Null hypothesis | All data values come from the same normal population |
---|---|
Alternative hypothesis | Smallest data value is an outlier |
Significance level | α = 0.05 |
Variable | N | Mean | StDev | Min | Max | G | P |
---|---|---|---|---|---|---|---|
BreakStrength | 14 | 123.4 | 46.3 | 12.4 | 193.1 | 2.40 | 0.044 |
Variable | Row | Outlier |
---|---|---|
BreakStrength | 10 | 12.38 |
In these results, the null hypothesis states that all data values come from the same normal population. Because the p-value is 0.044, which is less than the significance level of 0.05, the decision is to reject the null hypothesis and conclude that an outlier exists.
If the test identifies an outlier in the data, then Minitab displays an outlier table. Use the outlier table to determine the value of the outlier, and the row in the worksheet that contains the outlier.
Null hypothesis | All data values come from the same normal population |
---|---|
Alternative hypothesis | Smallest data value is an outlier |
Significance level | α = 0.05 |
Variable | N | Mean | StDev | Min | Max | G | P |
---|---|---|---|---|---|---|---|
BreakStrength | 14 | 123.4 | 46.3 | 12.4 | 193.1 | 2.40 | 0.044 |
Variable | Row | Outlier |
---|---|---|
BreakStrength | 10 | 12.38 |
In these results, the value of the outlier is 12.38, and it is in row 10.
Use the outlier plot to visually identify an outlier in the data. If an outlier exists, Minitab represents it on the plot as a red square. Try to identify the cause of any outliers. Correct any data–entry errors or measurement errors. Consider removing data values for abnormal, one-time events (also called special causes).