Question

Refer to Exercise\(33\)in Section\(7.3\). The cited article also gave the following observations on degree of polymerization for specimens having viscosity times concentration in a higher range:

\(\begin{array}{*{20}{l}}{429}&{430}&{430}&{431}&{436}&{437}\\{440}&{441}&{445}&{446}&{447}&{}\end{array}\)

a. Construct a comparative boxplot for the two samples, and comment on any interesting features.

b. Calculate a\(95\% \)confidence interval for the difference between true average degree of polymerization for the middle range and that for the high range. Does the interval suggest that\({\mu _1}\)and\({\mu _2}\)may in fact be different? Explain your reasoning.

Short answer

(a) Both distributions are roughly symmetric

(b) \(( - 7.8733,9.5525)\)

Step-by-step solution

a)Step 1: Find the quartile of the first data set

Given:

DATA SET 1: \(:{\rm{ }}418,{\rm{ }}421,{\rm{ }}421,{\rm{ }}422,{\rm{ }}425,{\rm{ }}427,{\rm{ }}431,{\rm{ }}434,{\rm{ }}437,{\rm{ }}439,{\rm{ }}446,{\rm{ }}447,{\rm{ }}448,{\rm{ }}453,{\rm{ }}454,{\rm{ }}463,{\rm{ }}465\)

DATA SET 2: \(429,{\rm{ }}430,{\rm{ }}430,{\rm{ }}431,{\rm{ }}436,{\rm{ }}437,{\rm{ }}440,{\rm{ }}441,{\rm{ }}445,{\rm{ }}446,{\rm{ }}447\)

(a) FIRST DATA SET

The minimum is \(418{\rm{ }}.\)

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

\(M = {Q_2} = 437\)The first quartile is the median of the data values below the median (or at \(25\% \)of the data):

\({Q_1} = \frac{{422 + 425}}{2} = 423.5\)

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

\({Q_3} = \frac{{448 + 453}}{2} = 450.5\)

The maximum is\(465.\)

Find the quartile of the second data set

SECOND DATA SET

The minimum is\(429.\)

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

\(M = {Q_2} = 437\)

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

\({Q_1} = 430\)

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

\({Q_3} = 445\)

The maximum is \(447.\)

Mapping the graph

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 two distributions appear to be roughly symmetric, because the boxed lie roughly in the middle between the whiskers and the vertical line of the median in the box of the boxplots lie roughly in the middle of the box.

B)Step 4: Find the mean and standard deviation

(b)Given:

\(c = 95\% = 0.95\)

The mean is the sum of all values divided by the number of values:

\(\begin{array}{l}{{\bar x}_1} = \frac{{418 + 421 + 421 + \ldots + 454 + 463 + 465}}{{17}} \approx 438.2941\\{{\bar x}_2} = \frac{{429 + 430 + 430 + \ldots + 445 + 446 + 447}}{{11}} \approx 437.4545\end{array}\)

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

\(\begin{array}{l}{s_1} = \sqrt {\frac{{{{(418 - 438.2941)}^2} + \ldots . + {{(465 - 438.2941)}^2}}}{{17 - 1}}} \approx 15.1442\\{s_2} = \sqrt {\frac{{{{(429 - 437.4545)}^2} + \ldots .. + {{(447 - 437.4545)}^2}}}{{11 - 1}}} \approx 6.8317\end{array}\)

Find the endpoint of confidence interval

Determine the degrees of freedom (rounded down to the nearest integer):

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{{{15.1442}^2}}}{{17}} + \frac{{{{6.8317}^2}}}{{11}}} \right)}^2}}}{{\frac{{{{\left( {{{15.1442}^2}/17} \right)}^2}}}{{17 - 1}} + \frac{{{{\left( {{{6.8317}^2}/11} \right)}^2}}}{{11 - 1}}}} \approx 23\)

Determine the\(t\)-value by looking in the row starting with degrees of freedom\(df = 23\)and in the column with\(1 - c/2 = 0.025\)in the Student's\(t\)distribution table in the appendix:

\({t_{\alpha /2}} = 2.069\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} = 2.069 \cdot \sqrt {\frac{{{{15.1442}^2}}}{{17}} + \frac{{{{6.8317}^2}}}{{11}}} \approx 8.7129\)

The endpoints of the confidence interval for\({\mu _1} - {\mu _2}\)are:

\(\begin{array}{l}\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - E = (438.2941 - 437.4545) - 8.7129 = 0.8396 - 8.7129 = - 7.8733\\\left( {{{\bar x}_1} - {{\bar x}_2}} \right) + E = (438.2941 - 437.4545) + 8.7129 = 0.8396 + 8.7129 = 9.5525\end{array}\)