Question

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 mean of their times to complete the maze and will use this information to test the hypotheses
H0: μ = 70, Ha: μ notequal 70

at significance level α = 0.05. If you observe a sample mean completion time of mean = 78.55, the P-value obtained

Answer 1

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.

Answer 2

Did you calculate the Z-score yet?

Answer 3

Yeah I got -4.06666667

Answer 4

How did you get that. It shouldn't be that extreme, and also should be positive.

Answer 5

[(9-70)/15\]

Answer 6

9 observations
70 seconds mean
15 seconds Standard dev

Answer 7

Ah. 9 is the sample size. you need to use the measured mean, x=78.55.
\[\large z=\frac{x-\mu}{\sigma/\sqrt{n}}\]

Answer 8

ohh! so 0.570?

Answer 9

A little closer.
The difference is 78.55-70=8.55.
You have to adjust the standard deviation by the sample size: 15/√9 = 5.

Z=8.55÷5 = 1.71

Answer 10

oh I forgot the \[\sqrt{9}\]

Answer 11

Thanks!

Answer 12

Don't forget that this is a two-tailed test when finding the p-value.