Question

When an introductory statistics course has been taught live, the average final numerical grade has historically been 71.6% with a standard deviation of 8.8%. Forty students were selected at random to take a new online version of the same course, and the average final numerical grade for these students was found to be 69.2%. Assume that the grades for the online students are normally distributed and that the population standard deviation is still 8.8%. In parts (a) and (b), use the p-value method to conduct an appropriate hypothesis test at the 5% level of significance.

(a) Is there sufficient evidence to indicate that students taking the online version of the course score lower, on average, than students taking the traditional live version of the course?

(b) Is there sufficient evidence to indicate that the average score for students taking the online version of the course differs from students taking the traditional live version of the course? (HINT: You do not need to recalculate the value of the test statistic.)

Answer 1

a) H₀: = 71.6

H₁: < 71.6

The test statistic z = ()/()

= (69.2 - 71.6)/(8.8/)

= -1.72

P-value = P(Z < -1.72)

= 0.0427

Since the p-value is less than the significance level (0.0427 < 0.05), so we should reject the null hypothesis.

At 5% significance level there is sufficient evidence to indicate that students taking the online version of the score lower, on average, than the students taking the traditional live version of the course.

b) H₀: = 71.6

H₁: 71.6

P-value = 2 * P(Z < -1.72)

= 2 * 0.0427 = 0.0854

Since the p-value is greater than the significance level (0.0854 > 0.05), so we should not reject the null hypothesis.

At 5% significance level there is not sufficient evidence to indicate that the average score for students taking the online version of the course differs from students taking the traditional live version of the course.