c) between 10 and 15 years; (d) never—you will continue to be in debt; (e) don’t know; and (f) prefer not to answer.

1. a. Only 354 of the 1000 respondents chose the correct answer of never. For purposes of this exercise, assume that the sample is representative of adult Americans. Is there convincing evidence that the proportion of adult Americans who can answer this question correctly is less than 0.40 (40%)? Use α = 0.05 to test the appropriate hypotheses. (Hint: See Example 10.10.)
2. b. The paper also reported that 37.8% of those in the sample chose one of the wrong answers (a, b, and c) as their response to this question. Is it reasonable to conclude that more than one-third of adult Americans would select a wrong answer to this question? Use α = 0.05.

To determine

Check whether the sample data provides convincing evidence that the proportion of adults of Country A who can answer the particular question correctly is less than 0.4.

Answer to Problem 31E

Yes, the sample data provides the evidence that the proportion of adults of Country A who can answer the particular question correctly is less than 0.4.

Explanation of Solution

Calculation:

In a sample of 1,000 adults of Country A, 354 were founded to answer the particular question correctly. It is assumed that the sample of adults American 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 can answer the particular question correctly.

Step 2:

Null hypothesis: $H_0:p=0.4$

That is, the proportion of adults of Country A who can answer the particular question correctly is 0.4.

Step 3:

Alternative hypothesis: $H_a:p<0.4$

That is, the proportion of adults of Country A who can answer the particular question correctly is less than 0.4.

Step 4:

Significance level, $\alpha$:

It is given 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 is the hypothesized proportion and n is the sample size.

Substituting the hypothesized proportion, $p=0.4$ in the test statistic,

$z=\frac{\widehat{p}-0.4}{\sqrt{\frac{0.4\left(1-0.4\right)}{n}}}$

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

$\begin{array}{c}n\left(\text{hypothesized value}\right)=np\\ =1,000\left(0.4\right)\\ =400\\ >10\end{array}$

$\begin{array}{c}n\left(1-\text{hypothesized value}\right)=n\left(1-p\right)\\ =1,000\left(1-0.4\right)\\ =1,000\left(0.6\right)\\ =600\\ >10\end{array}$

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 size of 1,000 acts as the representative for all adult American. It is also definite that the sample size is less than the 10% of the population of the adult American.

Step 7:

The value of the test statistic is obtained as follows:

$z=\frac{\widehat{p}-0.4}{\sqrt{\frac{0.4\left(1-0.4\right)}{n}}}$

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

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

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

Substitute the corresponding values to get the sample proportion:

$\begin{array}{c}\widehat{p}=\frac{354}{1,000}\\ =0.354\end{array}$

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

$\begin{array}{c}z=\frac{0.354-0.4}{\sqrt{\frac{0.4\left(1-0.4\right)}{1,000}}}\\ =\frac{-0.046}{0.01549}\\ \approx -2.97\end{array}$

Thus, the value of test statistic is –2.97.

Step 8:

P-value:

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

The P-value for the test statistic value of –2.97 is obtained as follows:

$\begin{array}{c}P\text{-value}=\text{Area to the left of }-2.97\\ =P\left(z<-2.97\right)\end{array}$

Use Table A in Appendix A: Standard Normal Cumulative Probabilities to find the z-value.

Procedure:

For z at –2.97:

•         Locate –2.9 in the left column of the table.

•         Obtain the value in the corresponding row below .07.

That is, $P\left(z<-2.97\right)=0.0015$.

Thus, the P-value for the test statistic of –2.97 is 0.0015.

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 0.0015 is less than the significance level 0.05.

That is, $P\text{-value}\left(=0.0015\right)<\text{Significance level}\left(=0.05\right)$.

The decision is that the null hypothesis is rejected.

Conclusion:

Hence, the sample data provides the evidence that the proportion of adults of Country A who can answer the particular question correctly is less than 0.4.

To determine

Check whether the sample data provides convincing evidence that more than one-third of the proportion of adults of Country A select wrong answer to the particular question.

Answer to Problem 31E

Yes, the sample data provides the evidence that more than one-third of adults of Country A select the wrong answer to the particular question is.

Explanation of Solution

Calculation:

The article also reports that 37.8% of adults from the sample of 1,000 chose the wrong answer to the particular question.

Step 1:

In this context, p denotes population proportion of adults of Country A who can select wrong answer to the particular question.

Step 2:

Null hypothesis: $H_0:p=\frac{1}{3}$

That is, the proportion of adults of Country A who select a wrong answer to the particular question is $\frac{1}{3}$.

Step 3:

Alternative hypothesis: $H_a:p>\frac{1}{3}$

That is, the proportion of adults of Country A who select a wrong answer to the particular question is greater than $\frac{1}{3}$.

Step 4:

Significance level, $\alpha$:

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

By substituting the hypothesized proportion, $p=\frac{1}{3}$ in the test statistic,

$z=\frac{\widehat{p}-\frac{1}{3}}{\sqrt{\frac{\frac{1}{3}\left(1-\frac{1}{3}\right)}{n}}}$

Here, the sample proportion $\widehat{p}$ is 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 sample is from a random sample.
• Check the conditions: $n\left(\text{hypothesized value}\right)\ge 10$ and $n\left(1-\text{hypothesized value}\right)\ge 10$.

$\begin{array}{c}n\left(\text{hypothesized value}\right)=np\\ =1,000\left(\frac{1}{3}\right)\\ =333.33\\ >10\end{array}$

$\begin{array}{c}n\left(1-\text{hypothesized value}\right)=n\left(1-p\right)\\ =1,000\left(1-\frac{1}{3}\right)\\ =1,000\left(\frac{2}{3}\right)\\ =666.67\\ >10\end{array}$

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 size of 1,000 acts as the representative for all adult American. It is also definite that the sample size is less than the 10% of the population of the adult American.

Step 7:

The value of the test statistic is obtained as follows:

$z=\frac{\widehat{p}-\frac{1}{3}}{\sqrt{\frac{\frac{1}{3}\left(1-\frac{1}{3}\right)}{n}}}$

Here, $\widehat{p}=0.378$.

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

$\begin{array}{c}z=\frac{0.378-0.3333}{\sqrt{\frac{0.3333\left(1-0.3333\right)}{1,000}}}\\ =\frac{0.0447}{0.0149}\\ \approx 3\end{array}$

Thus, the value of test statistic is 3.

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:

$\begin{array}{c}P\text{-value}=\text{Area to the right of }3\\ =P\left(z>3\right)\\ =1-P\left(z<3\right)\end{array}$

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

Procedure:

For z at 3.00:

•         Locate 3.0 in the left column of the table.

•         Obtain the value in the corresponding row below .00.

That is, $P\left(z<3\right)=0.9987$.

The P-value is obtained as follows:

$\begin{array}{c}P\text{-value}=1-P\left(z<3\right)\\ =1-0.9987\\ =0.0013\end{array}$

Thus, the P-value for the test statistic of 3 is 0.0013.

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 0.0013 is less than the significance level 0.05.

That is, $P\text{-value}\left(=0.0013\right)<\text{Significance level}\left(=0.05\right)$.

The decision is that the null hypothesis is rejected.

Conclusion:

Hence, the sample data provides the evidence that more than one-third of the adults of Country A select a wrong answer to the particular question.