The time needed for college students to complete a certain paper-and-pencil maze

Question

The time needed for college students to complete a certain paper-and-pencil maze follows a normal distri- bution with a mean of 30 seconds and a standard deviation of 3 seconds. You wish to see if the mean time is changed by vigorous exercise, so you have a group of nine college students exercise vigorously for 20 minutes and then complete the maze. Assume that remains unchanged at 3 seconds.

a) Carefully state the appropriate null hypothesis H0 and alternative hypothesis Ha for this study using the correct notation.

b) Suppose you compute the average time x that it takes these students to complete the maze and you find that the results are significant at the 5% level. What can you conclude?

Explanation / Answer

answer

H0: mu = 30, Ha: mu <> 30

at the 1% significance level The power of your test at mu = 28 seconds
is approximately

Power = 1 - P(Type II error) = 1 - P(Retain Ho | mu = 28)

We retain Ho when xbar is betwen 30-2.576(3/sqrt(9)) = 27.424
and 30+2.576(1) = 32.576.

Thus, under Ha: mu=28, we have

Power = 1 - P(27.424<xbar<32.576)
= 0.2823 [using 1-normalcdf(27.424,32.576,28,1)]

Thus, Choice C = Correct (with the slight difference because of
using the rounded 2.576).

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