/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Q43 E Cushing's disease is characteriz... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Q43 E (page 390)

Cushing's disease is characterized by muscular weakness due to adrenal or pituitary dysfunction. To provide effective treatment, it is important to detect childhood Cushing's disease as early as possible. Age at onset of symptoms and age at diagnosis (months) for 15 children suffering from the disease were given in the article "Treatment of Cushing's Disease in Childhood and Adolescence by Transphenoidal Microadenomectomy" (New Engl. J. of Med., 1984: 889). Here are the values of the differences between age at onset of symptoms and age at diagnosis:

\(\begin{array}{*{20}{l}}{ - 24}&{ - 12}&{ - 55}&{ - 15}&{ - 30}&{ - 60}&{ - 14}&{ - 21}\\{ - 48}&{ - 12}&{ - 25}&{ - 53}&{ - 61}&{ - 69}&{ - 80}&{}\end{array}\)

a. Does the accompanying normal probability plot cast strong doubt on the approximate normality of the population distribution of differences?

b. Calculate a lower\(95\% \)confidence bound for the population mean difference, and interpret the resulting bound.

c. Suppose the (age at diagnosis) - (age at onset) differences had been calculated. What would be a\(95\backslash \% \)upper confidence bound for the corresponding population mean difference?

Short Answer

Expert verified

a. Plausible;

b. \( - 49.14\)

c. \(49.14.\)

Step by step solution

01

A)Step 1: The difference of population difference

It can be said that the data follow a linear pattern, and thus that the data (differences) is from the normal population.

02

B)Step 2: Find the sample standard form

The paired \(t\) confidence interval for\({\mu _D}\) is

\(\left( {\bar d - {t_{\alpha /2,n - 1}} \cdot \frac{{{s_D}}}{{\sqrt n }},\bar d + {t_{\alpha /2,n - 1}} \cdot \frac{{{s_D}}}{{\sqrt n }}} \right)\)

A one-sided confidence bound can be obtained by retaining the relevant sing \(( + \) or\( - )\), and by replacing \({t_{\alpha /2,n - 1}}\) by\({t_{\alpha ,n - 1}}\).

The \(95\% \) lower confidence bound for the population mean difference is

\(\bar d - {t_{\alpha ,n - 1}} \cdot \frac{{{s_D}}}{{\sqrt n }}\)

03

Step 3:

The Sample Mean \(\bar x\) of observations\({x_1},{x_2}, \ldots ,{x_n}\) is given by

\(\bar x = \frac{{{x_1} + {x_2} + \ldots + {x_n}}}{n} = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} \)

The sample mean \(\bar d\) for the differences is

\(\bar d = \frac{1}{{15}} \cdot ( - 24 - 12 - \ldots - 80) = - 38.6\)

The Sample Variance \({s^2}\) is

\({s^2} = \frac{1}{{n - 1}} \cdot {S_{xx}}\)

where

\({S_{xx}} = \sum {{{\left( {{x_i} - \bar x} \right)}^2}} = \sum {x_i^2} - \frac{1}{n} \cdot {\left( {\sum {{x_i}} } \right)^2}\)

The Sample Standard Deviation \(s\) is

\(s = \sqrt {{s^2}} = \sqrt {\frac{1}{{n - 1}} \cdot {S_{xx}}} \)

The sample variance is

\(\begin{array}{l}s_D^2 = \frac{1}{{15 - 1}} \cdot \left( {{{( - 24 - ( - 38.6))}^2} + {{( - 12 - ( - 38.6))}^2} + \ldots + {{( - 80 - ( - 38.6))}^2}} \right)\\ = 537.3124\end{array}\)

and the sample standard deviation

\({s_D} = \sqrt {537.3124} = 23.18\)

The only missing value is

\({t_{\alpha /2,n - 1}} = {t_{0.05,14}} = 1.761\)

which was computed using a software (you can use the table in the appendix of the book).

The\(95\% \)lower confidence bound is

\(\bar d - {t_{\alpha ,n - 1}} \cdot \frac{{{s_D}}}{{\sqrt n }} = - 38.6 - 1.761 \cdot \frac{{23.18}}{{\sqrt {15} }} = \)

04

C)Step 4: Find the population mean difference

Similarly as in\((b)\), the \(95\% \)upper confidence bound is\(\bar d + {t_{\alpha ,n - 1}} \cdot \frac{{{s_D}}}{{\sqrt n }} = - 38.6 + 1.761 \cdot \frac{{23.18}}{{\sqrt {15} }} = 49.14\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A mechanical engineer wishes to compare strength properties of steel beams with similar beams made with a particular alloy. The same number of beams, \(n\), of each type will be tested. Each beam will be set in a horizontal position with a support on each end, a force of \(2500lb\) will be applied at the center, and the deflection will be measured. From past experience with such beams, the engineer is willing to assume that the true standard deviation of deflection for both types of beam is \(.05in\). Because the alloy is more expensive, the engineer wishes to test at level \(.01\) whether it has smaller average deflection than the steel beam. What value of n is appropriate if the desired type II error probability is \(.05\) when the difference in true average deflection favors the alloy by \(.04in\).?

The article "Flexure of Concrete Beams Reinforced with Advanced Composite Orthogrids"\((J\). of Aerospace Engr., 1997: 7-15) gave the accompanying data on ultimate load\((kN)\)for two different types of beams.

\( - 7.0944\)

a. Assuming that the underlying distributions are normal, calculate and interpret a\(99\% \)CI for the difference between true average load for the fiberglass beams and that for the carbon beams.

b. Does the upper limit of the interval you calculated in part (a) give a\(99\% \)upper confidence bound for the difference between the two\(\mu \)'s? If not, calculate such a bound. Does it strongly suggest that true average load for the carbon beams is more than that for the fiberglass beams? Explain.

Sometimes experiments involving success or failure responses are run in a paired or before/after manner. Suppose that before a major policy speech by a political candidate, n individuals are selected and asked whether \((S)\)or not (F) they favor the candidate. Then after the speech the same n people are asked the same question. The responses can be entered in a table as follows:

Before

After

S

F

S

\({{\bf{X}}_{\bf{1}}}\)

\({{\bf{X}}_{\bf{2}}}\)

F

\({{\bf{X}}_{\bf{3}}}\)

\({{\bf{X}}_{\bf{4}}}\)

Where\({{\bf{x}}_{\bf{1}}}{\bf{ + }}{{\bf{x}}_{\bf{2}}}{\bf{ + }}{{\bf{x}}_{\bf{3}}}{\bf{ + }}{{\bf{x}}_{\bf{4}}}{\bf{ = n}}\). Let\({{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}{\bf{,}}{{\bf{p}}_{\bf{3}}}\), and \({p_4}\)denote the four cell probabilities, so that \({p_1} = P(S\) before and S after), and so on. We wish to test the hypothesis that the true proportion of supporters (S) after the speech has not increased against the alternative that it has increased.

a. State the two hypotheses of interest in terms of\({p_1},{p_2}\),\({p_3}\), and \({p_4}\).

b. Construct an estimator for the after/before difference in success probabilities

c. When n is large, it can be shown that the rv \(\left( {{{\bf{X}}_{\bf{i}}}{\bf{ - }}{{\bf{X}}_{\bf{j}}}} \right){\bf{/n}}\) has approximately a normal distribution with variance given by\(\left[ {{{\bf{p}}_{\bf{i}}}{\bf{ + }}{{\bf{p}}_{\bf{j}}}{\bf{ - }}{{\left( {{{\bf{p}}_{\bf{i}}}{\bf{ - }}{{\bf{p}}_{\bf{j}}}} \right)}^{\bf{2}}}} \right]{\bf{/n}}\). Use this to construct a test statistic with approximately a standard normal distribution when \({H_0}\)is true (the result is called McNemar's test).

d. If\({{\bf{x}}_{\bf{1}}}{\bf{ = 350,}}\;\;\;{{\bf{x}}_{\bf{2}}}{\bf{ = 150,}}\;\;\;{{\bf{x}}_{\bf{3}}}{\bf{ = 200}}\), and\({x_4} = 300\), what do you conclude?

Persons having Reynaud鈥檚 syndrome are apt to suffer a sudden impairment of blood circulation in fingers and toes. In an experiment to study the extent of this impairment, each subject immersed a forefinger in water and the resulting heat output \((cal/c{m^2}/min)\) was measured. For \(m = 10\) subjects with the syndrome, the average heat output was \(\bar x = .64\), and for \(n = 10\) non-sufferers, the average output was \(2.05\). Let \({\mu _1}\) and \({\mu _2}\) denote the true average heat outputs for the two types of subjects. Assume that the two distributions of heat output are normal with \({\sigma _1} = .2\) and \({\sigma _2} = .4\).

a. Consider testing \({H_0}:{\mu _1} - {\mu _2} = - 1.0\) versus \({H_2}:{\mu _1} - {\mu _2} < - 1.0\)at level . \(01\). Describe in words what \({H_a}\) says, and then carry out the test.

b. What is the probability of a type II error when the actual difference between \({\mu _1}\) and \({\mu _2}\) is \({\mu _1} - {\mu _2} = - 1.2?\)

c. Assuming that \(m = n\), what sample sizes are required to ensure that \(\beta = .1\) when \({\mu _1} - {\mu _2} = - 1.2?\)

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?
See all solutions

Recommended explanations on Math Textbooks

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Pure Maths

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.