Question

Two methods were used to teach a high school algebra course. A sample of 75 scores was selected for method 1, and a sample of 60 scores was selected for method 2. The results are:

Method 1 Method 2
Sample mean 85 83
Sample s.d. 3 2

Required:
Test whether method 1 was more successful than method 2 at the 1% level.

Answer 1

Since the calculated value of t =4.629 falls in the critical region t> 2.355 we reject the null hypothesis and conclude that method 1 was more successful than method 2 at the 1% level.

Explanation:

Applying two sample t test as both methods are independent

The null and alternate hypotheses are

H0: u1= u2 against the claim Ha: u1> u2

The significance level is 0.01 and the critical region is t > t∝( n-1)

The degrees of freedom is calculated using the formula

υ = [s₁²/n1 + s₂²/n2]²/ (s₁²/n1 )²/ n1-1 + (s₂²/n2)²/n2-1

This formula is used when the the variances are unequal.

So as we do not know anything about variances we assume they are unequal.

The calculated degrees of freedom is 132

The critical region is t > 2.355 for one tailed test.

The test statistic is

t= x1`- x2`/ √s1²/n1 + s2²/n2

t= 85-83/ √3²/75+ 2²/60

t= 2/ 0.432

t= 4.629

Since the calculated value of t =4.629 falls in the critical region t> 2.355 we reject the null hypothesis and conclude that method 1 was more successful than method 2 at the 1% level.