Recall that Benford\'s Law claims that numbers chosen from very large data files
ID: 3350556 • Letter: R
Question
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 the auditor for a very large corporation. The revenue file contains millions of numbers in a large computer data bank. You draw a random sample of n = 223 numbers from this file and r = 86 have a first nonzero digit of 1. Let p represent the population proportion of all numbers in the computer file that have a leading digit of 1.)
a. What sampling distribution will you use?
The standard normal, since np > 5 and nq > 5.
The Student's t, since np > 5 and nq > 5.
The standard normal, since np < 5 and nq < 5.
The Student's t, since np < 5 and nq < 5.
b.What is the value of the sample test statistic? (Round your answer to two decimal places.)
c. Find the P-value of the test statistic. (Round your answer to four decimal places.)
Sketch the sampling distribution and show the area corresponding to the P-value.
-4 -2Explanation / Answer
Solution:-
n = 223, p = 0.301, r = 86
a) The standard normal, since np > 5 and nq > 5.
b)
State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.
Null hypothesis: P = 0.301
Alternative hypothesis: P 0.301
Note that these hypotheses constitute a two-tailed test. The null hypothesis will be rejected if the sample proportion is too big or if it is too small.
Formulate an analysis plan. For this analysis, the significance level is 0.05. The test method, shown in the next section, is a one-sample z-test.
Analyze sample data. Using sample data, we calculate the standard deviation () and compute the z-score test statistic (z).
= sqrt[ P * ( 1 - P ) / n ]
= 0.03072
z = (p - P) /
z = 2.76
where P is the hypothesized value of population proportion in the null hypothesis, p is the sample proportion, and n is the sample size.
c) Since we have a two-tailed test, the P-value is the probability that the z-score is less than - 2.76 or greater than 2.76. Thus, the P-value = 0.0058
Interpret results. Since the P-value (0.0058) is less than the significance level (0.05), we have to reject the null hypothesis.
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