In: Statistics and Probability
Five swimming floaters are woven using the fibre supplied by 4 different Suppliers. You perform the experiment of testing the strength of each of the 20 swimming floaters by collecting the tensile strength measurement of each swimming floaters. Results are organized by group along with the sample mean and the sample standard deviation of each group in the table below.
Supplier 1 |
Supplier 2 |
Supplier 3 |
Supplier 4 |
|
18.5 |
26.3 |
20.6 |
25.4 |
|
24.0 |
25.3 |
25.2 |
19.9 |
|
17.2 |
24.0 |
20.8 |
22.6 |
|
19.9 |
21.2 |
24.7 |
17.5 |
|
18.0 |
24.5 |
22.9 |
20.4 |
|
Sample mean |
19.52 |
24.26 |
22.84 |
21.16 |
Sample Standard Deviation |
2.69 |
1.92 |
2.13 |
2.98 |
At the 0.05 significance level, is there a difference in mean sales?
Five swimming floaters are woven using the fibre supplied by 4 different Suppliers. We have to perform the experiment of testing the strength of each of the 20 swimming floaters by collecting the tensile strength measurement of each swimming floaters. At 0.05 significamce level we have to check whetherv there is a difference in mean sales. Thus we consider the following hypothesis :
Null hypothesis
All means are equal
Alternative hypothesis At least one mean is different
Thus we perform ANOVA in MINITAB to get the desired result. To do
so we run the following steps:
1) Enter the given data of the four supplioers in four different columns.
2)Click stat then ANOVA the One Way
3) Select the four varieties in response (in seperate columns)
4)Click OK
Thus we get the following ANOVA Table:
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Factor 3 63.29
21.095 3.46 0.041
Error 16 97.50 6.094
Total 19 160.79
We shall reject the null at 0.05 significance level if p value<= 0.05
Thus here pvalue= 0.041< significance level so we reject the null hypothesis
So we conclude that there is a difference in mean sales.