Question 1125998: The accounts of a company show that on average, accounts receivable are $89.72. An auditor checks a random sample of 49 of these accounts, finding a sample mean of $85.11 and a standard deviation of $40.56. Based on these findings, can you conclude the mean accounts receivable is different from $89.72 at alpha=0.05?
1. Reject H0 if _____<______ or ______>_______
2. what is the test statistic?
3. If the t table is appropriate, _____ < p-value < _______
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! This is a one sample t-test, with the test statistic's being a t df=48,0.975, and the critical value is |t|>2.01
t=(xbar-mean)/s/sqrt(n)
this is -4.61*7/40.56=-0.79
p=0.43
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