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 2.8 seconds. You wish to see if the mean time is changed by vigorous exercise, so you have a group of 15 college students exercise vigorously for 30 minutes and then complete the maze. It takes them an average of x¯=30.2 seconds to complete the maze. Use this information to test the hypotheses

H0:=30

H1:30

Conduct a test using a significance level of =0.01.

(a) The test statistic=

(b) The positive critical value, z* =

(c) The final conclusion is
A. There is not sufficient evidence to reject the null hypothesis that =30.
B. We can reject the null hypothesis that =30 and accept that 30.

Explanation / Answer

here std error =std deviaiton/(n)1/2 =2.8/(15)1/2 =0.723

a) test statistic= (X-mean)/std deviation =(30.2-30)/0.723 =0.2766

b) positive critical value, z* =2.5758

c) A. There is not sufficient evidence to reject the null hypothesis that =30.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.