Question

A study was conducted to test the effectiveness of a juggling class. Before the class started, six subjects juggled as many balls as they could at once. After the class, the same six subjects juggled as many balls as they could. The differences in the number of balls are calculated. The differences have a normal distribution. Test at the 1% significance level. What conclusion can you draw about the juggling class including how to get p-value? Chart below.

Subject A B C D E F

Before 3 4 3 2 4 5

After 4 5 6 4 5 7

Answer 1

Let be the true mean difference between the number of balls after class and the number of balls before class.

The differences between the number of balls after class and the number of balls before class are

1 1 3 2 1 2

Sample mean difference, = (1 + 1 + 3 + 2 + 1 + 2)/6 = 1.666667

sample standard deviation of difference s =

= 0.8164966

Standard error of mean difference, se = s / = 0.8164966 / = 0.3333333

Degree of freedom = n-1 = 6 - 1 = 5

Test statistic = / se = 1.666667 / 0.3333333 = 5

P-value = P(t > 5, df = 5) =  0.0021

Since, p-value is less than 0.01 significance level, we reject null hypothesis H0 and conclude that there is significant evidence from the sample data at 0.01 significance level that the true mean difference between the number of balls after class and the number of balls before class.is greater than 0. Thus,  there is significant evidence that the juggling class is effective.