Benford\'s Law claims that numbers chosen from very large data files tend to hav

Question

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 = 247 numerical entries from the file and r = 60 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 = 0.1. Are the data statistically significant at the significance level? Based on your answers, will you reject or fail to reject the null hypothesis? The P-value is less than the level of significance so the data are statistically significant. Thus, we reject the null hypothesis. The P-value is greater than the level of significance so the data are not statistically significant. Thus, we reject the null hypothesis. The P-value is less than the level of significance so the data are statistically significant. Thus, we fail to reject the null hypothesis. The P-value is less than the level of significance so the data are statistically significant. Thus, we reject the null hypothesis. The P-value is less than the level of significance so the data are not statistically significant. Thus, we reject the null hypothesis.

Explanation / Answer

Given that,
possibile chances (x)=60
sample size(n)=247
success rate ( p )= x/n = 0.2429
success probability,( po )=0.301
failure probability,( qo) = 0.699
null, Ho:p=0.301  
alternate, H1: p<0.301
level of significance, = 0.1
from standard normal table,left tailed z /2 =1.28
since our test is left-tailed
reject Ho, if zo < -1.28
we use test statistic z proportion = p-po/sqrt(poqo/n)
zo=0.24291-0.301/(sqrt(0.210399)/247)
zo =-1.9902
| zo | =1.9902
critical value
the value of |z | at los 0.1% is 1.28
we got |zo| =1.99 & | z | =1.28
make decision
hence value of | zo | > | z | and here we reject Ho
p-value: left tail - Ha : ( p < -1.99017 ) = 0.02329
hence value of p0.1 > 0.02329,here we reject Ho
ANSWERS
---------------
null, Ho:p=0.301
alternate, H1: p<0.301
test statistic: -1.9902
critical value: -1.28
decision: reject Ho
p-value: 0.02329

we have enough evidence to support the claim that The P-value is less than the level of significance so the data are statistically significant

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