Question 1129223: The mean income per person in the United States is $50,000 and the distribution of incomes follows as normal distribution. A random sample of 10 residents of Wilmington, Delaware had a mean of $60,000 wit a standard deviation of $10,000 .05 SIG LEVEL Need the Ho HI one tail, critical value, sample mean, Z value
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Ho: mean income is <=50000
Ha: mean income is >50000
alpha is 0.05, critical value is 1.645
z=(x bar- mean)/sigma/sqrt(n) (I would use t here, although difference would not be great)
=10000/(10000/sqrt(10))
=10000*sqrt(10)/10000=3.16. This is the Z value, Sample mean is $60,000
One would reject Ho at the 0.05 level. p-value is 0.0008
conclude the mean income in Wilmington is > $50000.
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