Question 1096633: It has been reported that the average credit card debt for college seniors at the college book
store for a specific college is $3262 and the standard deviation is $1100. The student senate at
a large university feels that their seniors have debt much less than this, so it conducts a study of
50 randomly selected seniors and finds that average debt is $2995. Is the student senate correct
with 0.05.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Assuming a z test here with known sd.
the 95% CI, if this is used will be mean +/- 1.96*1100*sigma/sqrt(n)
this is +/- $304/90 or $305
The 95% CI is ($2690, $3300). Because $3262 is in the confidence interval, one cannot conclude that this is statistically significant at the 0.05 level
The z-test would be (2995-3262)/1100/sqrt(50)=-1.72
With Ho being that there is no difference
Ha being there is a difference
alpha being 0.05
test statistic of z=(bar-mean)/sigma/sqrt(n)
critical value is |z|>1.96
p-value is 0.086, which is greater than 0.05 so Ho is not rejected.
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