## Related questions with answers

The data in previous exercise follow a Normal distribution quite closely except for one patient with HAY angle 50 degrees, a high outlier. Data available below.

(a) Find the 95% confidence interval for the population mean based on the 37 patients who remain after you drop the outlier.

(b) Compare your interval in (a) with your interval from previous exercise. What is the most important effect of removing the outlier? In general, you shouldn't remove an outlier unless you have reason to believe it is an error or unless you are interested only in "typical" cases. For a more in-depth discussion of how to handle outliers in various situations, read the discussion on outliers in previous chapters.

28 32 25 34 38 26 25 18 30 26 28 13 20

21 17 16 21 23 14 32 25 21 22 20 18 26

16 30 30 20 50 25 26 28 31 38 32 21

Solution

Verified(a) The mean is the sum of all values divided by the number of values:

$\overline{x}=24.7568$

The variance is the expected value of the squared deviation from the mean. The standard deviation is the square root of the variance.

$s=6.3395$

The degrees of freedom is the sample size decreased by 1.

$df=n-1=37-1=36>30$

The critical t-value is the value in the row with $df=30$ and in the column with confidence level 95% of table C:

$t^*=2.042$

The confidence interval is:

$\overline{x}-t_{\alpha/2}\cdot \dfrac{s}{\sqrt{n}}\text{ to }\overline{x}+t_{\alpha/2}\cdot \dfrac{s}{\sqrt{n}}$

Fill in the known values:

$24.7568-2.042\cdot \dfrac{6.3395}{\sqrt{37}}\text{ to }24.7568+2.042\cdot \dfrac{6.3395}{\sqrt{37}}$

Simplify:

$22.6286\text{ to }26.8850$

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