The accounts of a company show that on average accounts rece

The accounts of a company show that on average, accounts receivable are $94.37. An auditor checks a random sample of 49 of these accounts, finding a sample mean of $88.61 and standard deviation of $40.56. Based on these findings, can you conclude the mean accounts receivable is different from $94.37 at =.10?

For the hypothesis stated above…

What is the decision rule?   Fill in only one of the following statements

Reject H0 if ______ ______ ______

Reject H0 if ______<_______OR______>________

What is the test statistic?__________

What is the p-value? Fill in only one of the following statements.

If the Z table is appropriate, P-Value=_________

If the t table is appropriate, _______< p-value <________

Solution

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   94.37  
Ha:    u   =/   94.37  
              
As we can see, this is a    two   tailed test.      
              
Thus, getting the critical z, as alpha =    0.1   ,      
alpha/2 =    0.05          
zcrit =    +/-   1.644853627      

Thus, we reject Ho if zstat < -1.644853627 OR z > 1.644853627. [ANSWER, DECISION RULE]

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Getting the test statistic, as              
              
X = sample mean =    88.61          
uo = hypothesized mean =    94.37          
n = sample size =    49          
s = standard deviation =    40.56          
              
Thus, z = (X - uo) * sqrt(n) / s =    -0.99408284   [ANSWER, TEST STATISTIC]

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Also, the p value is, as it is two tailed,              
              
p =    0.320182539   [ANSWER, P VALUE]

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