Question

Trimmed Mean Because the mean is very sensitive to extreme values, we say that it is not a resistant measure of center. By deleting some low values and high values, the trimmed mean is more resistant. To find the 10% trimmed mean for a data set, first arrange the data in order, then delete the bottom 10% of the values and delete the top 10% of the values, then calculate the mean of the remaining values. Use the axial loads (pounds) of aluminum cans listed below (from Data Set 30 “Aluminum Cans” in Appendix B) for cans that are 0.0111 in. thick. An axial load is the force at which the top of a can collapses. Identify any outliers, then compare the median, mean, 10% trimmed mean, and 20% trimmed mean.

247 260 268 273 276 279 281 283 284 285 286 288

289 291 293 295 296 299 310 504

Short answer

The summarized results of all measures are listed below.

Measures	Values
Outlier	504 pounds
Median	285.5 pounds
Mean	294.4 pounds
10% trimmed mean	285.4 pounds
20% trimmed mean	285.8 pounds

The mean is significantly different from other measures of the center as it is not resistant to extreme values.

Step-by-step solution

Given information

The axial loads of aluminum cans, which are 0.0111 inches thick, are known in pounds.

247 260 268 273 276 279 281 283 284 285 286 288 289 291 293 295 296 299 310 504

Find the outliers in the data

Outliers are those specific sets of observations that are quite different in measure compared to the rest of the data.

By observing the data, the value of 504 pounds is vastly different from other values.

Thus, the value 504 pounds is an outlier.

Find the median of the data

A median is computed using the following steps.

• Sort the data in ascending order.
• For even counts of observation, the average of two middle values gives the median.
• For odd counts of observation, the middle value gives the median.

The data is sorted as follows.

247
260
268
273
276
279
281
283
284
285
286
288
289
291
293
295
296
299
310
504

The number of observations is 20.

The middlemost observations are 285 and 286.

The median of the dataset is

$$\begin{gathered} M=\frac{285+286}{2} \\ =\frac{571}{2} \\ =285.5. \end{gathered}$$

Thus, the median is 285.5 pounds.

Find the mean of the data

The formula for the mean of n observations is shown below.

$\overline{x}=\frac{\sum x}{n}$

Substitute the values as follows.

$$\begin{gathered} \overline{x}=\frac{247+260+268+...+504}{20} \\ =\frac{5887}{20} \\ =294.35 \end{gathered}$$

Thus, the mean value is 294.4 pounds.

Find the 10% trimmed mean

The 10% trimmed mean is the mean value obtained after deleting the top and bottom 10% values in the data set.

The 10% of 20 observations is

$$\begin{gathered} 10\%\text{of}20=0.1\times 20 \\ =2. \end{gathered}$$

Delete the bottom and top 2 values from the data set.

268
273
276
279
281
283
284
285
286
288
289
291
293
295
296
299

The mean of these observations is computed as follows.

$$\begin{gathered} \overline{x}=\frac{268+273+...+299}{16} \\ =\frac{4566}{16} \\ =285.375 \end{gathered}$$

Thus, the 10% trimmed mean value is 285.4 pounds.

Find the 20% trimmed mean

The 20% trimmed mean is the mean value obtained after deleting the top and bottom 20% values in the data set.

The 20% of 20 observations is

$$\begin{gathered} 20\%\text{of}20=0.2\times 20 \\ =4. \end{gathered}$$

Delete the bottom and top four values from the data set.

276
279
281
283
284
285
286
288
289
291
293
295

The mean of the observations is computed as follows.

$$\begin{gathered} \overline{x}=\frac{276+279+...+295}{12} \\ =\frac{3430}{12} \\ =285.833 \end{gathered}$$

Thus, the 20% trimmed mean value is 285.8 pounds.

Compare the results

The summarized results are shown below.

Measures	Values
Outlier	504 pounds
Median	285.5 pounds
Mean	294.4 pounds
10% trimmed mean	285.4 pounds
20% trimmed mean	285.8 pounds

As per the results, there is only one outlier. The two measures of the center (median and mean) vary largely from the complete set of observations (285.5 and 294.34 ).

But the two trimmed mean measures are almost the same and close to the median value of 285.5.

Thus, it can be observed that the inclusion of the extreme value leads to significant diversion in the mean value. It means that the measure is not resistant to the outliers.

Also, the median value is resistant, so it does not change much and is comparable to the mean values, exclusive of extreme values.