Question

The U.S. Army commissioned a study to assess how deeply a bullet penetrates ceramic body armor. In the standard test, a cylindrical clay model is layered under the armor vest. A projectile is then fired, causing an indentation in the clay. The deepest impression in the clay is measured as an indication of survivability of someone wearing the armor. Here is data from one testing organization under particular experimental conditions; measurements (in mm) were made using a manually controlled digital caliper:

\(\begin{array}{l}{\rm{22}}{\rm{.4 23}}{\rm{.6 24}}{\rm{.0 24}}{\rm{.9 25}}{\rm{.5 25}}{\rm{.6 25}}{\rm{.8 26}}{\rm{.1 26}}{\rm{.4 26}}{\rm{.7 27}}{\rm{.4 27}}{\rm{.6 28}}{\rm{.3 29}}{\rm{.0}}\\{\rm{ 29}}{\rm{.1 29}}{\rm{.6 29}}{\rm{.7 29}}{\rm{.8 29}}{\rm{.9 30}}{\rm{.0 30}}{\rm{.4 30}}{\rm{.5 30}}{\rm{.7 30}}{\rm{.7 31}}{\rm{.0 31}}{\rm{.0 31}}{\rm{.4 31}}{\rm{.6 31}}{\rm{.7 31}}{\rm{.9 31}}{\rm{.9 }}\\{\rm{32}}{\rm{.0 32}}{\rm{.1 32}}{\rm{.4 32}}{\rm{.5 32}}{\rm{.5 32}}{\rm{.6 32}}{\rm{.9 33}}{\rm{.1 33}}{\rm{.3 33}}{\rm{.5 33}}{\rm{.5 33}}{\rm{.5 33}}{\rm{.5 33}}{\rm{.6 33}}{\rm{.6 33}}{\rm{.8 33}}{\rm{.9 }}\\{\rm{34}}{\rm{.1 34}}{\rm{.2 34}}{\rm{.6 34}}{\rm{.6 35}}{\rm{.0 35}}{\rm{.2 35}}{\rm{.2 35}}{\rm{.4 35}}{\rm{.4 35}}{\rm{.4 35}}{\rm{.5 35}}{\rm{.7 35}}{\rm{.8 36}}{\rm{.0 36}}{\rm{.0 36}}{\rm{.0 36}}{\rm{.1 36}}{\rm{.1 }}\\{\rm{36}}{\rm{.2 36}}{\rm{.4 36}}{\rm{.6 37}}{\rm{.0 37}}{\rm{.4 37}}{\rm{.5 37}}{\rm{.5 38}}{\rm{.0 38}}{\rm{.7 38}}{\rm{.8 39}}{\rm{.8 41}}{\rm{.0 42}}{\rm{.0 42}}{\rm{.1 44}}{\rm{.6 48}}{\rm{.3 55}}{\rm{.0}}\end{array}\)

a. Construct a box plot of the data and comment on interesting features.

b. Construct a normal probability plot. Is it plausible that impression depth is normally distributed? Is a normal distribution assumption needed in order to calculate a confidence interval or bound for the true average depth m using the foregoing data? Explain.

c. Use the accompanying Minitab output as a basis for calculating and interpreting an upper confidence bound for m with a confidence level of\({\rm{99\% }}\)

Variable Count Mean SE Mean StDev Depth\({\rm{83 33}}{\rm{.370 0}}{\rm{.578 5}}{\rm{.268}}\)

Q1 Median Q3 IQR

\({\rm{30}}{\rm{.400 33}}{\rm{.500 36}}{\rm{.000 5}}{\rm{.600}}\)

Short answer

a) Skewed to the right (or positively skewed)

b) Plausible No normal assumption needed

c) The boundaries of the confidence interval then become \((31.8808,34.8590)\)

Step-by-step solution

To construct a boxplot

Given:

\({\rm{n = 83}}\)

\(\begin{array}{l}{\rm{33}}{\rm{.6,33}}{\rm{.8,33}}{\rm{.9,34}}{\rm{.1,34}}{\rm{.2,34}}{\rm{.6,34}}{\rm{.6,35,35}}{\rm{.2,35}}{\rm{.2,35}}{\rm{.4,35}}{\rm{.4,35}}{\rm{.4,}}\\{\rm{35}}{\rm{.5,35}}{\rm{.7,35}}{\rm{.8,36,30,36,36}}{\rm{.1,36}}{\rm{.1,30}}{\rm{.2,36}}{\rm{.4,30}}{\rm{.6,37,37}}{\rm{.4,37}}{\rm{.5,37}}{\rm{.5,38,}}\\{\rm{38}}{\rm{.7,38}}{\rm{.8,39}}{\rm{.8,41,42,42}}{\rm{.1,44}}{\rm{.6,48}}{\rm{.3,55}}\end{array}\)

The minimum is\(22.4\)

Since the number of data values is odd, the median is the middle value of the sorted data set:

\({\rm{M - }}{{\rm{Q}}_{\rm{2}}}{\rm{ - 33}}{\rm{.5}}\)

The first quartile is the median of the data values below the median (or at $25 \%$ of the data):

\({{\rm{Q}}_{\rm{1}}}{\rm{ - 30}}{\rm{.4}}\)

The third quartile is the median of the data values above the median (or at $75 \%$ of the data):

\({{\rm{Q}}_{\rm{3}}}{\rm{ - 36}}\)

The maximum is 55 .

BOXPLOT

The whiskers of the boxplot are at the minimum and maximum value. The box starts at the first quartile, ends at the third quartile and has a vertical line at the median.

The first quartile is at \(25\% \)of the sorted data list, the median at \(50\% \)and the third quartile at \(75\% \)

The distribution is skewed to the right (or positively skewed), because the box in the boxplot is to the left between the whiskers.

Hence Skewed to the right (or positively skewed)

To construct a normal probability plot

(b)

Given:

\({\rm{n = 83}}\)

\(\begin{array}{l}33.0,33.8,33.9,34.1,34.2,34.6,34.6,35,35.2,35.2,35.4,\\35.4,35.4,35.5,35.7,35.8,36,36,36,30.1,36.1,30.2,30.4,36.6,37,37.4,\\37.5,37.5,38,38.7,38.8,39.8,41,42,42.1,44.6,48.3,55\end{array}\)

NORMAL PROBABILITY PLOT

The data values are on the horizontal axis and the standardized normal scores are on the vertical axis.\({\rm{\{ 1,2,3, \ldots }}{\rm{.,n\} }}\).

The smallest standardized score corresponds with the smallest data value, the second smallest standardized score corresponds with the second smallest data value, and so on.

If the pattern in the normal probability plot is roughly linear and does not contain strong curvature, then the distribution is approximately normal. impression depth is normally distributed, because the rest of the plot is roughly linear.

True average depth

Central limit theorem: If the sample size is large,then the sampling distribution of the sample mean\({\rm{\bar x}}\)is approximately normal. distribution assumption needed in order to calculate a confidence interval or bound for the true average depth\({\rm{\mu }}\)

Hence Plausible No normal assumption needed.

To calculate and interpret an upper confidence bound

Given:

\(\begin{array}{l}{\rm{n = 83 }}\\{\rm{c = 99\% = 0}}{\rm{.99}}\end{array}\)

\(\begin{array}{l}33.6,33.8,33.9,34.1,34.2,34.6,34.6,35,35.2,35.2,35.4,\\35.4,35.4,35.5,35.7,35.8,30,30,30,30.1,30.1,30.2,30.4,30.0,37,37.4,\\37.5,37.5,38,38.7,38.8,39.8,41,42,42.1,44.6,48.3,55\end{array}\)

Central limit theorem: If the sample size is large , then the sampling distribution of the sample mean\({\rm{\bar x}}\)is approximately normal.

LARGE-SAMPLE CONFIDENCE INTERVAL

The sample mean is the sum of all data values divided by the sample size:

\({\rm{\bar x - }}\frac{{{\rm{22}}{\rm{.4 + 23}}{\rm{.6 + 24}}{\rm{.0 + \ldots + 44}}{\rm{.6 + 48}}{\rm{.3 + 55}}{\rm{.0}}}}{{{\rm{83}}}}{\rm{ - }}\frac{{{\rm{2769}}{\rm{.7}}}}{{{\rm{83}}}}{\rm{\gg 33}}{\rm{.3699}}\)

The variance is the sum of squared deviations from the mean divided by\({\rm{n - 1}}\). The standard deviation is the square root of the variance:

\(\begin{array}{l}{\rm{s - }}\sqrt {\frac{{{{{\rm{(22}}{\rm{.4 - 33}}{\rm{.3699)}}}^{\rm{2}}}{\rm{ + \ldots }}{\rm{. + (55}}{\rm{.0 - 33}}{\rm{.3699}}{{\rm{)}}^{\rm{2}}}}}{{{\rm{43 - 1}}}}} {\rm{\gg 5}}{\rm{.268 }}\\{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ - 2}}{\rm{.575}}\\\end{array}\)

The margin of error is then:

\({\rm{E - }}{{\rm{z}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ - 2}}{\rm{.575 \times }}\frac{{{\rm{5}}{\rm{.2683}}}}{{\sqrt {{\rm{83}}} }}{\rm{\gg 1}}{\rm{.4891}}\)

The boundaries of the confidence interval then become:

\(\begin{array}{l}{\rm{\bar x - E - 33}}{\rm{.3699 - 1}}{\rm{.4891 - 31}}{\rm{.8808}}\\{\rm{\bar x + E - 33}}{\rm{.3699 + 1}}{\rm{.4891 - 34}}{\rm{.8590}}\end{array}\)

Hence the boundaries of the confidence interval then become \((31.8808,34.8590)\)