The time needed for college students to complete a certain paper and pencil maze
ID: 3130378 • Letter: T
Question
The time needed for college students to complete a certain paper and pencil maze follows a Normal distribution with a population mean of 80 seconds and a population standard deviation of 16 seconds. You wish to see if the mean time is changed by meditation, so you have a group of 10 college students meditate for 30 minutes and then complete the maze. It takes them an average of x= 74 seconds to complete the maze. Use this information to test the hypotheses H0:= 80,Ha:80at thesignificance level = 0.02.
a.Calculate the test statistic for this test.
b. Calculate the p-value for this test.
c.Is the null hypothesis rejected? Is this data statistically significant at the =0.05 level? Does meditation impact the mean time needed to complete the maze?
Explanation / Answer
a)
Formulating the null and alternative hypotheses,
Ho: u = 80
Ha: u =/ 80
As we can see, this is a two tailed test.
Thus, getting the critical z, as alpha = 0.02 ,
alpha/2 = 0.01
zcrit = +/- 2.326347874
Getting the test statistic, as
X = sample mean = 74
uo = hypothesized mean = 80
n = sample size = 10
s = standard deviation = 16
Thus, z = (X - uo) * sqrt(n) / s = -1.185854123 [ANSWER, TEST STATISTIC]
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b)
Also, the p value is, as this is two tailed
p = 0.235679913 [ANSWER, P VALUE]
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c)
As P > 0.05, we do not reject Ho.
Hence, this data is not significant at 0.05 level.
Hence, from our data, meditation does not impact the mean time needed to complete the maze. [CONCLUSION]
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