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. For the difference use Test A - Test B. Use at least 4 decimal places.

Person	A	B	C	D	E	F	G	H	I	J
Test A	121	90	96	71	100	115	107	111	83	102
Test B	122	95	95	72	99	111	113	110	86	102

1. Use a significance level to test the claim that people do better on the second test than they do on the first.
The test statistic is :

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

A. Yes
B. No

2. Construct a % confidence interval for the mean of the differences.
$< \mu <$

Answer 1

Differences(Test A - Test B): -1, -5, 1 -1, 1, 4, -6, 1, -3, 0 we calculate sample mean and std deviation from given data. Sample Mean, d^- = Sigma(d)/n = -9/10 = -0.9 Sample Variance, s^2 = Sigma(d - d^-)^2/n - 1 = 82.9/9 = 9.211111 Sample std dev, s = squareroot s^2 = squareroot 9.211111 almostequalto 3.034981 Test: mu_d < 0 with t dis^n Hypothesis test: H_0: mu_d = 0 H_a: mu_d < 0 (Null Hypothesis H_0: mu_d greaterthanorequalto 0 is also correct) (Alternative Hypothesis, also called H_1) this is lower tailed test. d^- = -0.9 s = 3.034981 n = 10 Significance Level, alpha = 0.01 degree of freedom df = n - 1 = 9 test statistic, t_observed = d^- mu_d/s squareroot n = -0.9 - 0/3.034981/squareroot 10 almostequalto - 0.938 Test statistic = -0.938 using t - table, we can observe that the p - value is between 0.15 and 0.20 Using excel we find p - value = P(t < - 0.938) = t.dist( - 0.938, 9, 1) = 0.1864273785 p - value = 0.1864