A teacher wishes to study the amount of time students in his statistics course s

Question

A teacher wishes to study the amount of time students in his statistics course spend each week in study for the course. He believes that the average should be the nominal 6 hours (two hours outside class for every hour in class). So he has the students keep track of and report the time spent in study during a typical week. A total of 9 students respond. The average time spent is 6.5 hours with a standard deviation of 2 hours. Is this enough evidence to show that the average time is not 6 hours of studying?

Explanation / Answer

Null Hypothesis H0: The average time spent by students is 6 hrs.

Alternative hypothesis H1: The average time spent by students is not 6 hrs.

We conduct two-tail t-test to test for the hypothesis.

standard error of the time spent = standard deviation / Sqrt(number of students)

= 2/sqrt(9) = 0.6667

t = (observed average - hypothesized average)/ standard error = (6.5 - 6)/0.6667 = 0.75

Degree of freedom = number of students - 1 = 9 - 1 = 8

p-value for t = 0.75 at degree of freedom, 8 is 0.2374

For two-tail test p-value = 2 * 0.2374 = 0.4748

As, the p-value is greater than 0.05, we fail to reject the null hypothesis and conclude that the average time spent by students is 6 hrs. So, there is not enough evidence to show that the average time is not 6 hours of studying.

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