Stats: Modeling the World Nasta Edition Grades 9-12
Stats: Modeling the World Nasta Edition Grades 9-12
3rd Edition
ISBN: 9780131359581
Author: David E. Bock, Paul F. Velleman, Richard D. De Veaux
Publisher: PEARSON
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Chapter 22, Problem 18E

(a)

To determine

To find: the type of design was used to gather this information.

(a)

Expert Solution
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Answer to Problem 18E

Prospective observational study

Explanation of Solution

This is a prospective study of observation. A prospective study defines and looks at a group of individuals (here, people with heart disease) in the future.

(b)

To determine

To Write: the suitable hypotheses.

(b)

Expert Solution
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Answer to Problem 18E

  H0:p1p2=0H1:p1p2>0

Explanation of Solution

Here want to evaluate how depressed patients are more likely to die than non-depressed patients among people who suffer from heart disease. Let p1 represent the percentage of patients who died with minor or major depression and let p2 represent the percentage of patients who died without depression. p1 should be higher than p2 if depressed patients are more likely to die than non-depressed patients. So, our hypotheses are going to be

  H0:p1p2=0H1:p1p2>0

(c)

To determine

To explain: to satisfy the assumptions and conditions required for conclusion.

(c)

Expert Solution
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Explanation of Solution

it is assuming that samples have been randomly chosen. The overall sample size is less than 10 percent of all heart disease sufferers. Since the samples were randomly chosen, it is fair to assume that the two groups are independent. The number of successes is 67 among the 361 patients without depression, and the number of failures is 361-67 = 294, and the number of successes is 26 among the 89 patients with mild or severe depression, and the number of failures is 89-26 = 63. These numbers, for each group, are at least 10.

The assumptions and conditions required to satisfy the inference.

(d)

To determine

To Test: the hypothesis and state the conclusion.

(d)

Expert Solution
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Explanation of Solution

Given:

  n1=89y1=26n2=361y2=67

Formula used:

  p^Pooled=y1+y2n1+n2SEPooled(p^1p^2)=p^Pooledq^Pooledn1+p^Pooledq^Pooledn2

Calculation:

  p^1=2689=0.2921p^2=67361=0.1856

Pool the sample data.

  p^Pooled=y1+y2n1+n2=26+6789+361=93450=0.2067

Estimating SD(p1p2) . Therefore,

  SEPooled(p^1p^2)=p^Pooledq^Pooledn1+p^Pooledq^Pooledn2=0.2067(10.2067)89+0.2067(10.2067)361=0.2067(0.7933)89+0.2067(0.7933)361=0.0479

The observed difference

  p^1p^2=0.29210.1856=0.1065

The z-score is

  z=(p^1p^2)0SEPooled(p^1p^2)=0.106500.0479=2.22

The P-value is

  P-value =P(z>2.22)=0.0131

P-value=0.0131 is very small; the null hypothesis should not be accepted. It might infer that depressed patients are more likely to death than nondepressed ones among people who suffer from heart disease.

(e)

To determine

To Explain: the means of P-value.

(e)

Expert Solution
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Explanation of Solution

The P-value is the likelihood that a test statistic would be as large as or greater than the observed value if the null hypothesis is valid. So, if the P-value is 0.0131, this means that there is only a 1.31 percent probability that normal variance in sampling will generate a 10.65 percent difference in the sample proportion when this difference is actually 0.0.

(f)

To determine

To find: the type of error committed if the statement is actually incorrect.

(f)

Expert Solution
Check Mark

Answer to Problem 18E

Type I

Explanation of Solution

It is rejected the null hypothesis here. If, indeed, the inference is false. This implies that, although it is valid, the null hypothesis is denied. So, it made an error with Type I.

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