Concept explainers
Eat your kale: Kale is a type of cabbage commonly found in salad and used in cooking in many parts of the world. Six measurements were made of the mineral content (in percent) of kale: with the following results-
26.1 17.5 15.4 164 15.1 12.8
It turns out that the value 26.1 came from a specimen that the investigator forgot to wash before measuring.
- The data contain an outlier that is clearly a mistake. Eliminate the outlier, then construct a 95% confidence interval for the
mean mineral content from the remaining values. - Leave the outlier in and construct the 95% confidence interval. Are the results noticeably different? Explain why it is important to check data for outliers.
a.
To find the confidence interval for the mean mineral content from the remaining values. (not including the outlier value)
Answer to Problem 37E
Explanation of Solution
Given information :
Data points:
26.1 17.5 15.4 16.4 15.1 12.8
Calculation:
x | ||
17.5 | 2.06 | 4.2436 |
15.4 | -0.04 | 0.0016 |
16.4 | 0.96 | 0.9216 |
15.1 | -0.34 | 0.1156 |
12.8 | -2.64 | 6.9696 |
Mean:
Sample standard deviation:
95% confidence interval can be computed as:
Since, sample size is less than 30 and population standard deviation is not known so , t-test will be used.
Degree of freedom = 5-1
= 4
t-value = 2.77
b.
To check the claim Whether online course and traditional course are same.
Answer to Problem 37E
CI =
Explanation of Solution
Calculation:
x | ||
26.1 | 8.89 | 79.0321 |
17.5 | 0.29 | 0.0841 |
15.4 | -1.81 | 3.2761 |
16.4 | -0.81 | 0.6561 |
15.1 | -2.11 | 4.4521 |
12.8 | -4.41 | 19.4481 |
Mean:
Sample standard deviation:
95% confidence interval can be computed as:
Since, sample size is less than 30 and population standard deviation is not known so, t-test will be used.
Degree of freedom = 6-1
= 5
t-critical value = 2.57 (t-value is taken from t-table)
As compare to part a) the confidence interval width has increased since, it is more likely to contain the true population mean.
Since, outlier is point which is far away from the other data points or mean value and it affects all descriptive parameters and normality of the distribution as well.
So, it is necessary to check the outlier in the data set
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Chapter 8 Solutions
Elementary Statistics
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