In: Statistics and Probability
Q1) How effective are laundry detergents at dissolving dirt? To answer this question, seven (N = 7) participants were asked to wash their laundry four different times using four (k = 4) different detergents. After every laundry wash, the amount of dirt present was measured in micrograms. These data are presented in the table below.
ID | No detergent | Purex | Downy | Tide |
---|---|---|---|---|
Participant 1 | 103 | 84 | 70 | 51 |
Participant 2 | 135 | 51 | 164 | 5 |
Participant 3 | 102 | 110 | 88 | 19 |
Participant 4 | 124 | 67 | 111 | 18 |
Participant 5 | 105 | 119 | 73 | 58 |
Participant 6 | 139 | 108 | 119 | 50 |
Participant 7 | 170 | 207 | 20 | 82 |
a. Carryout a one way repeated measures ANOVA to examine if there is any affect of detergent type on dirt collection. Please present the results in a source table. (assume compound sphericity).
b. Is using no detergent less affective compared to all other detergents combined?? Conduct the appropriate test and report your results.
c. Calculate how large the effect is in b.
a)
Null hypothesis: There is no significant difference in means of dirt collection on four treatment levels of Detergent.
Alternative hypothesis: There is a significant difference in means of dirt collection on four treatment levels of Detergent.
The p-value for the F-test is 0.003 and less than 0.05 level of significance. Hence, we can reject the null hypothesis and conclude that there is any effect of detergent type on dirt collection.
b)
Null Hypothesis: Mean dirt of detergent is equal to the Mean dirt of non-detergent.
Alternative Hypothesis: Mean dirt of detergent is less than the Mean dirt of non-detergent.
By dividing the data into two groups detergent and non-detergent, we conduct the two-sample t-test and get the result
The p-value is 0.014 and less than 0.05 level of significance. Hence, we can reject the null hypothesis and conclude that using no detergent is less effective compared to all other detergents combined.
c)
The effect of using Cohen's d= | -45.7/ 45.0905|=1.0135.