x:1,2,3,4,5,6,7,8,9 ---------------------------------------------------------------------------------- y:0.8,2.1,3.1,3.8,5.2,8.4,6.9,8.1,9 (which point would be considered an outlier? my ANS:(9,9)? IS IT RIGHT?
@Ms.Sue?
I don't see any outliers. All the values appear to lie near a sloping line.
Plot the points and see whether any is obviously more distant than the others.
To determine if a point is an outlier, you typically look for values that are significantly different from the rest of the data. One common method to identify outliers is by using the concept of the interquartile range (IQR) and the fences.
To identify the outlier, follow these steps:
1. Sort the x-values in ascending order: 1, 2, 3, 4, 5, 6, 7, 8, 9.
2. Sort the y-values in ascending order, while keeping their correspondence with the x-values: 0.8, 2.1, 3.1, 3.8, 5.2, 6.9, 8.1, 8.4, 9.
3. Find the first quartile (Q1) and the third quartile (Q3) of the y-values. Q1 is the median of the lower half of the data, and Q3 is the median of the upper half. In this case, Q1 ≈ 3.1 and Q3 ≈ 8.4.
4. Calculate the IQR by subtracting Q1 from Q3: IQR = Q3 - Q1 ≈ 8.4 - 3.1 ≈ 5.3.
5. Determine the lower inner fence (LIF) and upper inner fence (UIF): LIF = Q1 - 1.5 * IQR and UIF = Q3 + 1.5 * IQR. In this case, LIF ≈ -2.35 and UIF ≈ 13.85.
6. Check if any y-value falls above the UIF or below the LIF. If it does, it would be considered an outlier. In this case, none of the y-values fall outside the UIF or below the LIF, so there are no outliers.
Therefore, according to the calculations, the point (9, 9) would not be considered an outlier based on the given data.