SOLUTION: The amount of liquid a paper towel can absorb will be use to marked the paper towel. Suppose a sample of 15 new towels average 12.5 grams with standard deviation of 1.15 grams. the

Algebra ->  Probability-and-statistics -> SOLUTION: The amount of liquid a paper towel can absorb will be use to marked the paper towel. Suppose a sample of 15 new towels average 12.5 grams with standard deviation of 1.15 grams. the      Log On


   



Question 634174: The amount of liquid a paper towel can absorb will be use to marked the paper towel. Suppose a sample of 15 new towels average 12.5 grams with standard deviation of 1.15 grams. the previous version of this paper towel absorbed an average of 12 grams. With the =5%, we can conclude that, on average

the new towel absorbs more than the previous one since the critical test value is 1.645
the new towel absorbs more than the previous one since the critical test value is 1.761
the new towel does not absorb more than the previous one since the critical test value is 1.761
the new towel does not absorb more than the previous one since the critical test value is 1.645

Answer by stanbon(75887) About Me  (Show Source):
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The amount of liquid a paper towel can absorb will be used to mark the paper towel. Suppose a sample of 15 new towels average 12.5 grams with standard deviation of 1.15 grams. the previous version of this paper towel absorbed an average of 12 grams. With the =5%, we can conclude that, on average
a) the new towel absorbs more than the previous one since the critical test value is 1.645
b) the new towel absorbs more than the previous one since the critical test value is 1.761
c) the new towel does not absorb more than the previous one since the critical test value is 1.761
d) the new towel does not absorb more than the previous one since the critical test value is 1.645
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Ho: u <= 12
Ha: u > 12 (absorbs more)
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t(12.5) = (12.5-12)/[1.15/sqrt(15)] = 1.6839
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The critical t-value = 1.7613
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Since the test stat is less than the critical value, fail to reject Ho.
Ans: c
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Cheers,
Stan H.