In: Statistics and Probability
Consider Dataset C for answering the questions that follows below.
| A | Male | 4.80 | |
| 5 | A | Male | 4.24 | 
| 3 | A | Male | 3.45 | 
| 4 | A | Male | 1.78 | 
| 2 | A | Male | 1.06 | 
| 9 | B | Male | 4.12 | 
| 6 | B | Male | 3.91 | 
| 7 | B | Male | 3.42 | 
| 10 | B | Male | 2.92 | 
| 8 | B | Male | 1.25 | 
| 15 | C | Female | 4.84 | 
| 11 | C | Female | 4.45 | 
| 13 | C | Female | 4.13 | 
| 12 | C | Female | 1.84 | 
| 14 | C | Female | 1.64 | 
ANOVA
| A | B | C | 
| 4.8 | 4.12 | 4.84 | 
| 4.24 | 3.91 | 4.45 | 
| 3.45 | 3.42 | 4.13 | 
| 1.78 | 2.92 | 1.84 | 
| 1.06 | 1.25 | 1.64 | 
Using Excel, go to Data, select Data Analysis, choose Anova: Single Factor. Group by columns at alpha 0.05.
H0: µ1 = µ2 = µ3, Mean responses of the three teams are same
H1: Mean responses of the three teams are different
| ANOVA | ||||||
| Source of Variation | SS | df | MS | F | P-value | F crit | 
| Between Groups | 0.279 | 2 | 0.140 | 0.068 | 0.935 | 3.885 | 
| Within Groups | 24.699 | 12 | 2.058 | |||
| Total | 24.979 | 14 | 
p-value = 0.935
Since p-value is more than 0.05, we do not reject the null hypothesis and conclude that µ1 = µ2 = µ3.
So, mean responses of the three teams are same.
t-test
Using Excel, go to Data, select Data Analysis, choose t-Test: Two-Sample Assuming Unequal Variances at alpha 0.01.
| t-Test: Two-Sample Assuming Unequal Variances | ||
| Male | Female | |
| Mean | 3.095 | 3.380 | 
| Variance | 1.719 | 2.310 | 
| Observations | 10 | 5 | 
| Hypothesized Mean Difference | 0 | |
| df | 7 | |
| t Stat | -0.358 | |
| P(T<=t) one-tail | 0.365 | |
| t Critical one-tail | 2.998 | |
| P(T<=t) two-tail | 0.731 | |
| t Critical two-tail | 3.499 | 
H0: µ1 = µ2, Mean male responses is same as the mean female responses
H1: µ1 ≠ µ2, Mean male responses is different from the mean female responses
p-value = 0.731
Since p-value is more than 0.01, we do not reject the null hypothesis and conclude that µ1 = µ2.
So, mean male responses is same as the mean female responses.