[4 points] In a recent poll, 10 local students were asked how many beers they drink in a typical week. The responses, in no particular order, were:
4, 10, 6, 14, 12, 17, 30, 15, 12, 0
At a 90% confidence level, is there an outlier in the data given above? At a 95% confidence level?
To determine if there is an outlier in the data given above, we can use the concept of z-scores. A z-score measures the number of standard deviations a data point is away from the mean of a dataset.
To start, let's calculate the mean and standard deviation of the given data:
Mean:
To find the mean, we sum up all the values and divide it by the total number of students:
Mean = (4 + 10 + 6 + 14 + 12 + 17 + 30 + 15 + 12 + 0) / 10 = 120 / 10 = 12
Standard Deviation:
To calculate the standard deviation, we need to find the variance first. The variance is the average of the squared differences between each data point and the mean.
Variance = [(4 - 12)^2 + (10 - 12)^2 + (6 - 12)^2 + (14 - 12)^2 + (12 - 12)^2 + (17 - 12)^2 + (30 - 12)^2 + (15 - 12)^2 + (12 - 12)^2 + (0 - 12)^2] / 10
= (64 + 4 + 36 + 4 + 0 + 25 + 324 + 9 + 0 + 144) / 10
= 610 / 10
= 61
Standard Deviation = √(Variance) = √(61) ≈ 7.81
Now that we have the mean and standard deviation, we can calculate the z-score for each data point in the given dataset. The formula for calculating the z-score is:
z = (x - mean) / standard deviation
Using this formula, we can calculate the z-scores for the data points:
z1 = (4 - 12) / 7.81 ≈ -1.03
z2 = (10 - 12) / 7.81 ≈ -0.26
z3 = (6 - 12) / 7.81 ≈ -0.77
z4 = (14 - 12) / 7.81 ≈ 0.26
z5 = (12 - 12) / 7.81 = 0
z6 = (17 - 12) / 7.81 ≈ 0.64
z7 = (30 - 12) / 7.81 ≈ 2.31
z8 = (15 - 12) / 7.81 ≈ 0.38
z9 = (12 - 12) / 7.81 = 0
z10 = (0 - 12) / 7.81 ≈ -1.54
Now, let's consider the two confidence levels given: 90% and 95%.
At a 90% confidence level:
To determine if there is an outlier at a 90% confidence level, we need to calculate the critical z-score. The critical z-score corresponds to the desired confidence level and can be found in the z-table.
At a 90% confidence level, we will use a critical z-score of 1.645 (approximately). If any z-score exceeds this critical value, it would be considered an outlier.
In the given dataset, the z-scores for the individual data points do not exceed the critical z-score of 1.645. Therefore, at a 90% confidence level, there is no outlier in the data.
At a 95% confidence level:
Similarly, to determine if there is an outlier at a 95% confidence level, we need to calculate the critical z-score.
At a 95% confidence level, the critical z-score is 1.96 (approximately). If any z-score exceeds this critical value, it would be considered an outlier.
In the given dataset, the z-scores for the individual data points do not exceed the critical z-score of 1.96. Therefore, at a 95% confidence level, there is no outlier in the data either.
In conclusion, there are no outliers in the given data at both 90% and 95% confidence levels.