i Excel File Edit View Insert Format Tools Data Window Help 86%-. Thu 11:09 PM a

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i Excel File Edit View Insert Format Tools Data Window Help 86%-. Thu 11:09 PM a-E Q Search Sheet Insert Page Layout Formulas Data Review View cut AutoSum A Fill ClearSort a Calibri (Body) ,111 A-Ar , Wrap Text Copy -Merge & Center" | $, % , % 4% conditional Format Cell Formatting as Table Sty es EEE Insert Delete Format Format Blu--.-'A, Filter F31 MN 12 13 The accounts of a company show that on average, accounts receivable are $89.89. An auditor checks a random sample of 49 1s of these accounts, finding a sample mean of $85.96 and standard deviation of $40.56. Based on these findings, can you conclude the mean accounts receivable is different from $89.89 at =0.1? 19 21 For the hypothesis stated above 23 Question1 What is the dedision rule? Fill in only gne ofthe followine statements. If the hypethesis is one tailed: 26 Reject H lf If the hypathess is two ta led: 81 Reject Heif 32 Question 2 What is the test statistic? Question 3 What is the pvalue? Fill in only ane of the followine statements Il the 2 table is appropriate, p-value If thet table is appropriate, p-value Click to Grade Your Work Daily Problem+ Ready LEI + 100% 12

Explanation / Answer

Given that,
population mean(u)=89.89
standard deviation, =40.56
sample mean, x =85.96
number (n)=49
null, Ho: =89.89
alternate, H1: !=89.89
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
reject Ho, if zo < -1.645 OR if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 85.96-89.89/(40.56/sqrt(49)
zo = -0.67825
| zo | = 0.67825
critical value
the value of |z | at los 10% is 1.645
we got |zo| =0.67825 & | z | = 1.645
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != -0.67825 ) = 0.49761
hence value of p0.1 < 0.49761, here we do not reject Ho
ANSWERS
---------------
null, Ho: =89.89
alternate, H1: !=89.89
test statistic: -0.67825
critical value: -1.645 , 1.645
decision: do not reject Ho
p-value: 0.49761
we do not have enough evidence to support the claim

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