/*! 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} Q2E The National Health and Nutritio... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Q2E (page 1)

The National Health and Nutrition Examination Survey (NHANES) collects demographic, socioeconomic, dietary, and health related information on an annual basis. Here is a sample of \({\rm{20}}\) observations on HDL cholesterol level \({\rm{(mg/dl)}}\) obtained from the \({\rm{2009 - 2010}}\) survey (HDL is 鈥済ood鈥� cholesterol; the higher its value, the lower the risk for heart disease):

\(\begin{array}{l}{\rm{35 49 52 54 65 51 51}}\\{\rm{47 86 36 46 33 39 45}}\\{\rm{39 63 95 35 30 48}}\end{array}\)

a. Calculate a point estimate of the population mean HDL cholesterol level.

b. Making no assumptions about the shape of the population distribution, calculate a point estimate of the value that separates the largest \({\rm{50\% }}\) of HDL levels from the smallest \({\rm{50\% }}\).

c. Calculate a point estimate of the population standard deviation.

d. An HDL level of at least \({\rm{60}}\) is considered desirable as it corresponds to a significantly lower risk of heart disease. Making no assumptions about the shape of the population distribution, estimate the proportion \({\rm{p}}\) of the population having an HDL level of at least \({\rm{60}}\).

Short Answer

Expert verified

(a) A point estimate of the population mean HDL cholesterol level is\({\rm{\bar x = 49}}{\rm{.95 mg/dl}}\).

(b) A point estimate of the value that separates the largest\({\rm{50\% }}\)of HDL levels from the smallest\({\rm{50\% }}\)is\({\rm{M = 47}}{\rm{.5 mg/dl}}\).

(c) A point estimate of the population standard deviation is\({\rm{s = 16}}{\rm{.8100 mg/dl}}\).

(d) The proportion \({\rm{p}}\) of the population having an HDL level of at least \({\rm{60}}\) is \({\rm{\hat p = 20\% }}\).

Step by step solution

01

Concept Introduction

The average of the given numbers is computed by dividing the total number of numbers by the sum of the given numbers.

The median is the middle number in a list of numbers that has been sorted ascending or descending, and it might be more descriptive of the data set than the average. When there are outliers in the series that could affect the average of the numbers, the median is sometimes utilised instead of the mean.

The standard deviation is a statistic that measures the amount of variation or dispersion in a set of numbers.

02

Point of Estimate (Mean)

(a)

The value of\({\rm{n}}\)is given as\({\rm{n = 20}}\).

The data provided is 鈥�

\(\begin{array}{l}{\rm{35 49 52 54 65 51 51 47 86 36}}\\{\rm{46 33 39 45 39 63 95 35 30 48}}\end{array}\)

A point estimate of the population mean is the sample mean.

The sample mean is the sum of all values divided by the number of values 鈥�

\(\begin{array}{c}{\rm{\bar x = }}\frac{{{\rm{35 + 49 + 52 + 54 + 65 + \ldots + 63 + 95 + 35 + 30 + 48}}}}{{{\rm{20}}}}\\{\rm{ = }}\frac{{{\rm{999}}}}{{{\rm{20}}}}\\{\rm{ = 49}}{\rm{.95}}\end{array}\)

Therefore, the value is obtained as \({\rm{\bar x = 49}}{\rm{.95 mg/dl}}\).

03

Step 3:Point of Estimate (Median)

(b)

The value of\({\rm{n}}\)is given as\({\rm{n = 20}}\).

The data provided is 鈥�

\(\begin{array}{l}{\rm{35 49 52 54 65 51 51 47 86 36}}\\{\rm{46 33 39 45 39 63 95 35 30 48}}\end{array}\)

The median separates the largest \({\rm{50\% }}\)from the smallest \({\rm{50\% }}\).

Apoint estimate that separates the largest \({\rm{50\% }}\)from the smallest \({\rm{50\% }}\)is the(sample)median.

Order the data values from smallest to largest 鈥�

\(\begin{array}{c}{\rm{30, 33, 35, 35, 36, 39, 39, 45, 46, 47,}}\\{\rm{48, 49, 51, 51, 52, 54, 63, 65, 86, 95}}\end{array}\)

Since the number of data values is even,the sample median is the average of the two middle values of the sorteddata set 鈥�

\(\begin{array}{c}{\rm{M = }}{{\rm{Q}}_{\rm{2}}}\\{\rm{ = }}\frac{{{\rm{47 + 48}}}}{{\rm{2}}}\\{\rm{ = 47}}{\rm{.5}}\end{array}\)

Therefore, the value is obtained as \({\rm{M = 47}}{\rm{.5 mg/dl}}\).

04

Step 4:Point of Estimate (Standard Deviation)

(c)

The value of\({\rm{n}}\)is given as\({\rm{n = 20}}\).

The data provided is 鈥�

\(\begin{array}{l}{\rm{35 49 52 54 65 51 51 47 86 36}}\\{\rm{46 33 39 45 39 63 95 35 30 48}}\end{array}\)

A point estimate of the population mean is the sample mean.

The sample mean is the sum of all values divided by the number of values 鈥�

\(\begin{array}{c}{\rm{\bar x = }}\frac{{{\rm{35 + 49 + 52 + 54 + 65 + \ldots + 63 + 95 + 35 + 30 + 48}}}}{{{\rm{20}}}}\\{\rm{ = }}\frac{{{\rm{999}}}}{{{\rm{20}}}}\\{\rm{ = 49}}{\rm{.95}}\end{array}\)

Apoint estimate of the population standard deviation is the sample standard deviation.

Create the following table 鈥�

Find the sum of numbers in the last column to get 鈥�

\(\sum {{{{\rm{(}}{{\rm{x}}_{\rm{i}}}{\rm{ - \bar x)}}}^{\rm{2}}}{\rm{ = 5368}}{\rm{.95}}} \)

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

\(\begin{array}{c}{\rm{s = }}\sqrt {\frac{{{\rm{5368}}{\rm{.95}}}}{{{\rm{20 - 1}}}}} \\{\rm{ = }}\sqrt {\frac{{{\rm{5368}}{\rm{.95}}}}{{{\rm{19}}}}} \approx {\rm{16}}{\rm{.8100}}\end{array}\)

Therefore, the value is obtained as \({\rm{s = 16}}{\rm{.8100 mg/dl}}\).

05

Step 5:Point of Estimate (Proportion)

(d)

The value of\({\rm{n}}\)is given as\({\rm{n = 20}}\).

The data provided is 鈥�

\(\begin{array}{l}{\rm{35 49 52 54 65 51 51 47 86 36}}\\{\rm{46 33 39 45 39 63 95 35 30 48}}\end{array}\)

Apoint estimate of the population proportion is the sample proportion.

There are\({\rm{4}}\)data values in the sample that are greater than or equal to\({\rm{60}}\).

The sample proportion is the number of favourable outcomes divided by the sample size 鈥�

\(\begin{aligned} \hat p &= \frac{{\rm{4}}}{{{\rm{20}}}} = frac{{\rm{1}}}{{\rm{5}}}\\&= 0{\rm{.20}}\\ &= 20\%\end{aligned}\)

Therefore, the value is obtained as \({\rm{\hat p = 20\% }}\).

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

There is no nice formula for the standard normal cdf\({\rm{\Phi (z)}}\), but several good approximations have been published in articles. The following is from "Approximations for Hand Calculators Using Small Integer Coefficients" (Mathematics of Computation, 1977: 214-222). For \({\rm{0 < z}} \le {\rm{5}}{\rm{.5}}\)

\(\begin{array}{l}{\rm{P(Z}} \ge {\rm{z) = 1 - \Phi (z)}}\\ \Leftrightarrow {\rm{.5exp}}\left\{ {{\rm{ - }}\left( {\frac{{{\rm{(83z + 351)z + 562}}}}{{{\rm{703/z + 165}}}}} \right)} \right\}\end{array}\)

The relative error of this approximation is less than \({\rm{.042\% }}\). Use this to calculate approximations to the following probabilities, and compare whenever possible to the probabilities obtained from Appendix Table A.3.

a. \({\rm{P(Z}} \ge {\rm{1)}}\)

b. \({\rm{P(Z < - 3)}}\)

c. \({\rm{P( - 4 < Z < 4)}}\)

d. \({\rm{P(Z > 5)}}\)

Consider a sample \({x_1},{x_2},...,{x_n}\) and suppose that the values of \(\bar x\),\({s^2}\), and shave been calculated.

a. Let\({y_i} = {x_i} - \bar x\)for i=1,鈥�, n. How do the values of \({s^2}\)and sfor the\({y_i}'s\)compare to the corresponding values for the\({x_i}'s\)? Explain.

b. Let\({z_i} = \left( {{x_i} - \bar x} \right)/s\) for i=1,鈥�, n. What are the values of the sample variance and sample standard deviation for the \({z_i}'s\)?

A sample of 26 offshore oil workers took part in a simulated escape exercise, resulting in the accompanying data on time (sec) to complete the escape (鈥淥xygen Consumption and Ventilation During Escape from an Offshore Platform,鈥� Ergonomics, 1997: 281鈥�292):

389 356 359 363 375 424 325 394 402

373 373 370 364 366 364 325 339 393

392 369 374 359 356 403 334 397

a. Construct a stem-and-leaf display of the data. How does it suggest that the sample mean and median will compare?

b. Calculate the values of the sample mean and median.(Hint:\(\sum {{x_i} = } \)9638.)

c. By how much could the largest time, currently 424, be increased without affecting the value of the sample median? By how much could this value be decreased without affecting the value of the sample median?

d. What are the values of \(\bar x\)and \(\tilde x\), when the observations are re expressed in minutes?

Consider the following sample of observations on coating thickness for low-viscosity paint ("Achieving a Target Value for a Manufacturing Process: \({\rm{A}}\)Case Study, \({\rm{97}}\)J. of Quality Technology, \({\rm{1992:22 - 26}}\)):

\(\begin{array}{*{20}{r}}{{\rm{.83}}}&{{\rm{.88}}}&{{\rm{.88}}}&{{\rm{1}}{\rm{.04}}}&{{\rm{1}}{\rm{.09}}}&{{\rm{1}}{\rm{.12}}}&{{\rm{1}}{\rm{.29}}}&{{\rm{1}}{\rm{.31}}}\\{{\rm{1}}{\rm{.48}}}&{{\rm{1}}{\rm{.49}}}&{{\rm{1}}{\rm{.59}}}&{{\rm{1}}{\rm{.62}}}&{{\rm{1}}{\rm{.65}}}&{{\rm{1}}{\rm{.71}}}&{{\rm{1}}{\rm{.76}}}&{{\rm{1}}{\rm{.83}}}\end{array}\)

Assume that the distribution of counting thickness is normal (a normal probability plot strongly supports this assumption).

a. Calculate a point estimate of the mean value of coating thickness, and state which estimator you used.

b. Calculate a point estimate of the median of the coating thickness distribution, and state which estimator you used.

C. Calculate a point estimate of the value that separates the largest of\({\rm{10\% }}\) all values in the thickness distribution from the remaining \({\rm{90\% }}\)and state which estimator you used. (Hint Express what you are trying to estimate in terms of \({\rm{\mu }}\)and \({\rm{\sigma }}\)

d. Estimate \({\rm{P}}\left( {{\rm{X < 1}}{\rm{.5}}} \right){\rm{,}}\)i.e., the proportion of all thickness values less than 1.5. (Hint: If you knew the values of \({\rm{\mu }}\)and \({\rm{\sigma }}\), you could calculate this probability. These values are not available, but they can be estimated.)

e. What is the estimated standard error of the estimator that you used in part (b)?

Consider the strength data for beams given in Example

1.2.

a. Construct a stem-and-leaf display of the data. What appears to be a representative strength value? Do the observations appear to be highly

concentrated about the representative value or rather spread out?

b. Does the display appear to be reasonably symmetric about a representative value, or would you describe its shape in some other way?

c. Do there appear to be any outlying strength values?

d. What proportion of strength observations in this sample exceeds 10 MPa?
See all solutions

Recommended explanations on Math Textbooks

Calculus

Read Explanation

Mechanics Maths

Read Explanation

Applied Mathematics

Read Explanation

Logic and Functions

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

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.