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Chapter 9: Q38 E (page 388)

Adding computerized medical images to a database promises to provide great resources for physicians. However, there are other methods of obtaining such information, so the issue of efficiency of access needs to be investigated. The article "The Comparative Effectiveness of Conventional and Digital Image Libraries" J. of Audiovisual Media in Medicine, 2001: 8-15) reported on an experiment in which 13 computerproficient medical professionals were timed both while retrieving an image from a library of slides and while retrieving the same image from a computer database with a Web front end.

\(\begin{array}{*{20}{l}}{\;Subject\;\;\;\;\;1\;\;\;2\;\;3\;\;\;4\;\;\;5\;\;\;6\;\;\;\;\;7}\\{\;Slide\;\;\;\;\;\;\;\;30\;35\;40\;25\;20\;30\;\;35}\\{\;Digital\;\;\;\;\;\;\;25\;16\;15\;15 10\;20\;\;\;7}\\{\;Difference\;\;5\;19\;25\;10\;10\;10\;28}\\{\;Subject\;\;\;\;\;8\;\;9\;\;10\;11\;12\;13}\\{\;Slide\;\;\;\;\;\;\;\;\;62\;40\;51\;25\;42\;33\;}\\{Digital\;\;\;\;\;\;\;16\;15\;13\;11\;19\;19}\\{Difference\;46\;25\;38\;14\;23\;14}\end{array}\)

a. Construct a comparative boxplot of times for the two types of retrieval, and comment on any interesting features.

b. Estimate the difference between true average times for the two types of retrieval in a way that conveys information about precision and reliability. Be sure to check the plausibility of any assumptions needed in your analysis. Does it appear plausible that the true average times for the two types of retrieval are identical? Why or why not?

Short Answer

Expert verified

(a) The average time for "Slide" appears to be much higher than the average time for "Digital".

(b) \({\rm{(13}}{\rm{.3090,27}}{\rm{.7680)}}\)

Step by step solution

01

A)Step 1: Find the quartile for slide

Sort the data values from smallest to largest:

Slide: \(20,{\rm{ }}25,{\rm{ }}25,{\rm{ }}30,{\rm{ }}30,{\rm{ }}33,{\rm{ }}35,{\rm{ }}35,{\rm{ }}40,{\rm{ }}40,{\rm{ }}42,{\rm{ }}51,{\rm{ }}62\)

Digital: \(7,10,11,13,15,15,15,16,16,19,19,20,25\)

SLIDE

The minimum is \(20\) .

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

\(M = {Q_2} = 35\)

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

\({Q_1} = \frac{{25 + 30}}{2} = 27.5\)

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

\({Q_3} = \frac{{40 + 42}}{2} = 41\)

The maximum is \(62\) .

02

Find the quartile for Digital

DIGITAL

The minimum is \(7\) .

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

\(M = {Q_2} = 15\)

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

\({Q_1} = \frac{{11 + 13}}{2} = 12\)

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

\({Q_3} = \frac{{19 + 19}}{2} = 19\)

The maximum is \(25\)

03

Step 3: 

mapping the graph

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 of "Slide" appears to be skewed to the right (or positively skewed), because the box of the boxplot lies to the left between the whiskers.

The distribution of "Digital appears to be roughly symmetric, because the box of the boxplot lies roughly in the middle between the whiskers.

The average time for "Slide" appears to be much higher than the average time for "Digital", because almost the entire boxplot of "Slide" lies abo the boxplot of "Digital".

04

b)Step 4: Find the endpoint of confidence interval

Let us assume that we want to determine a\(95\% \)confidence interval (other confidence levels can be determined similarly).

\(c = 95\% = 0.95\)

Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.

\(\bar d = \frac{{5 + 19 + 25 + \ldots + 14 + 23 + 14}}{{13}} \approx 20.5385\)

Determine the sample standard deviation of the differences:

\({s_d} = \sqrt {\frac{{{{(5 - 20.5385)}^2} + \ldots + {{(14 - 20.5385)}^2}}}{{13 - 1}}} \approx 11.9625\)

Determine the\({t_{\alpha /2}}\)using the Student's T distribution table in the appendix with\(df = n - 1 = 13 - 1 = 12\):

\({t_{0.025}} = 2.179\)

The margin of error is then:\(E = {t_{\alpha /2}} \cdot \frac{{{s_d}}}{{\sqrt n }} = 2.179 \cdot \frac{{11.9625}}{{\sqrt {13} }} \approx 7.2295\)

The endpoints of the confidence interval for\({\mu _d}\)are:

\(\begin{array}{l}\bar d - E = 20.5385 - 7.2295 = 13.3090\\\bar d + E = 20.5385 + 7.2295 = 27.7680\end{array}\)

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Most popular questions from this chapter

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