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.
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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