In: Statistics and Probability
The table to the right shows the cost per ounce (in dollars) for a random sample of toothpastes exhibiting very good stain removal, good stain removal, and fair stain removal. At alpha=0.05, can you conclude that the mean costs per ounce are different? Perform a one-way ANOVA test by completing parts a through d. Assume that each sample is drawn from a normal population, that the samples are independent of each other, and that the populations have the same variances. |
Very good stain removal Good stain removal Fair
stain removal
0.33 0.64 0.34
0.35 2.76 1.16
0.41 0.60 0.60
1.51 0.98
0.48 0.75
0.51 1.34
Critical Values F=3.89
Rejection region F > 3.89
(c.) Calculate the test statistic
Decide to reject or fail to reject the null hypothesis and interpret the decision in the context of the original claim.
We solve this problem using excel Data Analysis toolpack.
Excel => Data => Data Analysis => Anova : Single factor => select input range => Lables in first row => output range => ok
Very good stain removal | Good stain removal | Fair stain removal |
0.33 | 0.64 | 0.34 |
0.35 | 2.76 | 1.16 |
0.41 | 0.6 | 0.6 |
1.51 | 0.98 | |
0.48 | 0.75 | |
0.51 | 1.34 |
k = number of group
N : total number of observation
Ho : The mean costs per ounce are same.
V/s
H1: The mean costs per ounce are different
Anova: Single Factor | ||||
SUMMARY | ||||
Groups | Count | Sum | Average | Variance |
Very good stain removal | 6 | 3.59 | 0.598333 | 0.204417 |
Good stain removal | 6 | 7.07 | 1.178333 | 0.674977 |
Fair stain removal | 3 | 2.1 | 0.7 | 0.1756 |
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Between Groups | 1.094327 | 2 | 0.547163 | 1.382841 | 0.288115 | 3.885294 |
Within Groups | 4.748167 | 12 | 0.395681 | |||
Total | 5.842493 | 14 |
Here test stat = F = 1.3828
Critical value = Fc = 3.89
Here F = 1.3828 < Fc = 3.89 at 0.05 level of significance then we fail to reject Ho.
Conclude that the mean costs per ounce are same.