SOLUTION: The average student-loan debt is reported to be $25,235.
A student believes that the student-loan debt us higher in her area.
She takes a random sample of 100 college students in
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Question 1183062: The average student-loan debt is reported to be $25,235.
A student believes that the student-loan debt us higher in her area.
She takes a random sample of 100 college students in her area and determines the mean student-loan debt is $27,524 and the standard deviation is $6,000.
Is there sufficient evidence to support the student's claim at a 5% significance level?
a) Is it safe to assume that n < 5% of all college students in the local area?
= Yes
b) Is n = 30? Yes
Test the claim:
a) Determine the null and alternative hypothesis.
Enter the correct symbol and value.
Ho: u = _________
Ha: u = > _________
b) Determine the test statistic.
Round to 2 decimals
t = ______
c) Find the p-value
Round to 4 decimals
p = ______
Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
It is safe to assume n is < 5% of all students if one thinks there are more than 2000 students in the area, since the sample is 100/2000 in that instance or 5%.
-
Ho:mu < = $25,235
Ha: mu >$25,235
Note: I prefer the greater than (without equals) in Ha and put the inequality and equal sign with Ho, since rejecting it means that single value postulated is rejected as well.
alpha=0.05 p{reject Ho|Ho true}
test stat is t(0.975, df=99) can use z, but it is a t-test and no reason not to use it in these days of calculators.
Critical value for a right tail test is t > 1.66
t=(x bar-mu)/s/sqrt(n)
=$2289*10/6000, inverting the denominator to multiply.
=3.815 or 3.82
This is > 1.66 so reject Ho and conclude that student debt is higher than the postulated value.
p-value=0.0001
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