6. Benford's Law claims that numbers chosen from very large data files tend to have "1" as the first nonzero digit disproportionately often. In fact, research has shown that if you randomly draw a number from a very large data file, the probability of getting a number with "1" as the leading digit is about 0.301. Suppose you are an auditor for a very large corporation. The revenue report involves millions of numbers in a large computer file. Let us say you took a random sample of n = 463 numerical entries from the file and r = 122 of the entries had a first nonzero digit of 1. Let p represent the population proportion of all numbers in the corporate file that have a first nonzero digit of 1. Test the claim that p is less than 0.301 by using What is the P-value of the test statistic?

-2.482

To find the P-value of the test statistic, we first need to determine the test statistic itself. In this case, since we are testing a proportion (p), we can use the z-test statistic.

The formula for the test statistic (z) is calculated as:

z = (p̂ - p₀) / √(p₀(1-p₀)/n)

Where:
- p̂ is the sample proportion (r/n)
- p₀ is the hypothesized proportion (0.301 in this case)
- n is the sample size (463 in this case)

Let's calculate the test statistic:

p̂ = r / n
= 122 / 463
p̂ ≈ 0.2637

z = (0.2637 - 0.301) / √(0.301(1-0.301)/463)

Now, we need to calculate the P-value using the standard normal distribution (Z-distribution). The P-value represents the probability of obtaining a test statistic as extreme as the observed one, assuming the null hypothesis (p ≤ 0.301) is true.

In this case, since we are testing the claim that p is less than 0.301, we are looking for the area to the left of the calculated z-value.

Using a Z-table or a calculator, we can find the area to the left of the calculated z-value. Let's call this area P(L):

P(L) = 0.121

However, since we are testing for p < 0.301, we need to consider the area in the left tail plus half of the area in the right tail. Let's call this total area P(T):

P(T) = 0.121 + (1 - 0.121)/2
= 0.5605

The P-value of the test statistic is approximately 0.5605.