font Styles F27 The accounts of a company show that on average, accounts receiva
ID: 2927041 • Letter: F
Question
font Styles F27 The accounts of a company show that on average, accounts receivable are $95.34. An auditor checks a random sample of 81 of these accounts, finding a sample mean of $91.40 and standard deviation of $40.56. Based on these findings, can you conclude the mean accounts receivable is different from $95.34 at -0.05? or the hypothesis stated above.. Question 1 What is the decision rule? Fill in only one of the following statements. If the hypothesis is one tailed Reject H, if If the hypothesis is two tailed: Reject H, if Question 2 What is the test statistic? Question 3 What is the p-value? Fill in only one of the following statements If the 2 table is appropriate, p-value If the t table is appropriate p-value Daly Problem92Explanation / Answer
Given that,
population mean(u)=95.34
sample mean, x =91.4
standard deviation, s =40.56
number (n)=81
null, Ho: =95.34
alternate, H1: !=95.34
level of significance, = 0.05
from standard normal table, two tailed t /2 =1.99
since our test is two-tailed
reject Ho, if to < -1.99 OR if to > 1.99
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =91.4-95.34/(40.56/sqrt(81))
to =-0.874
| to | =0.874
critical value
the value of |t | with n-1 = 80 d.f is 1.99
we got |to| =0.874 & | t | =1.99
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -0.8743 ) = 0.3846
hence value of p0.05 < 0.3846,here we do not reject Ho
ANSWERS
---------------
null, Ho: =95.34
alternate, H1: !=95.34
test statistic: -0.874
critical value: -1.99 , 1.99
decision: do not reject Ho
p-value: 0.3846
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