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Family Heights. In Exercises 1–5, use the following heights (in.) The data are matched so that each column consists of heights from the same family.

1. a. Are the three samples independent or dependent? Why?

b. Find the mean, median, range, standard deviation, and variance of the heights of the sons.

c. What is the level of measurement of the sample data (nominal, ordinal, interval, ratio)?

d. Are the original unrounded heights discrete data or continuous data?

Short Answer

Expert verified

a. The samples are dependent.

b. The mean, median, range, standard deviation, and variance of the heights of the sonsare given as follows:

\(\begin{array}{l}\bar x = 69.7\;{\rm{in}}{\rm{.}}\\M = 71.0\;{\rm{in}}{\rm{.}}\\{\rm{Range}} = 7.7\;{\rm{in}}{\rm{.}}\end{array}\)

\(\begin{array}{l}s = 2.6\;{\rm{in}}{\rm{.}}\\{s^2} = 6.6\;{\rm{i}}{{\rm{n}}^2}\end{array}\)

c. The data is measured on the ratio scale of measurement.

d. The data is continuous.

Step by step solution

01

Given information

The data for heights of three members of thesame family is recorded.

02

Identify independent anddependent samples

a.

Certain groups of sampled observations are independent if they are not related to each other. In case the observations are related to one another, the samples become dependent. The paired or matched observations are also considered as dependent.

In this case, the heights of the members of the same family are observed.The three groups of observations from father, mother, and son are matched. Thus, the three samples are independent.

03

Compute the statistical measures for the heights of sons

b.

For n observations,\({x_i}\)the formulae for different measures are shown below:

Mean:

\(\bar x = \frac{{\sum {{x_i}} }}{n}\)

Median:The middle observation is the median, where the central location obtained by the formula:

\(M = \left\{ \begin{array}{l}{\left( {\frac{n}{2}} \right)^{th}};\;\;\;\;\;\;\;\;\;\;\;\;\;\;n\;{\rm{is}}\;{\rm{odd}}\\\frac{{{{\left( {\frac{n}{2}} \right)}^{th}} + {{\left( {\frac{{n + 1}}{2}} \right)}^{th}}}}{2};\;\;\;\;n\;{\rm{is}}\;{\rm{even}}\end{array} \right.\)

Range:The range is the difference between the maximum and minimum values of the dataset.

Standard deviation:

\(s = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({x_i} - \bar x)}^2}} }}{{n - 1}}} \)

Variance is\({s^2}\).

Compute the values for the heights of sons.

Mean:

\(\begin{aligned}{c}\bar x &= \frac{{71 + 64 + 71 + ... + 71}}{8}\\ &= 69.7\;{\rm{in}}\end{aligned}\)

Median:

Since n is even, the location of the middlemost observation is obtainedas shown below:

\(\begin{aligned}{c}M &= \frac{{{{\left( {\frac{8}{2}} \right)}^{th}} + {{\left( {\frac{8}{2} + 1} \right)}^{th}}}}{2}\\ &= \frac{{4th + 5th}}{2}\\ &= \frac{{71 + 71}}{2}\\ &= 71.0\;{\rm{in}}\end{aligned}\)

Range:

\(\begin{aligned}{c}Range &= {\rm{Max}} - {\rm{Min}}\\ &= 71.7 - 64\\ &= 7.7\;{\rm{in}}\end{aligned}\)

Standard deviation:

\(\begin{aligned}{c}s &= \sqrt {\frac{{{{\left( {71 - 69.71} \right)}^2} + {{\left( {64 - 69.71} \right)}^2} + ... + {{\left( {71 - 69.71} \right)}^2}}}{{8 - 1}}} \\ &= \sqrt {\frac{{1.66 + 32.63 + ... + 1.66}}{7}} \\ &= 2.57\;{\rm{in}}{\rm{.}}\\ &\approx {\rm{2}}{\rm{.6 in}}{\rm{.}}\end{aligned}\)

Variance:

\(\begin{aligned}{c}{s^2} &= {\left( {2.57} \right)^2}\\ &= 6.6\;{\rm{i}}{{\rm{n}}^2}\end{aligned}\).

04

identify the level of measurement

c.

The four levels of measurements are listed below:

Nominal: The data can only be categorized.

Ordinal: The data can be categorized and ordered.

Interval: The data can be categorized, ordered, and have a meaningful interpretation of differences between any two values.

Ratio: The data consists of all three properties of interval and have a meaningful interpretation of 0.

Since the variable height has all four properties (categorization, rank, meaningful differences, and true meaning of zero measure), it has a ratio scale of measurement.

05

Identify the type of data

d.

The continuous form of data is the quantitative variable thatcan take any value within a range, including all decimal numbers. A discrete data can take certain values of a distinct set.

The height of the family members is a quantitative variable where heights can take any decimal value in a range of numbers. Thus, it is continuous data.

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Most popular questions from this chapter

Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\))

Regular Coke and Diet Coke Data Set 26 “Cola Weights and Volumes” in Appendix B includesweights (lb) of the contents of cans of Diet Coke (n= 36,\(\overline x \)= 0.78479 lb, s= 0.00439 lb) and of the contents of cans of regular Coke (n= 36,\(\overline x \)= 0.81682 lb, s= 0.00751 lb).

a. Use a 0.05 significance level to test the claim that the contents of cans of Diet Coke have weights with a mean that is less than the mean for regular Coke.

b. Construct the confidence interval appropriate for the hypothesis test in part (a).

c. Can you explain why cans of Diet Coke would weigh less than cans of regular Coke?

Using Confidence Intervals

a. Assume that we want to use a 0.05 significance level to test the claim that p1 < p2. Which is better: A hypothesis test or a confidence interval?

b. In general, when dealing with inferences for two population proportions, which two of the following are equivalent: confidence interval method; P-value method; critical value method?

c. If we want to use a 0.05 significance level to test the claim that p1 < p2, what confidence level should we use?

d. If we test the claim in part (c) using the sample data in Exercise 1, we get this confidence interval: -0.000508 < p1 - p2 < - 0.000309. What does this confidence interval suggest about the claim?

Determining Sample Size The sample size needed to estimate the difference between two population proportions to within a margin of error E with a confidence level of 1 - a can be found by using the following expression:

E=zα2p1q1n1+p2q2n2

Replace n1andn2 by n in the preceding formula (assuming that both samples have the same size) and replace each of role="math" localid="1649424190272" p1,q1,p2andq2by 0.5 (because their values are not known). Solving for n results in this expression:

n=zα222E2

Use this expression to find the size of each sample if you want to estimate the difference between the proportions of men and women who own smartphones. Assume that you want 95% confidence that your error is no more than 0.03.

Before/After Treatment Results Captopril is a drug designed to lower systolic blood pressure. When subjects were treated with this drug, their systolic blood pressure readings (in mm Hg) were measured before and after the drug was taken. Results are given in the accompanying table (based on data from “Essential Hypertension: Effect of an Oral Inhibitor of Angiotensin-Converting Enzyme,” by MacGregor et al., British Medical Journal, Vol. 2). Using a 0.01 significance level, is there sufficient evidence to support the claim that captopril is effective in lowering systolic blood pressure?

Subject

A

B

C

D

E

F

G

H

I

J

K

L

Before

200

174

198

170

179

182

193

209

185

155

169

210

After

191

170

177

167

159

151

176

183

159

145

146

177

Hypothesis Test Use a 0.05 significance level to test the claim that differences between heights of fathers and their sons have a mean of 0 in.

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