<PRE><B>Question 1058357</B>
I am assuming the mean is 20 and the standard deviation is 3
the interval is +/-t(0.95,df=24)s/sqrt(n)
for 90% interval, it is +/- 1.711*3/5=1.03 so the interval is (18.97,21.03)
for 95%, t changes to 2.064 and the interval width 1.24, so (18.76,21.24)
t=(18-21)*5/3=-5. Virtually the entire t-distribution would be greater than 18, so that one could say with extremely high confidence that the value is &gt;18 euros.
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t(0.95, df=48) is 1.677
s/sqrt (n)=2/7
interval width is 0.48.  Interval itself is (17.52,18.48)
t(0.99 df=48) is 2.682, width 0.77, Interval (17.23,18.77)
t for 15 euros is (15-18)/(2/7)=-10.5 and the whole distribution would be greater than 15 euros.
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