Males | Females |
620 | 660 |
570 | 590 |
540 | 540 |
580 | 560 |
590 | 610 |
580 | 590 |
480 | 610 |
620 | 650 |
570 | 600 |
610 | 620 |
590 | 630 |
570 | 640 |
610 | 590 |
590 | 640 |
570 | 580 |
550 | 560 |
530 | 570 |
560 | 560 |
620 | 600 |
520 | 600 |
560 | 590 |
620 | 590 |
580 | 590 |
610 | 630 |
530 | 560 |
480 | 600 |
590 | 560 |
620 | 560 |
590 | 560 |
580 | 560 |
NOTE: use Excel and submit screenshots and directions on how to do it on excel
These are the scores of males and females SAT scores
1. Construct the null and the alternative hypotheses to test if females outscore males on writing tests.
2. Assuming the difference in scores is normally distributed, calculate the value of the test statistic. Do not assume that the population variances are equal.
3. Implement the test at alpha=0.01 and interpret your results.
Ho:means are not significantly different.
H1:means are significantly different.
Data-----data analysis toolpack------t test -two sample assuming unequal varriances
t-Test: Two-Sample Assuming Unequal Variances | ||
Males | Females | |
Mean | 574.3333 | 593.3333 |
Variance | 1466.782 | 988.5057 |
Observations | 30 | 30 |
Hypothesized Mean Difference | 0 | |
df | 56 | |
t Stat | -2.10021 | |
P(T<=t) one-tail | 0.020113 | |
t Critical one-tail | 1.672522 | |
P(T<=t) two-tail | 0.040227 | |
t Critical two-tail | 2.003241(5%) |
The absolute value of the calculated t is smaller than critical value (2.1002<2.678(1%)), so the means are not significantly different.
Get Answers For Free
Most questions answered within 1 hours.