The time needed for college students to complete a certain paper and pencil maze follows a Normal distribution, with a mean of 70 seconds and a standard deviation of 15 seconds. You wish to see if the mean time μ is changed by meditation, so you have a group of 9 college students meditate for 30 minutes and then complete the maze. You compute the average of their times to complete the maze and will use this information to test the hypotheses H0: μ = 70, Ha: μ 70 at significance level α = 0.05. If you observe a sample mean completion time of = 78.55, the P-value obtained is A. less than 0.01. B. between 0.01 and 0.025. C. between 0.025 and 0.05. D. between 0.05 and 0.10.
i got the p value of 1.91 but i dont know how to apply that to the choices
pV can only be between 0 and 1, its a probability value
1.91, assuming you dod it right, is a zvalue associated with it
*did
consider how the hypothesis is arranged. is it Ho not= Ha? then the pValue is 2 tailed. If its < or > then we have a 1 tail value
im so lost. H0: μ = 70, Ha: μ 70( here it has the equal sign withthe line through so im assuming that means two tailed. at significance level α = 0.05. .....does this have something to do with my equation. Like i said i got 1.91 thats none of the answers above.
.05 is extra stuff and is not used to evaluate the pValue its 2 tailed since it uses the \(\ne\) notation. You are essentially finding the zscore of Ha within Ho
\[z=\frac{H_a-H_o}{\sigma}\] and adjusted for the sample size we get \[z\sqrt n=\frac{H_a-H_o}{\sigma/\sqrt n}\]
z = 3*8.55/15 z = 8.55/5 = 1.71
the empirical rule gives us that about 95% falls inside of 2sd, giving us about 5% outside of it ... and since this isnt quite 2, we are a little bigger than 5% outside right?
right and when you look at the table 1.71=.9564 and if two sided its 1.91. but how would i apply that in to the prior problem you posted
by table, it looks like a left tail table from that information ... in other words, 1 tail = 1-.9564 = .0436, x2 for 2 tails of course
in either case, we are between 5 and 10 % :)
oooo so i was suppose to subtract from one first and then times by two.....thankyou
or since the table is symetric, we could have looked at the -1.71 to find the .0436
good luck ;)
i was thinking i should look at -1.71 but i was second guessing myself . thanks again
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