17. Suppose we are interested in the mean annual income of all university studen
ID: 3135001 • Letter: 1
Question
17. Suppose we are interested in the mean annual income of all university students who graduated last year. More specifically, we want to know if the population mean differs from $27,000. The population of the incomes is approximately normal with a standard deviation of $2000. A random sample of 25 graduates was taken and the mean was calculated to be $24,500. (28 pts). a. Parameter (2 pts) = b. Statistic (2 pts) = c. Construct a 95% Confidence Interval-Make sure to check your conditions! (8 pts) d. Using an of 0.05, do the data provide sufficient evidence to conclude that the population mean differs from $27,000. Use the p-value method, keeping 4 decimal places for your p-value. (12 pts) i. Hypothesis (2 pts) ii. Conditions (2 pts) iii. Calculation (4 pts) iv. Decision (2 pts) v. Conclusion (2 pts) e. Do your results from letters c and d agree? Why? (4 pts)
Explanation / Answer
A)
A parameter is a quantity that decribes a population. Here, a parameter is
$2000 [POPULATION STANDARD DEVIATION]
*******************************
b)
A statistic is a quantity that describes a sample. Hence, here, a statistic is
$24500 [SAMPLE MEAN]
******************************
C)
The population standard deviation is given, and the distirbution is approximately normal, n = 25, so we can use the z distribution.
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 24500
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 2000
n = sample size = 25
Thus,
Margin of Error E = 783.9855938
Lower bound = 23716.01441
Upper bound = 25283.98559
Thus, the confidence interval is
( 23716.01441 , 25283.98559 ) [ANSWER]
******************************
d)
Formulating the null and alternative hypotheses,
Ho: u = 27000
Ha: u =/ 27000 [HYPOTHESES, I]
As sigma is known and the distrbution is approximately normal, we can use z distribution. [II]
As we can see, this is a two tailed test.
Getting the test statistic, as
X = sample mean = 24500
uo = hypothesized mean = 27000
n = sample size = 25
s = standard deviation = 2000
Thus, z = (X - uo) * sqrt(n) / s = -6.25
Also, the p value is
p = 4.10453E-10 [P VALUE, III]
As P < 0.05, we REJECT THE NULL HYPOTHESIS. [DECISION, IV]
Hence, there is significant evidnece that the population mean differs from $27,000. [CONCLUSION, V]
***************************
e)
Yes, because part c interval does not include 27000 inside it, so it is consistent with rejecting Ho in d. They should agree because we basically used the same distribution to calculate both.
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.