Chapter 8.3, Problem 7P

(i)

To determine

(a)

Find the level of significance.

State the null and alternative hypothesis.

(b)

Identify the sampling distribution to be used.

Identify whether the sample size is sufficiently large or not.

Find the value of the sample test statistic.

Find the z value.

(c)

Find the P-value of the test statistic.

Draw the sampling distribution by showing the area corresponding to the P-value.

(d)

Check whether the null hypothesis is rejecting or fail to reject.

Identify whether the data statistically significant at level 0.01 or not.

(e)

Interpret the conclusion in the context of the application.

Answer to Problem 7P

(a)

The level of significance is 0.01.

The null hypothesis is $H_0:p=0.301$.

The alternative hypothesis is $H_1:p<0.301$.

(b)

The sampling distribution to be used is normal distribution.

Yes, the sample size is sufficiently large.

The value of the sample test statistic is 0.214.

The z value is –2.78.

(c)

The P-value is 0.0027.

(d)

The null hypothesis is rejected.

The data is statistically significant at level 0.01.

Explanation of Solution

Calculation:

(a)

Let p denotes the population proportion of all numbers in the corporate file that have a first nonzero digit of 1.

From the given information the value of $\alpha$ is 0.01, and the probability of getting a number with “1” as the leading digit is less than 0.301.

Hence, the level of significance is 0.01.

The null and alternative hypothesis is,

Null hypothesis:

$H_0:p=0.301$

Alternative hypothesis:

$H_1:p<0.301$

(b)

Conditions:

The conditions for using the normal distribution to test a proportion p with proportion of success $\left(p\right)$ and proportion of failure $\left(q=1-p\right)$ for a sufficiently large n (number of trails) are,

• $np>5$
• $nq>5$

Test statistic for proportion z:

The z statistic value for population proportion p is,

$z=\frac{\widehat{p}-p}{\sqrt{\frac{pq}{n}}}$

In the formula $\widehat{p}=\frac{r}{n}$ is the sample test statistic, r is the number of successes, n is number of trails, p is proportion specified in $H_0$, and $q=1-p$.

The number of trails is $n=215$, and the probability of success for each trail is $p=0.301$.

Checking conditions:

$\begin{array}{c}np=215\left(0.301\right)\\ =64.7\\ >5\end{array}$

$\begin{array}{c}nq=n\left(1-p\right)\\ =215\left(1-0.301\right)\\ =215\left(0.699\right)\\ =150.3\\ >5\end{array}$

It can be observed that two of the conditions $np>5$, $nq>5$ are satisfied. It is appropriate to use normal distribution to the $\widehat{p}$ distribution and also the sample size is sufficiently large.

The value of $\widehat{p}$ is,

$\begin{array}{c}\widehat{p}=\frac{46}{215}\\ =0.214\end{array}$

Hence, the value of $\widehat{p}$ is 0.214.

Test statistic:

Substitute $\widehat{p}$ as 0.214, p as 0.301, q as 0.699, and n as 215 in the test statistic formula

$\begin{array}{c}z=\frac{0.214-0.301}{\sqrt{\frac{0.301\times 0.699}{215}}}\\ =\frac{-0.087}{0.0313}\\ =-2.78\end{array}$

Hence, the standardized sample test statistic is –2.78.

(c)

Step by step procedure to obtain P-value using MINITAB software is given below:

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Left Tail, for the region of the curve to shade.
• Enter the X value as –2.78.
• Click OK.

Output using MINITAB software is given below:

From Minitab output, the P-value is 0.0027.

Hence, the P-value of the test statistic is 0.0027.

(d)

Rejection rule:

• If the P-value is less than or equal to $\alpha$, then reject the null hypothesis and the test is statistically significant. That is, $P\text{-value}\le \alpha$.

Conclusion:

The P-value is 0.0027 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, $0.0027\left(=P\text{-value}\right)<0.01\left(=\alpha \right)$.

By the rejection rule, the null hypothesis is rejected.

Hence, the data is statistically significant at level 0.01.

(e)

The null hypothesis is rejected. This shows that, the probability of getting a number with “1” as the leading digit is less than 0.301 at level of significance 0.01.

(ii)

To determine

Explain whether the proportion p is less than 0.301 can be suspected that there are not enough numbers in the data file with leading 1’s.

Identify whether the books have been ‘cooked’ by ‘pumping up’ or inflating the numbers or not.

Comment from the viewpoint of a stockholder.

Comment from the perspective of the Federal Bureau of Investigation.

Explanation of Solution

If the probability that of getting a number with 1 as the leading digit is less than 0.301, then it can be suspected that there are less numbers in the data file with leading 1’s. Also, it indicates that the numbers of books are cooked by pumping up more files with leading 1’s (inflating) the numbers. That is, there are more numbers in the data files with higher first nonzero digits than predicted by Benford’s law.

If the numbers are inflated, then the stockholder might think that the company is in high profits and would invest more in the company.

The Federal Bureau of Investigation (FBI) looks for money laundering, but the company has shown the false profits. From point of FBI the company has committed fraud accounting based on profits.

(iii)

To determine

Explain whether further investigation can be recommended or not before accusing the company of fraud.

Explanation of Solution

When the null hypothesis $H_0$ is rejected at level of significance $\alpha$, it does not mean that $H_0$ is completely proved to be false. There is $\alpha \%$ chance of making mistake in rejecting null hypothesis.

In the conducted hypothesis test, the null hypothesis is rejected indicating that the probability of getting a number with “1” as the leading digit is less than 0.301. That is, there are few numbers with leading digit as 1. But there might be chance of 1% for the conclusion being false. This shows that, a further investigation can be recommended before accusing the company as fraud.