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In Exercises 5鈥16, use the listed paired sample data, and assume that the samples are simple random samples and that the differences have a distribution that is approximately normal.

Friday the 13th Researchers collected data on the numbers of hospital admissions resulting from motor vehicle crashes, and results are given below for Fridays on the 6th of a month and Fridays on the following 13th of the same month (based on data from 鈥淚s Friday the 13th Bad for Your Health?鈥 by Scanlon et al., British Medical Journal, Vol. 307, as listed in the Data and Story Line online resource of data sets). Construct a 95% confidence interval estimate of the mean of the population of differences between hospital admissions on days that are Friday the 6th of a month and days that are Friday the 13th of a month. Use the confidence interval to test the claim that when the 13th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are not affected.

Friday the 6th

9

6

11

11

3

5

Friday the 13th

13

12

14

10

4

12

Short Answer

Expert verified

The 95% confidence interval is equal to (-6.49,-0.17).

There is sufficient evidence to reject the claim that when the 13th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are not affected.

Step by step solution

01

Given information

The data are given about the hospital admissions resulting from motor vehicle crashes on two days of the month, Friday the 6th and Friday the 13th.

02

Hypotheses

Null hypothesis: The mean of the population of differences in the number of hospital admissions on Friday the 6th and Friday the 13thequals 0.

\({H_{0\;}}:\;{\mu _d} = 0\)

Alternative hypothesis: The mean of the population of differences in the number of hospital admissions on Friday the 6th and Friday the 13th is not equal to 0.

\({H_1}\;:\;{\mu _d} \ne 0\)

Here,\({\mu _d}\;\)is the mean of the population of differences in the number of hospital admissions for the two days.

03

Table of differences

The following table shows thedifferences in the number of hospital admissions on Friday the 6th and Friday the 13th :

Friday the 6th

9

6

11

11

3

5

Friday the 13th

13

12

14

10

4

12

Differences (d)

-4

-6

-3

1

-1

-7

04

Mean of differences and standard deviation of differences

The value of the mean of the differences in the number of hospitaladmissions on Friday the 6th and Friday the 13this computed below:

\(\begin{array}{c}\bar d = \frac{{\left( { - 4} \right) + \left( { - 6} \right) + ..... + \left( { - 7} \right)}}{6}\\ = - 3.33\end{array}\)

The value of the standard deviation of the differences in the number ofhospital admissions on Friday the 6th and Friday the 13th is computed below:

\(\begin{array}{c}{s_d} = \sqrt {\frac{{\sum\limits_{i = 1}^n {{{({d_i} - \bar d)}^2}} }}{{n - 1}}} \\ = \sqrt {\frac{{{{\left( {\left( { - 4} \right) - \left( { - 3.33} \right)} \right)}^2} + {{\left( {\left( { - 6} \right) - \left( { - 3.33} \right)} \right)}^2} + ...... + {{\left( {\left( { - 7} \right) - \left( { - 3.33} \right)} \right)}^2}}}{{6 - 1}}} \\ = 3.01\end{array}\)

05

Confidence Interval

The formula of the confidence interval is given below:

\(CI = \bar d - E < {\mu _d} < \bar d + E\;\)

06

Compute the margin of error

The level of significance is equal to 0.05.

The degrees of freedom are computed below:

\(\begin{array}{c}df = n - 1\\ = 6 - 1\\ = 5\end{array}\)

The value of the margin of error is equal tothe following:

\(\begin{array}{c}E = {t_{\frac{\alpha }{2},df}} \times \frac{{{s_d}}}{{\sqrt n }}\\ = {t_{0.025,5}} \times \frac{{3.01}}{{\sqrt 6 }}\\ = 2.5706 \times \frac{{3.01}}{{\sqrt 6 }}\\ = 3.15997\end{array}\)

07

Compute the confidence interval

Substituting the required values, the following value is obtained:

\(\begin{array}{c}\bar d - E < {\mu _d} < \bar d + E\;\\\left( { - 3.33 - 3.15997} \right) < {\mu _d} < \left( { - 3.33 + 3.15997} \right)\\ - 6.49 < {\mu _d} < - 0.17\end{array}\)

Thus, the 95% confidence interval equals (-6.49,-0.17).

08

Conclusion based on the confidence interval

Since the confidence interval does not contain the value 0, it can be said that the mean of the difference in the number of hospital admissions cannot be equal to 0.

Thus, there is sufficient evidence to reject the claim that when the 13th day of a month falls on a Friday, the numbers of hospital admissions from motor vehicle crashes are not affected.

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Most popular questions from this chapter

IQ and Lead Exposure Data Set 7 鈥淚Q and Lead鈥 in Appendix B lists full IQ scores for a random sample of subjects with low lead levels in their blood and another random sample of subjects with high lead levels in their blood. The statistics are summarized on the top of the next page. Use a 0.05 significance level to test the claim that IQ scores of people with low lead

levels vary more than IQ scores of people with high lead levels.

Low Lead Level: n = 78, \(\bar x\) = 92.88462, s = 15.34451

High Lead Level: n = 21, \(\bar x\) = 86.90476, s = 8.988352

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 鈥淓ssential Hypertension: Effect of an Oral Inhibitor of Angiotensin-Converting Enzyme,鈥 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

True?For the methods of this section, which of the following statements are true?

a.When testing a claim with ten matched pairs of heights, hypothesis tests using the P-valuemethod, critical value method, and confidence interval method will all result in the same conclusion.

b.The methods of this section are robustagainst departures from normality, which means that the distribution of sample differences must be very close to a normal distribution.

c.If we want to use a confidence interval to test the claim that\({{\bf{\mu }}_{\bf{d}}}{\bf{ < 0}}\)with a 0.01 significancelevel, the confidence interval should have a confidence level of 98%.

d.The methods of this section can be used with annual incomes of 50 randomly selected attorneysin North Carolina and 50 randomly selected attorneys in South Carolina.

e.With ten matched pairs of heights, the methods of this section require that we use n= 20.

Testing Claims About Proportions. In Exercises 7鈥22, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim

Question:Headache Treatment In a study of treatments for very painful 鈥渃luster鈥 headaches, 150 patients were treated with oxygen and 148 other patients were given a placebo consisting of ordinary air. Among the 150 patients in the oxygen treatment group, 116 were free from head- aches 15 minutes after treatment. Among the 148 patients given the placebo, 29 were free from headaches 15 minutes after treatment (based on data from 鈥淗igh-Flow Oxygen for Treatment of Cluster Headache,鈥 by Cohen, Burns, and Goads by, Journal of the American Medical Association, Vol. 302, No. 22). We want to use a 0.01 significance level to test the claim that the oxygen treatment is effective.

a. Test the claim using a hypothesis test.

b. Test the claim by constructing an appropriate confidence interval.

c. Based on the results, is the oxygen treatment effective?

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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