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 distribution 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 30 minutes and then complete the maze. Assume that remains unchanged at 3 seconds.

Suppose it takes the nine students an average of 32.05 seconds to complete the maze. At the 2.5% significance level, what can you conclude?

Select one or more:

a. H0 should be rejected because the P-value is less than 0.025.

b. H0 should not be rejected because the P-value is greater than 0.025.

c. Ha should be rejected because the P-value is less than 0.025.

d. Ha should not be rejected because the P-value is greater than 0.025.

e. Cannot be determined

Explanation / Answer

Formulating the null and alternative hypotheses,              
              
Ho:   u   =   30  
Ha:    u   =/   30  
              
As we can see, this is a    two   tailed test.      
              
Thus, getting the critical t,              
df = n - 1 =    8          
tcrit =    +/-   2.751523596      
              
Getting the test statistic, as              
              
X = sample mean =    32.05          
uo = hypothesized mean =    30          
n = sample size =    9          
s = standard deviation =    3          
              
Thus, t = (X - uo) * sqrt(n) / s =    2.05          
              
Also, the p value is              
              
p =    0.074508172          
              
As P > 0.025, we   FAIL TO REJECT THE NULL HYPOTHESIS.          

Hence,

OPTION B. H0 should not be rejected because the P-value is greater than 0.025. [ANSWER, B]

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