In: Statistics and Probability
A savvy business owner wanted to assess whether the type of fragrance influenced the amount of money spent. He tried peppermint, lavender, male cologne, and a floral perfume in his four stores. Amount of money spent (in hundreds) is reported for each type of fragrance. Conduct a one-way repeated measures ANOVA to determine whether fragrance influences total amount of money spent.
Peppermint |
Lavender |
Cologne |
Floral |
4.2 |
3.3 |
5.1 |
3.9 |
5.1 |
1.8 |
4.9 |
4.3 |
4.8 |
3.0 |
3.2 |
3.5 |
6.2 |
3.2 |
4.0 |
3.7 |
3.1 |
2.3 |
3.8 |
2.1 |
4.5 |
2.9 |
4.7 |
2.3 |
4.8 |
3.5 |
3.8 |
1.0 |
3.7 |
4.7 |
4.1 |
2.8 |
2.8 |
3.1 |
3.5 |
4.2 |
Go To data tab in excel choose data analysis and choose Anova: Two-Factor Without Replication.
Rows refers to test scores for each of the flavours. Notice that the p-value or probability of obtaining an F statistic of 2.355 or larger when the null hypothesis is true is 0.478. Since the p-value is larger than the specified alpha of 0.05, the null hypothesis is not rejected; there is a no significant statistical difference between the means of each fragrance influences total amount of money spent.
Columns refers to the four categories of test amount: mathematics, reading, science and social studies. The p-value in cell M24 is very close to 0. This means that the probability of obtaining an F statistic of 3.01 or larger when the null hypothesis is true is also very close to 0. Since the p-value is less than the specified alpha of 0.05 and the calculated F statistic is larger than the value for F crit, the null hypothesis is rejected. There is a significant statistical difference in the calculated means of the four categories.