H0

Significance level: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif

Group with the larger mean: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif

(b) 99% confidence interval for m1-m2: -2.5 to 5.5

Conclusion: http://edugen.wileyplus.com/edugen/art2/common/pixel.gifH0

Significance level: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif

Group with the larger mean: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif

(c) 90% confidence interval for m1-m2: -10.8 to -3.3

Conclusion: http://edugen.wileyplus.com/edugen/art2/common/pixel.gifH0

Significance level: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif

Group with the larger mean: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif

Chapter 4, Section 5, Exercise 156

Are you “In a Relationship”?

A new study1 shows that relationship status on Facebook matters to couples. The study included 58college-age heterosexual couples who had been in a relationship for an average of 19 months. In 45 of the 58 couples, both partners reported being in a relationship on Facebook. In 31 of the 58 couples, both partners showed their dating partner in their Facebook profile picture. Men were somewhat more likely to include their partner in the picture than vice versa. However, the study states: “Females’ indication that they are in a relationship was not as important to their male partners compared with how females felt about male partners indicating they are in a relationship.” Using a population of college-age heterosexual couples who have been in a relationhip for an average of 19 months:

 

(a) A 95% confidence interval for the proportion with both partners reporting being in a relationship on Facebook is about 0.66 to 0.88. What is the conclusion in a hypothesis test to see if the proportion is different from 0.5? What significance level is being used?

Conclusion: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif H0

Significance level: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif

 

(b) A 95% confidence interval for the proportion with both partners showing their dating partner in their Facebook profile picture is about 0.40 to 0.66 . What is the conclusion in a hypothesis test to see if the proportion is different from 0.5? What significance level is being used?

Conclusion: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif H0

Significance level: http://edugen.wileyplus.com/edugen/art2/common/pixel.gif

 

1 Roan, Shari, “The true meaning of Facebook’s ‘in a relationship'”, Los Angeles Times, February 23, 2012, reporting on a study in Cyberpsychology, Behavior, and Social Networking.

Chapter 5, Section 1, Exercise 013

Find the specified areas for a normal density.

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(a) The area below 77 on a N(75,10) distribution

Round your answer to three decimal places.

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the tolerance is +/-2%

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(b) The area above 26 on a N(20,6) distribution

Round your answer to three decimal places.

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the tolerance is +/-2%

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(c) The area between 11 and 14 on a N(12.2, 1.6) distribution

Round your answer to three decimal places.

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the tolerance is +/-2%

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Chapter 5, Section 1, Exercise 020

Find endpoint(s) on the given normal density curve with the given property.

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(a) The area to the left of the endpoint on a N(5,4) curve is about 0.10.

Round your answer to two decimal places.

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the tolerance is +/-2%

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(b) The area to the right of the endpoint on a N(500,24) curve is about 0.05.

Round your answer to the nearest integer.

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the tolerance is +/-2%

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Chapter 5, Section 1, Exercise 028

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Choose the graph for the middle 80% for a standard normal distribution.

 

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Choose the graph for the the middle 80% for a standard normal distribution converted to a N(100,15) distribution.

 

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Chapter 5, Section 1, Exercise 031

Random Samples of College Degree Proportions

The distribution of sample proportions of US adults with a college degree for random samples of size n=500 is N(0.275,0.02). How often will such samples have a proportion, Ùp, that is more than 0.300?

Round your answer to one decimal place.

http://edugen.wileyplus.com/edugen/art2/common/pixel.gif[removed]% of samples of 500 US adults will contain more than 30.0% with at least a bachelor’s degree.

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the absolute tolerance is +/-0.1

 

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Chapter 5, Section 1, Exercise 040

Curving Grades on an Exam

A statistics instructor designed an exam so that the grades would be roughly normally distributed with mean m=75 and standard deviation δ=11. Unfortunately, a fire alarm with ten minutes to go in the exam made it difficult for some students to finish. When the instructor graded the exams, he found they were roughly normally distributed, but the mean grade was 60 and the standard deviation was 18. To be fair, he decides to ‘‘curve” the scores to match the desired N(75,11) distribution. To do this, he standardizes the actual scores to z-scores using the N(60,18) distribution and then ‘‘unstandardizes” those z-scores to shift to N(75,11).

What is the new grade assigned for a student whose original score was 45?

Round your answer to the nearest integer.

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How about a student who originally scores an 87?

Round your answer to the nearest integer.

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Chapter 5, Section 2, Exercise 044

Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed.

A 95% confidence interval for a proportion p if the sample has n=200 with Ùp=0.34, and the standard error is SE=0.03

Round your answers to three decimal places.

The 95% confidence interval is http://edugen.wileyplus.com/edugen/art2/common/pixel.gif[removed]to http://edugen.wileyplus.com/edugen/art2/common/pixel.gif[removed].

Chapter 5, Section 2, Exercise 046

Find the indicated confidence interval. Assume the standard error comes from a bootstrap distribution that is approximately normally distributed.

A 90% confidence interval for a mean mif the sample has n=80 with `x=22.9 and s=5.8, and the standard error is SE=0.65

Round your answers to three decimal places.

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Chapter 5, Section 2, Exercise 059

Where Is the Best Seat on the Plane?

A survey of 1000 air travelers1 found that 60% prefer a window seat. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error is SE=0.015.

Use a normal distribution to find a 90% confidence interval for the proportion of air travelers who prefer a window seat.

Round your answers to three decimal places.

The 90% confidence interval is http://edugen.wileyplus.com/edugen/art2/common/pixel.gif[removed]to http://edugen.wileyplus.com/edugen/art2/common/pixel.gif[removed].

1Willingham, A., ‘‘And the best seat on a plane is… 6A!,” HLNtv.com, April 25, 2012.

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Chapter 5, Section 2, Exercise 061

Smoke-Free Legislation and Asthma

Hospital admissions for asthma in children younger than 15 years was studied1 in Scotland both before and after comprehensive smoke-free legislation was passed in March 2006. Monthly records were kept of the annualized percent change in asthma admissions, both before and after the legislation was passed. For the sample studied, before the legislation, admissions for asthma were increasing at a mean rate of 5.2% per year. The standard error for this estimate is 0.7% per year. After the legislation, admissions were decreasing at a mean rate of 18.2% per year, with a standard error for this mean of 1.79%. In both cases, the sample size is large enough to use a normal distribution.

1Mackay, D., et. al., ‘‘Smoke-free Legislation and Hospitalizations for Childhood Asthma,” The New England Journal of Medicine, September 16, 2010; 363(12):1139-45.

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(a) Find a 95% confidence interval for the mean annual percent rate of change in childhood asthma hospital admissions in Scotland before the smoke-free legislation.

Round your answers to one decimal place.

The 95% confidence interval is http://edugen.wileyplus.com/edugen/art2/common/pixel.gif[removed]to http://edugen.wileyplus.com/edugen/art2/common/pixel.gif[removed].

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(b) Find a 95% confidence interval for the same quantity after the legislation.

Round your answers to one decimal place.

The 95% confidence interval is http://edugen.wileyplus.com/edugen/art2/common/pixel.gif[removed]to http://edugen.wileyplus.com/edugen/art2/common/pixel.gif[removed].

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(c) Is this an experiment or an observational study?

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Observational study

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(d) The evidence is quite compelling. Can we conclude cause and effect?

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Chapter 5, Section 2, Exercise 064

Penalty Shots in World Cup Soccer

A study1 of 138 penalty shots in World Cup Finals games between 1982 and 1994 found that the goalkeeper correctly guessed the direction of the kick only 41% of the time. The article notes that this is ‘‘slightly worse than random chance.” We use these data as a sample of all World Cup penalty shots ever. Test at a 5% significance level to see whether there is evidence that the percent guessed correctly is less than 50%. The sample size is large enough to use the normal distribution. The standard error from a randomization distribution under the null hypothesis is SE=0.043.

1St.John, A., ‘‘Physics of a World Cup Penalty-Kick Shootout – 2010 World Cup Penalty Kicks,” Popular Mechanics, June 14, 2010.

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State the null and alternative hypotheses.

 

What is the test statistic?

Round your answer to two decimal places.

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the tolerance is +/-2%

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What is the p-value?

Round your answer to three decimal places.

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the tolerance is +/-2%

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What is the conclusion?

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Reject H0 and find evidence that the proportion guessed correctly is less than half.

 

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Do not reject H0 and do not find evidence that the proportion guessed correctly is less than half.

 

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Do not reject H0and find evidence that the proportion guessed correctly is less than half.

 

Chapter 5, Section 2, Exercise 065

How Often Do You Use Cash?

In a survey1 of 1000 American adults conducted in April 2012, 43% reported having gone through an entire week without paying for anything in cash. Test to see if this sample provides evidence that the proportion of all American adults going a week without paying cash is greater than 40%. Use the fact that a randomization distribution is approximately normally distributed with a standard error of SE=0.016. Show all details of the test and use a 5% significance level.

1‘‘43% Have Gone Through a Week Without Paying Cash,” Rasmussen Reports, April 11, 2011.

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State the null and alternative hypotheses.

 

What is the test statistic?

Round your answer to two decimal places.

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the tolerance is +/-2%

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What is the p-value?

Round your answer to two decimal places.

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the tolerance is +/-2%

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What is the conclusion?

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Do not reject H0 and find evidence that the proportion is greater than 40%.

 

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Do not reject H0 and do not find evidence that the proportion is greater than 40%.

 

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Reject H0 and find evidence that the proportion is not greater than 40%.

 

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Reject H0 and find evidence that the proportion is greater than 40%.

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