Recall that 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. Now 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 = 218 numerical entries from the file and r = 52 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.
α = 0.10.
(a) What is the value of the sample test statistic? (Round your answer to two decimal places.)
(b) (c) Find the P-value of the test statistic. (Round your answer to four decimal places.)
I'm just having trouble with A and C. I don't know how to go about doing them. I'm trying to use the equation z=phat-p/squ p*q/n but I think the probably is that I don't really know how to find p in this problem. Thanks.
p = 52/218 q = 166/218
n = 218.
Your test will be comparing your p-value with .301 to see if it is the same or different.
Ho: p = .031
Ha: p does not = .301
To find the values of the sample test statistic (a) and the P-value (c), you need to calculate the sample proportion (phat) and the standard deviation (s).
(a) Finding the sample test statistic:
The sample proportion (phat) is the ratio of the number of entries with a first nonzero digit of 1 (r = 52) to the total sample size (n = 218):
phat = r/n
Plugging in the given values:
phat = 52/218
Calculate the value of phat by dividing 52 by 218:
phat ≈ 0.2385
Now, we need to find the standard deviation (s). The standard deviation is calculated using the formula √(pq/n), where p and q are the probabilities of success and failure, respectively.
Since p is the probability of a number having "1" as the first nonzero digit according to Benford's Law (p = 0.301), we can assume q = 1 - p. Thus, q ≈ 1 - 0.301 ≈ 0.699.
Substituting the values into the standard deviation formula:
s = √(p*q/n)
= √(0.301*0.699/218)
≈ √(0.209499/218)
≈ √0.000960325
≈ 0.030997
Now, we can calculate the sample test statistic (z) using the formula:
z = (phat - p) / s
Substituting the values:
z = (0.2385 - 0.301) / 0.030997
≈ (-0.0625) / 0.030997
≈ -2.014
Therefore, the value of the sample test statistic is approximately -2.014.
(c) Finding the P-value of the test statistic:
To find the P-value, we need to determine the probability of observing a test statistic as extreme or more extreme than the calculated z-value (-2.014) under the null hypothesis.
Using a standard normal distribution table or calculator, we can find that the probability associated with a z-value of -2.014 is approximately 0.0222.
However, since it's a two-sided test, we need to consider both tails. Therefore, we double the calculated probability:
P-value ≈ 2 * 0.0222
≈ 0.0444
Hence, the P-value of the test statistic is approximately 0.0444, rounded to four decimal places.
To find the value of the sample test statistic, you'll need to calculate the sample proportion (phat) and the population proportion (p). Let's calculate them step by step:
Step 1: Calculate the sample proportion (phat):
phat = r/n
= 52/218
≈ 0.2385
Step 2: Calculate the population proportion (p) based on Benford's Law:
p = 0.301
Now you can substitute these values into the formula for the sample test statistic:
z = (phat - p) / sqrt(p*(1-p)/n)
Step 3: Substitute the values into the formula:
z = (0.2385 - 0.301) / sqrt(0.301 * (1 - 0.301) / 218)
Calculating this expression will give you the value of the sample test statistic. Round your answer to two decimal places.
For part (c), you need to find the P-value of the test statistic. The P-value represents the probability of observing a test statistic as extreme as (or more extreme than) the one calculated, assuming the null hypothesis is true.
To find the P-value, you can use a standard normal distribution table or a statistical software. The P-value is the probability of observing a test statistic value as extreme as the calculated value in either tail of the standard normal distribution.
Since α = 0.10, compare the P-value to α to determine whether to reject or fail to reject the null hypothesis.
I hope this helps! Let me know if you have further questions.