Answer the next two questions based on the following: A water utility manager worried that demand has changed from its former 110 million gallons monthly average samples the most recent 60 months (millions of gallons/month).
TEST OF MU = 110.00 VS MU NOT = 110.00
N MEAN STDEV SE MEAN T P VALUE
DEMAND 60 117.70 14.42 1.86 4.14 0.0001
1. A test is conducted at the ( = 0.01 level to determine if average monthly demand now is significantly different from 110 million gallons. We may conclude that
a. p > ( and therefore reject H0 b. p < ( and therefore reject H0
c. p < ( and therefore cannot reject H0 d. p > ( ...view middle of the document...
is 115.1 c. is 14.1 d. is 2.39 e. is 0.019
5. This tests significant at any ( level equal to or greater than
a. ( = .20 b. ( = .10 c. ( = .05 d. ( = .01
6. If a one-tailed test had been conducted instead, the computer output above would have been the same except
a. the t-ratio would have been twice as large b. the t-ratio would have been half as large
c. the p-value would have been twice as large d. the p-value would have been half as large
e. both the p-value and t-ratio would be the same as before
Answer the next three questions using the following case and output: For the previous case, only one-bedroom apartments are surveyed (n = 33) to estimate the average size in square feet.
One-Sample T: squ.feet
Variable N Mean StDev SE Mean 90.0% CI
squ.feet 33 675.3 93.8 16.3 ( 647.6, 702.9)
7. Standard error of the mean in the above output is a little over sixteen because that is
a. about half the sample size, and we use about 2 standard errors to calculate margins of error
b. the standard deviation divided by the square root of the sample size
c. the number of standard deviations from $675 d. the result of converting apartment size to square yards
e. None of the above
8. What does the interval (647.6 square feet, 702.9 square feet) tell us about one-bedroom...