Ten randomly selected people took an IQ test A, and next day they took a very si

Question

Ten randomly selected people took an IQ test A, and next day they took a very similar IQ test B. Their scores are shown in the table below.
Person   A   B   C   D   E   F   G   H   I   J
Test A   72   96   125   107   113   110   94   103   86   91
Test B   70   95   127   106   113   114   97   108   89   96

1. Consider (Test A - Test B). Use a 0.01 significance level to test the claim that people do better on the second test than they do on the first. (Note: You may wish to use software.)
(a) What test method should be used?

A. Two Sample z
B. Matched Pairs
C. Two Sample t

(b) The test statistic is

(c) The critical value is

(d) Is there sufficient evidence to support the claim that people do better on the second test?

A. No
B. Yes

2. Construct a 99% confidence interval for the mean of the differences. Again, use (Test A - Test B).

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Explanation / Answer

SOLUTION

Since the question is whether Test B score is significantly greater than the Test A score, the decision variable, say D = Test B score - Test A score

Assumption: D follows Normal Distribution with mean µ and variance 2. Since 2 is not known, it needs be estimated by the sample variance, s2 = {Sum(d - meanD)2}/(n - 1), where n = sample size (10 in this case).

Null Hypothesis H0 : µ = 0   vs Alternative H1: µ > 0

Test Statistics: t = (n)(meanD - 0)/s

Under H0, t follows t-distribution with (n - 1) degrees of freedom.

Decision Criterion: Reject H0 at % level of significance if calculated value of t, say tcal, is greater than tn – 1, , where = level of significance, 1% in this case.and tn – 1, = upper % point of t-distribution with (n - 1) degrees of freedom.

Calculations

MeanD = (-2 + 0 + 2 + -1 + 0 + 4 + 0 + 5 + 3 + 5)/10 = 16/10 = 1.6 (meanD)

Sample Variance: s2 = (84 – 25.6)/9 = 58.4/9 = 6.4889 or s = 6.4889 = 2.54 = s

Test Statistic: t = (10)(1.6 - 0)/2.54 = 1.96 = tcal

Upper 1% point of t distribution with (n – 1) degrees of freedom = 2.821 = tn – 1, .01

Inference: Since tcal< tn – 1, .01, H0is accepted.

Decision: There is no statistical evidence to infer that Test B scores are more than Test A scores

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