Question

"Use z* value of either 2 or 1.96, round to 3 decimals."

(2 points) 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 102 92 74 126 11083 104 94 117 99 Test B 106 94 76 129 113 82 108 97 120 103 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. Two Sample t C. Matched Pairs (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. Yes B. No 2. Construct a 99% confidence interval for the mean of the differences. Again, use (Test A-Test B)

Thank you!

Answer 1

Solution:-

1)

a)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: u_d> 0

Alternative hypothesis: u_d < 0

Note that these hypotheses constitute a one-tailed test.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a matched-pairs t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard deviation of the differences (s), the standard error (SE) of the mean difference, the degrees of freedom (DF), and the t statistic test statistic (t).

Test A Test B A- B (d dbar)*2 106 102 92 0.49 74 0.49 126 110 0.09 13.69 104 108 1.69 94 120 1.69 Sum 1001 102林 1 Mean 100.1 102.8 2.7 2.01

s = sqrt [ (\sum (d_i - d)² / (n - 1) ]

s = 1.4944

SE = s / sqrt(n)

S.E = 0.4726

DF = n - 1 = 10 -1

D.F = 9

b)

t = [ (x₁ - x₂) - D ] / SE

t = - 5.71

c)

t_critical = - 2.822

Rejection region is t < - 2.822

where d_i is the observed difference for pair i, d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs.

Interpret results. Since the t-value (- 5.71) lies in the rejection region, hence we have to reject the null hypothesis.

Reject H0.

d) Yes, there is sufficient evidence that people do better on the second test.

2) 99% confidence interval for the mean of the differences is C.I = (- 4.236, - 1.164).

S.D

C.I = - 2.7 + 3.25 × 0.4726

C.I = - 2.7 + 1.53595

C.I = (- 4.236, - 1.164)