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Refer to Exercise 10.83 and find a 90 % confidence interval for the difference between the mean numbers of acute postoperative days in the hospital with the dynamic and static systems.

Short Answer

Expert verified

The interval is 0.00264 to 0.01299

Step by step solution

01

:Given information

With the dynamic and static systems, calculate a 90%confidence interval for the difference in the mean numbers of acute postoperative days in the hospital.

02

calculation

The degree of freedom is computed using the formula:

df=[(s12n1)+(s22n2)]2(s1n1)2n11+(s22n2)2n21=[(0.00514210)+(0.00470215)]2(0.00614210)2101+(0.00470215)215118

The interval's end point is,

x1x2ta2s12n1+s22n2=(0.242460.01643)2.552.0.00514210+0.00470215

=0.00264to0.01299

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Most popular questions from this chapter

In Exercises 5鈥20, 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 鈥淭able鈥 answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\).)Blanking Out on Tests Many students have had the unpleasant experience of panicking on a test because the first question was exceptionally difficult. The arrangement of test items was studied for its effect on anxiety. The following scores are measures of 鈥渄ebilitating test anxiety,鈥 which most of us call panic or blanking out (based on data from 鈥淚tem Arrangement, Cognitive Entry Characteristics, Sex and Test Anxiety as Predictors of Achievement in Examination Performance,鈥 by Klimko, Journal of Experimental Education, Vol. 52, No. 4.) Is there sufficient evidence to support the claim that the two populations of scores have different means? Is there sufficient evidence to support the claim that the arrangement of the test items has an effect on the score? Is the conclusion affected by whether the significance level is 0.05 or 0.01?

Questions Arranged from Easy to Difficult

24.64

39.29

16.32

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28.02

33.31

20.60

21.13

26.69

28.9

26.43

24.23

7.10

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30.02

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25.49

38.81

27.85

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Questions Arranged from Difficult to Easy

33.62

34.02

26.63

30.26

35.91

26.68

29.49

35.32

27.24

32.34

29.34

33.53

27.62

42.91

30.20

32.54

Eyewitness Accuracy of Police Does stress affect the recall ability of police eyewitnesses? This issue was studied in an experiment that tested eyewitness memory a week after a nonstressful interrogation of a cooperative suspect and a stressful interrogation of an uncooperative and belligerent suspect. The numbers of details recalled a week after the incident were recorded, and the summary statistics are given below (based on data from 鈥淓yewitness Memory of Police Trainees for Realistic Role Plays,鈥 by Yuille et al., Journal of Applied Psychology, Vol. 79, No. 6). Use a 0.01 significance level to test the claim in the article that 鈥渟tress decreases the amount recalled.鈥

Nonstress: n = 40,\(\bar x\)= 53.3, s = 11.6

Stress: n = 40,\(\bar x\)= 45.3, s = 13.2

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:

\({\bf{E = }}{{\bf{z}}_{\frac{{\bf{\alpha }}}{{\bf{2}}}}}\sqrt {\frac{{{{\bf{p}}_{\bf{1}}}{{\bf{q}}_{\bf{1}}}}}{{{{\bf{n}}_{\bf{1}}}}}{\bf{ + }}\frac{{{{\bf{p}}_{\bf{2}}}{{\bf{q}}_{\bf{2}}}}}{{{{\bf{n}}_{\bf{2}}}}}} \)

Replace \({{\bf{n}}_{\bf{1}}}\;{\bf{and}}\;{{\bf{n}}_{\bf{2}}}\) by n in the preceding formula (assuming that both samples have the same size) and replace each of \({{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{q}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}\;{\bf{and}}\;{{\bf{q}}_{\bf{2}}}\)by 0.5 (because their values are not known). Solving for n results in this expression:

\({\bf{n = }}\frac{{{\bf{z}}_{\frac{{\bf{\alpha }}}{{\bf{2}}}}^{\bf{2}}}}{{{\bf{2}}{{\bf{E}}^{\bf{2}}}}}\)

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.

A sample size that will ensure a margin of error of at most the one specified.

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?

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