Q. 68 - The Practice of Statistics for AP

Question

Hallux abducto valgus (call it HAV) is a deformation of the big toe that is fairly uncommon in youth and often requires surgery. Doctors used $X$-rays to measure the angle (in degrees) of deformity in a random sample of patients under the age of $21$ who came to a medical center for surgery to correct HAV. The angle is a measure of the seriousness of the deformity. For these $21$ patients, the mean HAV angle was $24.76$ degrees and the standard deviation was $6.34$ degrees. A dot plot of the data revealed no outliers or strong skewness.

(a) Construct and interpret a $90\%$ confidence interval for the mean HAV angle in the population of all such patients.

(b) Researchers omitted one patient with an HAV angle of $50$ degrees from the analysis due to a measurement issue. What effect would including this outlier have on the confidence interval in (a)? Justify your answer. 

Step-by-Step Solution

Answer

(a) We are $90\%$ confident that the true population mean is between $22.3735$ degrees and $27.1465$ degrees.

(b) The confidence interval from part (a) could also not be generalized for the population, because the normal requirement for the use of the $t$-distribution is not satisfied.

Part (a) Step 1: Given Information

Given

$\overline{x}=24.76$

$s=6.34$

$n=21$

$c=90\%$

Part (b) Step 2: Explanation

Determine the t-value by looking in the row starting with degrees of freedom  and in the row with $c=90\%$ in table B :

$t^{*}=1.725$

The margin of error is then:

$$\begin{gathered} E=t^{*}·\frac{s}{\sqrt{n}} \\ =1.725\times \frac{6.34}{\sqrt{21}} \\ \approx 2.3865 \end{gathered}$$

Then the confidence interval becomes:

$$\begin{gathered} 22.3735=24.76-2.3865 \\ =\overline{x}-E<\mu <\overline{x}+E \\ =24.76+2.3865 \\ =27.1465 \end{gathered}$$

$22.3735<\mu <27.1465$

Part (b) Step 2: Given Information

Given

$\overline{x}=24.76$

$s=6.34$

$n=21$

$c=90\%$

Part (b) Step 2: Explanation

The mean and standard deviation are strongly influenced by outliers and thus adding the outlier to the other data value will significantly increase both the mean and the standard deviation.