Question 1058357: Question
A. A survey on the Cypriot market on the mean amount spent per customer for lunch meals at restaurant, data were collected for 25 customers in Nicosia. Assume a sample standard deviation of 3 and a sample of 20.
1. Find the 90% confidence interval.
2. Find the 95% confidence interval.
3. Test that the mean amount spent is larger than 18euros (value from tables=2.131).
B. For the same project data were also collected for 49 customers in Larnaca. They found that they spent on average 18 Euros with standard deviation of 2 Euros.
4. Find the 90% confidence interval.
5. Find the 99% confidence interval.
6. Test that the mean amount spent is indifferent to 15 Euros (value from tables=1.96).
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 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 >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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