The Bureau of Economic Analysis in the U.S. Department of Commerce reported that
ID: 3049167 • Letter: T
Question
The Bureau of Economic Analysis in the U.S. Department of Commerce reported that the mean annual income for a resident of North Carolina is $18,688 (USA Today, August 24, 1995). A researcher for the state of North Carolina wants to test the following hypothesis: Where is the mean annual income for a resident of North Carolina. The researcher gathers information from a sample of 625 residents of North Carolina and finds a sample mean of 17,076 with a sample standard deviation equal to 15,500.
a. What is the appropriate conclusion pertaining to the hypothesis formulated above? Use a 0.01 level of significance.
b. What is the -value for this test?
c. Construct a 99% Confidence Interval for the value of the mean pertaining to the North Carolina residents’ income.
d. How would your answer to part a. change if the sample size, instead of 625, were 25? Demonstrate and Explain.
e. How would your answer to part d. change if you KNEW the value of the standard deviation and it was equal to 16,000? Demonstrate and Explain.
In all these problems, show the area under the curve to graphically demonstrate your answers.
Explanation / Answer
a)
TS = (Xbar - mu)/(s/sqrt(n))
= (17076 - 18688)/(15500/sqrt(625))
= -2.6
critical value = 2.576
since |TS| > critical value we reject the null hypothesis
b)
p-value = 2 P(T > 2.6 ) = 2*0.0047 = 0.0094
c)
99 % confidence level
(15474, 18678)
d)
if sample size is 25 , confidence level increase
as margin of error increases
One-Sample T
N Mean StDev SE Mean 99% CI
25 17076 15500 3100 (8405, 25747)
e)
we will use z-test instead of T-test when we know population standard deviation
One-Sample Z
The assumed standard deviation = 16000
N Mean SE Mean 99% CI
25 17076 3200 (8833, 25319)
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