Question

A. 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 = 217 numerical entries from the file and r = 50 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.

(i) Test the claim that p is less than 0.301. Use ? = 0.05.

(a) What is the level of significance?

(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.)

B. Is the national crime rate really going down? Some sociologists say yes! They say that the reason for the decline in crime rates in the 1980s and 1990s is demographics. It seems that the population is aging, and older people commit fewer crimes. According to the FBI and the Justice Department, 70% of all arrests are of males aged 15 to 34 years†. Suppose you are a sociologist in Rock Springs, Wyoming, and a random sample of police files showed that of 39 arrests last month, 22 were of males aged 15 to 34 years. Use a 5% level of significance to test the claim that the population proportion of such arrests in Rock Springs is different from 70%.

(a) What is the level of significance?

(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.)

Answer 1

Solution A:

Null Hypothesis (Ho): p 0.301

Alternative Hypothesis (Ha): p < 0.301

a. Level of significance, a = 0.05

b. Sample proportion, p' = r/n

Sample proportion, p' = 50/217

Sample proportion, p' = 0.2304

Test Statistics

Z =

Z =

Z = -0.07059/0.031138

Z = -2.27

c. Using Z-tables, the p-value is

P [Z < -2.27] =1 - P [Z -2.27

= 1 - 0.9884

= 0.0116

Since p-value is less than 0.05 significance level, we reject Ho.

Hence, we can conclude that claim holds true.

Solution B.

Null Hypothesis (Ho): p =0.70

Alternative Hypothesis (Ha): p 0.70

a. Level of significance, a = 0.05

b. Sample proportion, p' = x/n

Sample proportion, p' = 22/39

Sample proportion, p' = 0.5641

Test Statistics

Z =

Z =

Z = -1.85

c. Using Z-tables, the p-value is

P [Z -1.85] = P [Z < -1.85] + P [Z > 1.85]

= 0.0322 + 0.0322

= 0.0644

Since p-value is greater than 0.05 level of significance, we fail to reject Ho.

Hence, we cannot conclude that the population proportion of such arrests in Rock Springs is different from 70%.