A large university offers several statistics sections for business students. Two

Question

A large university offers several statistics sections for business students. Two of the sections are taught by graduate teaching assistants. The assistants teach each of their sections using the curriculum created by the department chair. The chair would like to determine if the mean grades differ in the class taught by the first teaching assistant versus the mean grades taught by the second teaching asssistant. To test if grade performance in the classes differ, the chair randomly samples 40 students from both. Determine whether the mean exam grade for the TA 1 class differs from the mean exam grade for the TA 2 class using the full procedure for testing two means (independent samples). Show all steps. Below is their sample data.

TA 1 Class:

89, 98, 85, 70, 24, 60, 45, 47, 66, 70, 53, 87, 80, 64, 81, 78, 96, 94, 99, 58, 92, 92, 70, 90, 73, 71, 82, 67, 97, 87, 100, 70, 62, 85, 73, 92, 86, 78, 83, 95

TA 2 Class:

98, 93, 88, 73, 25, 86, 73, 86, 86, 77, 62, 95, 82, 49, 80, 63, 80, 89, 95, 82, 94, 93, 91, 72, 77, 84, 74, 72, 85, 87, 100, 74, 63, 96, 76, 90, 87, 90, 94, 93

Explanation / Answer

Step 1 : Hypothesis

Null hypothesis : H0 : Mean exam grade for TA1 class is same as Mean exam grade for TA2 Class. 1 =2

Alternative Hypothesis : Ha : Mean exam grade for TA2 class is same as Mean exam grade for TA2 Class. 1 2

Step : 2 significance level alpha = 0.05

Step : 3 Test statistic calculation

Here for class TA1

Sample mean x1 = 77.225

Sample standard deviation s1 = 16.908

For class TA2

Sample mean x2 = 81.35

Sample standard deviation s2 =14.498

Pooled standard deviation sp = sqrt [{(n1 -1)s12 + (n2 -1)s22}/(n1 +n2-2)] = sqrt[(16.9082 + 14.4982)/2] = 15.749

Standard error of difference sed = sp * sqrt [1/n1 + 1/n2] = 15.749 * sqrt (1/40 + 1/40) = 3.522

Here test statistic

t = ( x2 -  x1)/sed = (81.35 - 77.225)/3.522 = 1.1712

Step :4

Critical test statistic for dF = 40 + 40 - 2 = 78 and two tailed test

tcritical = t0.05,78 = 1.991

Step 5 : Conclusion Here t < tcritical so we failed to reject the null hypothesis and can conclude that there is not significant difference between means of class TA1 and class TA2.

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