Could anyone please help me with Question #3 and please show me the steps. Thank

Question

Could anyone please help me with Question #3 and please show me the steps. Thank you very much!

If the t table is appropriate, _______ < p-value< _____1_____ (I guessed it)

The accounts of a company show that on average, accounts receivable are $95.72. An auditor checks a random sample of 49 of these accounts, finding a sample mean of $92.23 and standard deviation of $40.56. Based on these findings, can you conclude the mean accounts receivable is different from $95.72 at =0.05? For the hypothesis stated above Question 1 hat is the decision rule? Fillin only one of the following statements. f the hypothesis is one tailed Reject Ho if If the hypothesis is two tailed: Reject Ho if stat 2.011 stat 2.011 Question2 What is the test statistic? 0.602 Question 3 What is the p-value? Fill in only one of the following statements. If the Ztable is appropriate . p-value If thet table is appropriate p-value

Explanation / Answer

Given that,
population mean(u)=95.72
sample mean, x =92.23
standard deviation, s =40.56
number (n)=49
null, Ho: =95.72
alternate, H1: !=95.72
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.011
since our test is two-tailed
reject Ho, if to < -2.011 OR if to > 2.011
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =92.23-95.72/(40.56/sqrt(49))
to =-0.602
| to | =0.602
critical value
the value of |t | with n-1 = 48 d.f is 2.011
we got |to| =0.602 & | t | =2.011
make decision
hence value of |to | < | t | and here we do not reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != -0.6023 ) = 0.5498
hence value of p0.05 < 0.5498,here we do not reject Ho
ANSWERS
---------------
null, Ho: =95.72
alternate, H1: !=95.72
test statistic: -0.602
critical value: -2.011 , 2.011
decision: do not reject Ho
p-value: 0.5498
we do not have enough evidence to support the claim

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