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Q3: Consider the Math and Writing test scores for twelve students: Student 1: 540 M, 474...

Q3: Consider the Math and Writing test scores for twelve students:
Student 1: 540 M, 474 W; Student 2: 432 M, 380 W; Student 3: 528 M, 463 W; Student 4: 574 M, 612 W; Student 5: 448 M, 420 W; Student 6: 502 M, 526 W; Student 7: 480 M, 430 W; Student 8: 499 M, 459 W; Student 9: 610 M, 615 W; Student 10: 572 M, 541 W; Student 11: 390 M, 335 W; Student 12: 593 M, 613 W.
Make two assumptions, for convenience, that are not actually satisfied here:
Assumption 1: the sample sizes involved are large enough that our test statistics generally follow a normal distribution when the population variance is known;
Assumption 2: the sample sizes are large enough to imply that the relevant t-distribution and the standard normal distribution are identical.
a. Assume a significance level of 0.05. Formally state and conduct a two-sided hypothesis test about the difference in Math and Writing test scores using a test statistic ignoring the fact that the samples are matched.
b. Assume a significance level of 0.05. Formally state and conduct a two-sided hypothesis test about the difference in Math and Writing test scores using a test

statistic for matched samples.

Homework Answers

Answer #1

data

Student M W
1 540 474
2 432 380
3 528 463
4 574 612
5 448 420
6 502 526
7 480 430
8 499 459
9 610 615
10 572 541
11 390 335
12 593 613

a)

Using Excel

data -> data analysis ->

t-Test: Two-Sample Assuming Unequal Variances
M W
Mean 514 489
Variance 4661.2727 8735.8182
Observations 12 12
Hypothesized Mean Difference 0
df 20
t Stat 0.7482
P(T<=t) one-tail 0.2315
t Critical one-tail 1.7247
P(T<=t) two-tail 0.4630
t Critical two-tail 2.0860

2-tailed p-value = 0.4630 > alpha

hence we fail to reject the null hypothesis

b)

t-Test: Paired Two Sample for Means
M W
Mean 514 489
Variance 4661.2727 8735.8182
Observations 12 12
Pearson Correlation 0.9422
Hypothesized Mean Difference 0
df 11
t Stat 2.3374
P(T<=t) one-tail 0.0197
t Critical one-tail 1.7959
P(T<=t) two-tail 0.0394
t Critical two-tail 2.2010

p-value = 0.0394 < 0.05

hence we reject the null hypothesis

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