Arnie Pizer Rochester Problem Library Fall 2005 WeBWorK ...webwork.maa.org/viewvc/rochester/trunk/rochester_problib/set... · WeBWorK assignment Statistics1Data due 07/01/2008 at

Arnie Pizer Rochester Problem Library Fall 2005WeBWorK assignment Statistics1Data due 07/01/2008 at 02:00am EDT.

1. (1 pt) rochesterLibrary/setStatistics1Data/ur stt 1 1.pg

Determine whether the following examples of data are quanti-tative or qualitative. Write ”QUANTITATIVE” for quantitativeand ”QUALITATIVE” for qualitative. (without quotations)

(a) The marital status of your coworkers.answer:

(b) The amount of bacteria on a piece of moldy bread.answer:

(c) Your college GPA.answer:

(d) The occupation of your neighbors.answer:


Determine whether the following examples are discrete or con-tinuous data sets. Write ”DISCRETE” for discrete and ”CON-TINUOUS” for continuous. (without quotations)

(a) The number of errors found on a student’s research paper.answer:

(b) The number of customers waiting in line at the grocerystore.answer:

(c) The number of students applying to graduate schools.answer:

(d) The length of time it takes to fill up your gas tank.answer:


Determine whether the follow descriptions correspond to an ob-servational study or an experiment. Write ”EXPERIMENT”for experiment and ”OBSERVATION” for observational study.(without quotations)

(a) The effectiveness of lecture teaching is tested with asample of students who has completed numerous lecture stylecourses.answer:

(b) Studying how patients respond when given a placebo.answer:

(c) A new antibiotic is tested in effectiveness by recordinghow the drug works on patients that already take the drug.answer:


Grade on Statistics ExamFrequencyBelow 50 350−59 160−69 1470−79 2080−89 2090−100 12

Given the frequency table above, construct the following:(a) The relative frequency table that corresponds with the

above table.

Grade on Statistics ExamRelative FrequencyBelow 5050−5960−6970−7980−8990−100

(b) The cumulative frequency table that corresponds with theabove table.

Grade on Statistics ExamCumulative FrequencyBelow 5050−5960−6970−7980−8990−100


Complete the table below.

Books read within the past yearFrequency Relative Frequencynone 80−4 75−9 19

10−14 0.20967741935483915−19 1320−25 2

total 62 1

Generated by the WeBWorK systemc©WeBWorK Team, Department of Mathematics, University of Rochester

1

Arnie Pizer Rochester Problem Library Fall 2005WeBWorK assignment Statistics2Measures due 07/02/2008 at 02:00am EDT.

1. (1 pt) rochesterLibrary/setStatistics2Measures/urstt 2 1.pg

The length (pgs) of math research projects is given below. Us-ing this information, calculate the range, variance, and standarddeviation.

49, 11, 25, 29, 42, 38, 26, 35, 30, 55, 25

range=variance=standard deviation=


Calculate the mode, mean, and median of the following data:

20, 7, 14, 12, 14, 16, 9, 14

Mode =Mean =Median =

3. (1 pt) rochesterLibrary/setStatistics2Measures/urstt 2 3a.pg

Given the data set below, calculate the range, variance, and stan-dard deviation.

19, 16, 16, 35, 42, 25, 30, 48, 9

range= variance= standard devia-tion =


For each of the given the data sets below, calculate the mean,variance, and standard deviation.

(a) 91, 32, 46, 28, 37, 49, 91, 5, 82mean= variance= standard deviation=

(b) 55, 43, 44, 57, 46, 42mean= variance= standard deviation=

(c) 4.3, 3.1, 4, 4.2, 4mean= variance= standard deviation=


Calculate the mean and median of the following grades on amath test:

84, 83, 80, 78, 77, 76, 74, 68, 65, 44, 27

Mean =Median =Is this data set skewed to the right, symmetric, or skewed to

the left?(Enter SR, SYM, or SL);


If the average low temperature of a winter month in Rochester,NY is 18◦ and the standard deviation is 4.3, then according toChebyshev’s theorem, the percentage of averages low tempera-tures in Rochester, NY between 9.4◦ and 26.6◦ is %.


Calculate the mean and median of the following data:

−9, −5, −3, −2, 4

Mean =Median =Is this data set skewed to the right, symmetric, or skewed to

the left?(Enter SR, SYM, or SL.)


Find the indicated decile of the following data set

28, 24, 24, 63, 31, 55, 35, 48, 26, 41, 14, 38

D8 =


IQ scores have a mean of 100 and a standard deviation of 15.John has an IQ of 124.What is the difference between John’s IQ and the mean?Convert John’s IQ score to a z score:


Joe took 4 courses last semester: Biology, Calculus, History,and Spanish. The means and standard deviations for the finalexams, and Joe’s scores are given in the table below. ConvertJoe’s score into z scores.

Subject Mean Stand. dev. Joe’s score Joe’s z scoreBiology 77 10 97Calculus 70 12 64History 53 16 57Spanish 44 12 50

On what exam did Joe have the highest relative score?(Enter the suject.)


Here is a list of 25 scores on a Math midterm exam:38.5, 41.5, 52, 52.5, 61, 63, 63.5, 68, 69, 69,78.5, 79, 80, 83, 87, 88.5, 88.5, 91, 91.5, 92,92.5, 94, 94, 97, 97

FindP82:

1


Here is a list of 27 scores on a Statistics midterm exam:

20, 30, 31, 32, 46, 48, 49, 52, 54,59, 61, 69, 71, 73, 74, 79, 81, 81,81, 85, 86, 87, 88, 91, 94, 96, 97

FindQ2:


2

Arnie Pizer Rochester Problem Library Fall 2005WeBWorK assignment Statistics3Estimates due 07/03/2008 at 02:00am EDT.

1. (1 pt) rochesterLibrary/setStatistics3Estimates/urstt 3 1.pg

Match the confidence level with the confidence interval forµ.

1. x±1.645(

σ√n

)2. x±2.575

(σ√n

)3. x± .99

(σ√n

)A. 90%B. 67.78%C. 99%


Starting salaries of 115 college graduates who have taken a sta-tistics course have a mean of $43,148 and a standard deviationof $9,646.Using a 0.96 degree of confidence, find both of the following:

A. The margin of errorE

B. The conficence interval for the meanµ:< µ<


A random sample of 80 observations produced a mean ofx =22.9 and a standard deviationx = 3.32.

(a) Find a 99% confidence interval forµ≤ µ≤

(b) Find a 95% confidence interval forµ≤ µ≤

(c) Find a 90% confidence interval forµ≤ µ≤


Listed below are the lenths (in minutes) of randomly selectedmusic CDs. Construct a 95% confidence interval for the meanlength of all such CDs.50.21 53.99 61.38 60.03 64.93 70.97 57.7750.42 41.76 47.05 65.24 58.02 55.68 76.8464.92 67.74 25.3 62.83 45.47 72.3 47.4759.84 35.72 56.15 47.06 54.44 49.16 52.1564.91 54.89 61.25 47.38 54.58 59.02 38.0657.38 39.75 60.95 45.14 47.12

< µ<


A random sample ofn measurements was selected from a pop-ulation with unknown meanµ and standard deviationσ. Calcu-late a 90 % confidence interval forµ for each of the followingsituations:

(a) n = 75, x = 52.1, s= 3.11≤ µ≤

(b) n = 75, x = 92.5, s= 3.75≤ µ≤

(c) n = 95, x = 18, s= 2.41

≤ µ≤(d) n = 95, x = 93.5, s= 4.58

≤ µ≤


Studies have suggusted that twins, in their early years, tend tohave lower IQs and pick up language more slowly than non-twins. The slower intellectual growth might be caused by be-nign parental neglect. Suppose it is desired to estimate the meanattention time given to twins per week by their parents. A sam-ple of 40 sets of 2 year old boys is taken, and after 1 week theattention time recieved was recorded. The data (in hours) calcu-lated the mean at 24.3 and the standard deviation at 11.9. Usethis information to contruct a 98% confidence interval for themean attention time given to al twin boys by their parents.

< µ<


Use the given data to find the 95% confidence interval esti-ate of the population meanµ. Assume that the population has anormal distribution.

IQ scores of professional athletes:Sample sizen = 20Meanx = 103Standard deviations= 14

< µ<


Suppose you have selected a random sample ofn= 13 measure-ments from a normal distribution. Compare the standard normalzvalues with the correspondingt values if you were forming thefollowing confidence intervals.

(a) 95% confidence intervalz=t =

(b) 90% confidence intervalz=t =

(c) 80% confidence intervalz=t =


Weights of 10 red and 36 brown randomly chosen M&Mplain candies are listed below.

Red:0.898 0.924 0.912 0.936 0.8770.907 0.871 0.933 0.92 0.923

1

Brown:

0.871 0.955 0.93 0.928 0.929 0.8670.858 0.876 0.912 0.985 0.902 0.9130.923 0.988 1.001 0.856 0.889 0.9090.857 0.986 0.921 0.904 0.936 0.8720.909 0.875 0.966 0.931 0.866 0.930.931 0.897 0.905 0.898 0.914 0.861

1. To construct a 90% confidence interval for the meanweight of red M&M plain candies, you have to use

• A. The t distribution with 11 degrees of freedom• B. The normal distribution• C. The t distribution with 10 degrees of freedom• D. The t distribution with 9 degrees of freedom• E. None of the above

2. A 90% confidence interval for the mean weight of redM&M plain candies is

< µ<3. To construct a 90% confidence interval for the mean

weight of brown M&M plain candies, you have to use

• A. The normal distribution• B. The t distribution with 35 degrees of freedom• C. The t distribution with 37 degrees of freedom• D. The t distribution with 36 degrees of freedom• E. None of the above

4. A 90% confidence interval for the mean weight of brownM&M plain candies is

< µ<


The following random sample was selected from a normal dis-tribution:

3 3 15 5 4 9 3 4 5 9

(a) Construct a 90% confidence interval for the populationmeanµ.

≤ µ≤(b) Construct a 95% confidence interval for the population

meanµ.≤ µ≤


The scientific productivity of major world cities was the subjectof a recent study. The study determined the number of scientificpapers published between 1994 and 1997 by researchers fromeach of the 20 world cities, and is shown below.

City Number of papers City Number of papersCity 1 2 City 11 22City 2 10 City 12 9City 3 25 City 13 28City 4 24 City 14 25City 5 25 City 15 2City 6 30 City 16 17City 7 22 City 17 5City 8 30 City 18 21City 9 24 City 19 6City 10 5 City 20 18

Construct a 80 % confidence interval for the average numberof papers published in major world cities.

< µ<


The standard IQ test is designed so that the mean is 100 andthe standard deviation is 15 for the population of all adults. Wewish to find the sample size necessary to estimate the mean IQscore of statistics students. Suppose we want to be 96% con-fident that our sample mean is within 1.5 IQ points of the truemean. The mean for this population is clearly greater than 100. The standard deviation for this population is probably lessthan 15 because it is a group with less variation than a grouprandomly selected from the general population; therefore, if weuseσ = 15, we are being conservative by using a value that willmake the sample size at least as large as necessary. Assume thenthatσ = 15 and determine the required sample size.

Answer:


Periodically, the county Water Department tests the drinkingwater of homeowners for contminants such as lead and copper.The lead and copper levels in water specimens collected in 1998for a sample of 10 residents of a subdevelopement of the countyare shown below.

lead (µg/L) copper (mg/L)3.7 0.0282.2 0.2914.8 0.0891.1 0.1973.7 0.4840 0.406

1.3 0.1850.8 0.6595.7 0.112 0.732

(a) Construct a 99% confidence interval for the mean leadlevel in water specimans of the subdevelopment.

≤ µ≤(b) Construct a 99% confidence interval for the mean copper

level in water specimans of the subdevelopment.≤ µ≤

2


Suppose that the minimum and maximum ages for typical text-books currently used in college courses are 0 and 8 years. Usethe range rule of thumb to estimate the standard deviation.

Standard deviation =Find the size of the sample required to estimage the mean age

of textbooks currently used in college courses. Assume that youwant 93% confidence that the sample mean is within 0.25 yearof the population mean.

Required sample size =


A poll is taken in which 327 out of 550 randomly selected vot-ers indicated their preference for a certain candidate. Find a80% confidence interval forp.

≤ p≤16. (1 pt) rochesterLibrary/setStatistics3Estimates/urstt 3 14.pg

Use the given confidence interval limits to find the point esti-matep and the margin of errorE.

0.79< p < 0.97p = E =


Astronaunts often report that there are times when they becomedisoriented as they move around in zero-gravity. Therefore, theyususally rely on bright colors and other visual information tohelp them estabish a top-down orientation. A study was con-ducted to assses the potential of using color as body orienting.80 college students, reclining on their backs in the dark, foundit difficult to establish orientation when positioned on under arotating disk. This rotating disk was painted half black and halfwhite. Out of the 80 students, 64 believed they were right sideup when the white was on top.

Use this information to estimate the true proportion of sub-jects who use the white color as a cue for right-side-up orien-tation. That is, construct a 90% confidence interval for the trueproportion.

≤ p≤18. (1 pt) rochesterLibrary/setStatistics3Estimates/urstt 3 16.pg

Construct the 95% confidence interval estimate of the popula-tion proportionp if the sample size isn = 200 and the numberof successes in the sample isx = 50.

< p <


The EPA wants to test a randomly selected sample ofn waterspecimens and estimate the mean daily rate of pollution pro-duced by a mining operation. If the EPA wants a 99% confi-dence interval with a bound of error of 1.5 milligram per liter

(mg/L), how many water specimens are required in the sam-ple? Assume prior knowledge indicates that pollution readingsin water samples taken during a day have been approximatelynormally distributed with a standard deviation of 4 (mg/L).

n =


College officials want to estimate the percentage of studentswho carry a gun, knife, or other such weapon. How many ran-domly selected students must be surveyed in order to be 91%confident that the sample percentage has a margin of error of1.5 percentage points?

(a) Assume that there is no available information that couldbe used as an estimate of ˆp.

Answer:(b) Assume that another study indicated that 7% of college

students carry weapons.Answer:


A random sample of elementary school children in New Yorkstate is to be selected to estimate the proportionp who have re-ceived a medical examination during the past year. An intervalestimate of the proportionp with a bound of 0.075 and 99%confidence is required.

(a) Assuming no prior information aboutp is available, ap-proximately how large of a sample size is needed?n =

(b) If a planning study indicates thatp is around 0.4, ap-proximately how large of a sample size is needed?n =


Find the critical valuesχ2L = χ2

1−α/2 andχ2R = χ2

α/2 that corre-spond to 95% degree of confidence and the sample sizen = 13.

χ2L = χ2

R =


According to the Food and Drug Administration (FDA), a cupof coffee contains on average 115 miligrams (mg) of caffeine,with the amount per cup ranging from 60 to 180 mg. Supposeyou want to repeat the FDA experiment to obtain an estimate ofthe mean caffeine content in a cup of coffee correct to witin 3.2mg with 95% confidence. How many cups of coffee would haveto be included in your sample?

n =


Find the minimum sample size needed to be 95% confident thatthe sample variance is within 30% of the population variance.


3

Arnie Pizer Rochester Problem Library Fall 2005WeBWorK assignment Statistics4HypothesisTesting due 07/04/2008 at 02:00am EDT.

1. (1 pt) rochesterLibrary/setStatistics4HypothesisTesting-

/ur stt 4 1.pg

Type I error is:

• A. Deciding null hypothesis is true when it is false• B. Deciding null hypothesis is false when it is true• C. Deciding alternative hypothesis is true when it is

false• D. Deciding alternative hypothesis is true when it is true• E. All of the above• F. None of the above

Type II error is:

• A. Deciding null hypothesis is true when it is false• B. Deciding alternative hypothesis is false when it is

true• C. Deciding null hypothesis is false when it is true• D. Deciding alternative hypothesis is true when it is true• E. All of the above• F. None of the above


/ur stt 4 2.pg

For each statement, express the null hypothesisH0 and alterna-tive hypothesisH1 in symbolic form.

1. The mean salary of statistics professors is greater than70,000 dollars.

• A. H0 : µ≥ 70,000, H1 : µ< 70,000• B. H0 : µ< 70,000, H1 : µ≥ 70,000• C. H0 : µ≤ 70,000, H1 : µ> 70,000• D. H0 : µ> 70,000, H1 : µ≤ 70,000

2. At most one-half of all Internet users make on-line pur-chases.

• A. H0 : µ≤ 0.5, H1 : µ> 0.5• B. H0 : µ≥ 0.5, H1 : µ< 0.5• C. H0 : p≤ 0.5, H1 : p > 0.5• D. H0 : p≥ 0.5, H1 : p < 0.5

3. IQ scores of statistics students have a standard deviationat most 15.

• A. H0 : σ≤ 15, H1 : σ > 15• B. H0 : µ≤ 15, H1 : µ> 15• C. H0 : µ< 15, H1 : µ≥ 15• D. H0 : σ≥ 15, H1 : σ < 15


/ur stt 4 3.pg

Given the significance levelα = 0.03 find the following:(a) lower-tailedz value

z=(b) right-tailedz value

z=

(c) two-tailedz value|z|=


/ur stt 4 4.pg

Find the criticalzvalue for a right-tailed test using a significancelevel of α = 0.1.


/ur stt 4 6.pg

Find the criticalzvalue using a significance level ofα = 0.06 ifthe alternative hypothesisH0 is µ< 42.


/ur stt 4 5.pg

A random sample of 100 observations from a population withstandard deviation 17.1962203310272 yielded a sample meanof 92.9.(a) Given that the null hypothesis isµ= 90 and the alternativehypothesis isµ> 90 usingα = .05, find the following:

(i) critical z score(ii) test statistic=

(b) Given that the null hypothesis isµ= 90 and the alternativehypothesis isµ 6= 90 usingα = .05, find the following:

(i) the positive criticalz score(ii) the negative criticalz score(iii) test statistic=The conclusion from part (a) is:

• A. Reject the null hypothesis• B. There is insufficient evidence to reject the null hy-

pothesis• C. None of the above

The conclusion from part (b) is:

• A. Reject the null hypothesis• B. There is insufficient evidence to reject the null hy-

pothesis• C. None of the above


/ur stt 4 7.pg

It is necessary for an automobile producer to estimate the num-ber of miles per gallon achieved by its cars. Suppose that thesample mean for a random sample of 110 cars is 29.5 miles andassume the standard deviation is 2.3 miles. Now suppose the carproducer wants to test the hypothesis thatµ, the mean numberof miles per gallon, is 26.6 against the alternative hypothesisthat it is not 26.6. Conduct a test usingα = .05 by giving thefollowing:

(a) positive criticalz score(b) negative criticalz score(c) test statistic

1

The final conclustion is

• A. We can reject the null hypothesis thatµ = 26.6 andaccept thatµ 6= 26.6.

• B. There is not sufficient evidence to reject the null hy-pothesis thatµ= 26.6.


/ur stt 4 8.pg

The contents of 32 cans of Coke have a mean ofx= 12.15 and astandard deviation ofs= 0.09. Find the value of the test statisticz for the claim that the population mean isµ= 12.


/ur stt 4 9.pg

Golf-course designers have become concerned that old coursesare becoming obsolete since new technology has given golfersthe ability to hit the ball so far. Designers, therefore, have pro-posed that new golf courses need to be built expecting that theaverage golfer can hit the ball more than 235 yards on average.Suppose a random sample of 150 golfers be chosen so that theirmean driving distance is 238.4 yards, with a standard deviationof 40.6.

Conduct a hypothesis test whereH0 : µ≤ 235 andH1 : µ >235 by computing the following:(a) test statistic(b) p-valuep =(c) If this was a two-tailed test, then the p-value is


/ur stt 4 9a.pg

Golf-course designers have become concerned that old coursesare becoming obsolete since new technology has given golfersthe ability to hit the ball so far. Designers, therefore, have pro-posed that new golf courses need to be built expecting that theaverage golfer can hit the ball more than 255 yards on average.Suppose a random sample of 122 golfers be chosen so that theirmean driving distance is 255.9 yards, with a standard deviationof 47.6.

Conduct a hypothesis test whereH0 : µ = 255 andH1 : µ >255 by computing the following:(a) test statistic(b) p-valuep =(c) If this was a two-tailed test, then the p-value is


/ur stt 4 10.pg

Assume you are using a significance level ofα = 0.05 to testthe claim thatµ< 9 and that your sample is a random sample of47 values. Findβ, the probability of making a type II error (fail-ing to reject a false null hypothesis), given that the populationactually has a normal distribution withµ= 5 andσ = 7.

β =


/ur stt 4 11.pg

Physicians at a clinic gave what they thought were drugs to 990asthma, ulcer, and herpes patients. Although the doctors laterlearned that the drugs were really placebos, 53 % of the patientsreported an improved condition. Assume that if the placebo isineffective, the probability of a patients condition improving is0.51. For the hypotheses that the proportion of improving is0.51 against that it is> 0.51, find the p-value.

p =


/ur stt 4 12.pg

Test the claim that the population of sophomore college studentshas a mean grade point average greater than 2.2. Sample statis-tics includen = 180,x = 2.38, ands= 0.6. Use a significancelevel of α = 0.03.

The test statistic isThe critical value isThe P-Value isThe final conclustion is

• A. There is sufficient evidence to support the claim thatthe mean grade point average is greater than 2.2.

• B. There is not sufficient evidence to support the claimthat the mean grade point average is greater than 2.2.


/ur stt 4 14.pg

50 people are randomly selected and the accuracy of their wrist-watches is checked, with positive errors representing watchesthat are ahead of the correct time and negative errors represent-ing watches that are behind the correct time. The 50 values havea mean of 104sec and a standard deviation of 200sec. Use a0.01 significance level to test the claim that the population of allwatches has a mean of 0sec.

The test statistic isThe P-Value isThe final conclustion is

• A. There is not sufficient evidence to warrant rejectionof the claim that the mean is equal to 0

• B. There is sufficient evidence to warrant rejection ofthe claim that the mean is equal to 0


/ur stt 4 13.pg

A sample of 7 measurments, randomly selected from a normallydistributed population, resulted in a sample mean,x = 5.6 andsample standard deviations = 1.83. Usingα = 0.05, test thenull hypothesis that the mean of the population is 5.5 against thealternative hypothesis that the mean of the population,µ < 5.5by giving the following:(a) the degree of freedom(b) the criticalt value(c) the test statistic

2


• A. There is not sufficient evidence to reject the null hy-pothesis thatµ= 5.5.

• B. We can reject the null hypothesis thatµ = 5.5 andaccept thatµ< 5.5.


/ur stt 4 15.pg

The effectiveness of a new bug repellent is tested on 18 subjectsfor a 10 hour period. Based on the number and location of thebug bites, the percentage of surface area exposed protected frombites was calculated for each of the subjects. The results wereas follows:

x = 87 %, s= 14%The new repellent is considered effective if it provides a per-

cent repellency of at least 96. Usingα = 0.01, construct a hy-pothesis test with null hypothesisµ = 0.96 and alternative hy-pothesisµ > 0.96 to determine whether the mean repellency ofthe new bug relellent is greater than 96 by computing the fol-lowing:

(a) the degree of freedom(b) the criticalt value(c) the test statisticsThe final conclustion is

• A. We can reject the null hypothesis thatµ = 0.96 andaccept thatµ > 0.96, that is, the bug repellent is effec-tive.

• B. There is not sufficient evidence to reject the null hy-pothesis thatµ= 0.96.


/ur stt 4 16.pg

Test the claim that for the population of statistics final exams,the mean score isµ = 72. Sample statistics includen = 26,x = 74, ands= 13. Use a significance level ofα = 0.02.

The test statistic isThe positive critical value isThe negative critical value isThe conclustion is

• A. There is sufficient evidence to warrant rejection ofthe claim that the mean score is equal to 72

• B. There is not sufficient evidence to warrant rejectionof the claim that the mean score is equal to 72


/ur stt 4 18.pg

When a poultry farmer uses his regular feed, the newborn check-ens have normally distributed weights with a mean of 63 oz. Inan experiment with an enriched feed mixture, ten chickens areborn with the following weights (in ounces).

67.8, 65.8, 65.9, 66.6, 64, 66, 66.1, 68.1, 61, 63.7

Use theα = 0.05 significance level to test the claim that themean weight is higher with the enriched feed.

The sample mean isx =The sample standard deviation iss=The test statistic ist =The critical value ist =The conclusion is

• A. There is sufficient evidence to support the claim thatwith the enriched feed, the mean weight is greater than63.

• B. There is not sufficient evidence to support the claimthat with the enriched feed, the mean weight is greaterthan 63.


/ur stt 4 24.pg

One of the most feared predators in the ocean is the great whiteshark. It is known that the white shark grows to a mean lengthof 18 feet; however, one marine biologist believes that greatwhite sharks off the Bermuda coast grow much longer. To testthis claim, full-grown white sharks were captured, measured,and then set free. However, this was a difficult, costly andvery dangerous task, so only four sharks were actually sampled.Their lengths were 24, 24, 26,and 26 feet. Do the data providesufficient evidence to support the claim? Useα = 0.01

test statistict =rejection regiont >The final conclustion is

• A. There is not sufficient evidence to reject the null hy-pothesis that the average length of the shark is 18.

• B. We can reject the null hypothesis that the averagelength of the shark is 18, and accept that the averagelength of the shark is greater than 18.


/ur stt 4 17.pg

A random sample of 100 observations is selected from a bino-mial population with unknown probability of successp. Thecomputed value of ˆp is 0.77.(1) TestH0 : p≤ 0.65 againstH1 : p > 0.65. Useα = 0.01.

test statisticz=critical z score


• A. There is not sufficient evidence to reject the null hy-pothesis thatp≤ 0.65.

• B. We can reject the null hypothesis thatp≤ 0.65 andaccept thatp > 0.65.

(2) TestH0 : p≥ 0.5 againstH1 : p < 0.5. Useα = 0.01.test statisticz=critical z score


• A. We can reject the null hypothesis thatp≥ 0.5 andaccept thatp < 0.5.

• B. There is not sufficient evidence to reject the null hy-pothesis thatp≥ 0.5.

3

(3) TestH0 : p = 0.65 againstH1 : p 6= 0.65. Useα = 0.01.test statisticz=positive criticalz scorenegative criticalz score


• A. There is not sufficient evidence to reject the null hy-pothesis thatp = 0.65.

• B. We can reject the null hypothesis thatp = 0.65 andaccept thatp 6= 0.65.


/ur stt 4 19.pg

According to a recent marketing campaign, 110 drinkers of ei-ther Diet Coke or Diet Pepsi participated in a blind taste test tosee which of the drinks was their favorite. In one Pepsi televi-sion commercial, an anouncer states that ”in recent blind tastetests, more than one half of the surveyed preferred Diet Pepsiover Diet Coke.” Suppose that out of those 110, 44 preferredDiet Pepsi. Test the hypothesis, usingα = 0.01 that more thanhalf of all participants will select Diet Pepsi in a blind taste testby giving the following:(a) the test statistic(b) the criticalz score


• A. There is not sufficient evidence to reject the null hy-pothesis thatp≤ 0.5.

• B. We can reject the null hypothesis thatp≤ 0.5 andaccept thatp > 0.5.


/ur stt 4 20.pg

A survey of 1180 people who took trips revealed that 124 ofthem included a visit to a theme park. Based on those surveryresults, a management consultant claims that less than 12 %of trips include a theme park visit. Test this claim using theα = 0.05 significance level.

The test statistic isThe critical value isThe conclusion is

• A. There is not sufficient evidence to support the claimthat less than 12 % of trips include a theme park visit.

• B. There is sufficient evidence to support the claim thatless than 12 % of trips include a theme park visit.


/ur stt 4 23.pg

A new cream that advertises that it can reduce wrinkles and im-prove skin was subject to a recent study. A sample of 41 women

over the age of 50 used the new cream for 6 months. Of those41 women, 34 of them reported skin improvement(as judged bya dermatologist). Is this evidence that the cream will improvethe skin of more than 60% of women over the age of 50? Testusingα = 0.01.

test statisticsz=rejection regionz>The final conclustion is

• A. There is not sufficient evidence to reject the null hy-pothesis thatp = 0.6. That is, there is not sufficientevidence to reject that the cream can improve the skinof more than 60% of women over 50.

• B. We can reject the null hypothesis thatp = 0.6 andaccept thatp > 0.6. That is, the cream can improve theskin of more than 60% of women over 50.


/ur stt 4 21.pg

A random sample ofn = 7 observations from a normal popula-tion produced the following measurements:

2 3 4 9 0 8 7Do the data provide sufficient evidence to indicate thatσ2 < 3?Useα = 0.01, and compute the following:(a) sample standard deviations=(b) test statisticχ2 =(c) critical χ2

α =The final conclustion is

• A. There is not sufficient evidence to reject the null hy-pothesis thatσ2 = 3.

• B. We can reject the null hypothesis thatσ2 = 3 andaccept thatσ2 < 3.


/ur stt 4 22.pg

Use aα = 0.01 significance level to test the claim thatσ = 12 ifthe sample statistics includen = 12, x = 97, ands= 19.

The test statistic isThe smaller critical number isThe bigger critical number isWhat is your conclusion?

• A. There is sufficient evidence to warrant the rejectionof the claim that the population standard deviation isequal to 12

• B. There is not sufficient evidence to warrant the rejec-tion of the claim that the population standard deviationis equal to 12


4

Arnie Pizer Rochester Problem Library Fall 2005WeBWorK assignment Statistics5Inferences2Samples due 07/05/2008 at 02:00am EDT.

1. (1 pt) rochesterLibrary/setStatistics5Inferences2Samples-

/ur stt 5 1.pg

In order to compare the means of two populations, independentrandom samples of 430 observations are selected from eachpopulation, with the following results:

Sample 1 Sample 2x1 = 5295 x2 = 5083s1 = 130 s2 = 145

(a) Use a 96 % confidence interval to estimate the differencebetween the population means(µ1−µ2).

≤ (µ1−µ2)≤(b) Test the null hypothesis:H0 : (µ1−µ2) = 0 versus the al-ternative hypothesis:Ha : (µ1−µ2) 6= 0. Usingα = 0.04, givethe following:

(i) the test statisticz=(ii) the positive criticalz score(iii) the negative criticalz scoreThe final conclustion is

• A. We can reject the null hypothesis that(µ1−µ2) = 0and accept that(µ1−µ2) 6= 0.

• B. There is not sufficient evidence to reject the null hy-pothesis that(µ1−µ2) = 0.

(c) Test the null hypothesis:H0 : (µ1−µ2) = 24 versus the al-ternative hypothesis:Ha : (µ1−µ2) 6= 24. Usingα = 0.04, givethe following:

(i) the test statisticz=(ii) the positive criticalz score(iii) the negative criticalz scoreThe final conclustion is




/ur stt 5 2.pg

Test the claim that the two samples described below come frompopulations with the same mean. Assume that the samples areindependent simple random samples. Use a significance levelof 0.05.Sample 1:n1 = 70, x1 = 17, s1 = 2.Sample 2:n2 = 93, x2 = 19, s2 = 3.

The test statistic isThe P-Value isThe conclusion is

• A. There is sufficient evidence to warrant rejection ofthe claim that the two populations have the same mean.

• B. There is not sufficient evidence to warrant rejectionof the claim that the two populations have the samemean.


/ur stt 5 3.pg

The purpose of this question is to compare the variability ofx1

andx2 with the variability of(x1−x2).(a) Suppose the first sample of 100 observations is selected

from a population with meanµ1 = 170 and varianceσ21 = 1380.

Construct an interval extending 2 standard deviations ofx1 oneach side ofµ1.

≤ µ1 ≤(b) Suppose the second sample of 100 observations is se-

lected from a population with meanµ2 = 170 and varianceσ2

2 = 1200. Construct an interval extending 2 standard devia-tions ofx2 on each side ofµ2.

≤ µ2 ≤(c) Consider the difference between the two sample means

(x1− x2). Compute the mean and the standard deviation of thesampling distribution of(x1−x2).mean =standard deviation =

(d) Based on 100 observations, construct an interval extend-ing 2 standard deviations of(x1−x2) on each side of(µ1−µ2)

≤ (µ1−µ2)≤4. (1 pt) rochesterLibrary/setStatistics5Inferences2Samples-

/ur stt 5 4.pg

Randomly selected 110 student cars have ages with a mean of7.5 years and a standard deviation of 3.4 years, while randomlyselected 65 faculty cars have ages with a mean of 5 years and astandard deviation of 3.7 years.

1. Use a 0.03 significance level to test the claim that studentcars are older than faculty cars.

The test statistic isThe critical value isIs there sufficient evidence to support the claim that student

cars are older than faculty cars?

• A. No• B. Yes

2. Construct a 97% confidence interval estimate of the dif-ferenceµ1−µ2, whereµ1 is the mean age of student cars andµ2

is the mean age of faculty cars.< (µ1−µ2) <


/ur stt 5 5.pg

Two independent samples have been selected, 93 observationsfrom population 1 and 67 observations from population 2. Thesample means have been calculated to bex1 = 9.2 andx2 = 5.4.From previous experience with these populations, it is knownthat the variances areσ2

1 = 28 andσ22 = 36.

1

(a) Findσ(x1−x2).answer:

(b) Determine the rejection region for the test ofH0 :(µ1−µ2) = 3.03 andHa : (µ1−µ2) > 3.03 Useα = 0.01.z>

(c) Compute the test statistic.z=


• A. There is not sufficient evidence to reject the null hy-pothesis that(µ1−µ2) = 3.03.

• B. We can reject the null hypothesis that(µ1− µ2) =3.03 and accept that(µ1−µ2) > 3.03.

(d) Construct a 99 % confidence interval for(µ1−µ2).≤ (µ1−µ2)≤


/ur stt 5 7.pg

Test the claim that the two samples described below come frompopulations with the same mean. Assume that the samples areindependent simple random samples. Use a signifcance level ofα = 0.01Sample 1:n1 = 17, x1 = 21, s1 = 6.11Sample 2:n2 = 12, x2 = 28.8, s2 = 4.76(a) The degree of freedom is(b) The test statistic is(c) Determine the rejection region for the test ofH0 : (µ1−µ2) = 0 andHa : (µ1−µ2) 6= 0|t|>





/ur stt 5 16.pg

Test the given claim using theα = 0.05 significance level andassuming that the populations are normally distributed.

Claim: The treatment population and the placebo populationhave the same mean.

Treatment group:n = 7, x = 67, s= 5.8.Placebo group:n = 15, x = 76, s= 5.6.The test statistic isThe positive critical value isThe negative critical value isIs there sufficient evidence to warrant the rejection of the

claim that the treatment and placebo populations have the samemean?

• A. Yes• B. No


/ur stt 5 18.pg

Randomly selected students were given five seconds to estimatethe value of a product of numbers with the results shown below.

Estimates from students given 1×2×3×4×5×6×7×8:

40320, 150, 200, 500, 2040, 5635, 4000, 842, 200, 5000

Estimates from students given 8×7×6×5×4×3×2×1:

50000, 1500, 23410, 225, 2500, 400, 2000, 1876, 40320, 200

Use a 0.05 significance level to test the following claims:1. Claim: the two populations have equal variances.The test statistic isThe larger critical value isThe conclusion is

• A. There is sufficient evidence to warrant the rejectionof the claim that the two populations have equal vari-ances

• B. There is not sufficient evidence to warrant the rejec-tion of the claim that the two populations have equalvariances

2. Claim: the two populations have the same mean.The test statistic isThe positive critical value isThe negative critical value isThe conclusion is

• A. There is not sufficient evidence to warrant the rejec-tion of the claim that the two populations have the samemean

• B. There is sufficient evidence to warrant the rejectionof the claim that the two populations have the samemean


/ur stt 5 6.pg

Suppose you want to test the claim the the paired sample datagiven below come from a population for which the mean differ-ence isµd = 0.

x 85 86 73 83 64 70 57y 92 92 83 68 94 56 69

Use a 0.05 significance level to find the following:(a) The mean value of the differncesd for the paired sample

datad =

(b) The standard deviation of the differencesd for the pairedsample datasd =

(c) The t test statistict =

(d) The positive critical valuet =

(e) The negative critical valuet =

(f) Does the test statistic fall in the critical region?

• A. Yes• B. No

2

(g) Construct a 95% conficence interval for the populationmean of all differencesx−y.

< µd <


/ur stt 5 8.pg

Ten randomly selected people took an IQ test A, and next daythey took a very similar IQ test B. Their scores are shown in thetable below.

Person A B C D E F G H I JTest A 111 71 98 90 118 96 102 126 83 106Test B 111 71 97 87 117 98 101 128 83 112

1. Use a 0.05 significance level to test the claim that peopledo better on the second test than they do on the first.

The test statistic isThe critical vaue isIs there sufficient evidence to support the claim that people

do better on the second test?

• A. Yes• B. No

2. Construct a 95% confidence interval for the mean of thedifferences.

< µ<


/ur stt 5 9.pg

A paired difference experiment yieldednD pairs of observa-tions. In each case described below, what is the rejection regionfor testingH0 : µ= 7 againstHa : µ> 7? UsesD = 10.8.(a) nD = 34, α = 0.05

z>(b) nD = 26, α = 0.09

t >(c) nD = 25, α = 0.06

t >


/ur stt 5 11.pg

A paired difference experiment produced the following results:

nD = 42, x1 = 179, x2 = 171, xD = 8, sD = 58,

(a) Determine the rejection region for the hypothesisH0 : µD =0 if Ha : µD > 0. Useα = 0.09.

z>(b) Conduct a paired difference test described above.

The test statistic isThe final conclustion is

• A. There is not sufficient evidence to reject the null hy-pothesis thatµD = 0.

• B. We can reject the null hypothesis thatµD = 0 andaccept thatµD > 0.


/ur stt 5 10.pg

In a study of red/green color blindness, 850 men and 2550women are randomly selected and tested. Among the men, 73have red/green color blindness. Among the women, 6 havered/green color blindness. Test the claim that men have a higherrate of red/green color blindness.

The test statistic isIs there sufficient evidence to support the claim that men have

a higher rate of red/green color blindness than women?

• A. Yes• B. No

Construct the 96% confidence interval for the difference be-tween the color blindness rates of men and women.

< (p1− p2) <


/ur stt 5 13.pg

Independent random samples, each containing 900 observa-tions, were selected from two binomial populations. The sam-ples from populations 1 and 2 produced 823 and 681 successes,respectively.(a) TestH0 : (p1− p2) = 0 againstHa : (p1− p2) 6= 0. Useα = 0.09

test statistic=rejection region|z|>The final conclustion is

• A. There is not sufficient evidence to reject the null hy-pothesis that(p1− p2) = 0.

• B. We can reject the null hypothesis that(p1− p2) = 0and accept that(p1− p2) 6= 0.

(b) TestH0 : (p1− p2) = 0 againstHa : (p1− p2) > 0. Useα = 0.05

test statistic=rejection regionz>The final conclustion is

• A. We can reject the null hypothesis that(p1− p2) = 0and accept that(p1− p2) > 0.

• B. There is not sufficient evidence to reject the null hy-pothesis that(p1− p2) = 0.


/ur stt 5 15.pg

Suppose a group of 800 smokers (who all wanted to give upsmoking) were randomly assigned to recieve an antidepres-sant drug or a placebo for six weeks. Of the 318 patientswho recieved the antidepressant drug, 26 were not smok-ing one year later. Of the 482 patients who recieved theplacebo, 42 were not smoking one year later. Given the nullhypothesisH0 : (p1− p2) = 0 and the alternative hypothesisHa : (p1− p2) 6= 0, conduct a test to see if taking an antidepres-sant drug can help smokers stop smoking. Useα = 0.08(a) The rejection region is|z|>

3

(b) The test statistic isz=The final conclustion is

• A. There is not sufficient evidence to reject the null hy-pothesis that(p1− p2) = 0.

• B. We can reject the null hypothesis that(p1− p2) = 0and accept that(p1− p2) 6= 0.


/ur stt 5 14.pg

Test the given claim using theα = 0.02 significance level andassuming that the populations are normally distributed.

Claim: The treatment population and the placebo populationhave different variances.

Treatment group:n = 9, x = 72.8, s= 20.1.Placebo group:n = 15, x = 84.7, s= 13.5.The test statistic isThe larger critical value isWhat is your conclusion?

• A. There is sufficient evidence to support the claim thatthe treatment and placebo populations have differentvariances.

• B. There is not sufficient evidence to support the claimthat the treatment and placebo populations have differ-ent variances.


/ur stt 5 17.pg

Suppose you wanted to estimate the difference between twopopulation means correct to within 3.2 with probability 0.93.If prior information suggests that the popluation variances areapproximately equal toσ2

1 = σ22 = 14 and you want to select

independent random samples of equal size from the poulations,how large should the sample sizes,n1 andn2 be?

answer:n1 = n2 =


/ur stt 5 12.pg

Find the size of each sample needed to estimate the differencebetween the proportions of boys and girls under 10 years oldwho are afraid of spiders. Assume that we want 97% confidencethat the error is smaller than 0.03.

n =


/ur stt 5 12a.pg

The sample size needed to estimate the difference betweentwo population proportions to within a margin of errorE with asignificance level ofα can be found as follows. In the expres-sion

E = zα/2

√p1q1

n1+

p2q2

n2

we replace bothn1 andn2 by n (assuming that both sampleshave the same size) and replace each ofp1, p2, q1, andq2 by0.5 (because their values are not known). Then we solve for n,and get

n =(zα/2)2

2E2 .

Finally, increase the value ofn to the next larger integer num-ber.

Use the above formula to find the size of each sample neededto estimate the difference between the proportions of boys andgirls under 10 years old who are afraid of spiders. Assume thatwe want 97% confidence that the error is smaller than 0.05.

n =


4

Arnie Pizer Rochester Problem Library Fall 2005WeBWorK assignment Statistics6CorrelationRegression due 07/06/2008 at 02:00am EDT.

1. (1 pt) rochesterLibrary/setStatistics6CorrelationRegression-

/ur stt 6 2.pg

Use a scatterplot and the linear correlation coefficientr to determine whether there is a correlation between the two variables.

x 0.2 1.3 2.6 3.1 4.3 5.2 6.5 7.5 8 9.1 10 11.2 12.2 13 14.4y 14.2 10.3 6.4 5.2 2.7 1.4 0.3 0 0.1 0.7 1.7 3.7 5.9 8.1 12.7

r =There is

• A. a positive correlation betweenx andy• B. a negative correlation betweenx andy• C. a nonlinear correlation betweenx andy• D. a perfect positive correlation betweenx andy• E. a perfect negative correlation betweenx andy• F. no correlation betweenx andy


/ur stt 6 1.pg

Match the following sample correlation coefficients with the ex-plaination of what that correlation coeffiecient means.

1. r =−12. r = 13. r = .924. r =−.97

A. a perfect positive relationship betweenx andyB. a strong negative relationship betweenx andyC. a strong positive relationship betweenx andyD. a perfect negative relationship betweenx andy


/ur stt 6 3.pg

Given the following data set,

x −2 1 −1 1 1 3 0y 2 2 3 2 3 5 4

Compute the coefficient of correlationrr =


/ur stt 6 4.pg

Heights (in centimeters) and weights (in kilograms) of 7 super-models are given below. Find the regression equation, lettingthe first variable be the independent(x) variable, and predict theweight of a supermodel who is 171 cm tall.

Height 178 178 176 174 172 176 166Weight 57 58 55 54 52 56 47

The regression equation isy = + x.

The best predicted weight of a supermodel who is 171 cmtall is .


/ur stt 6 5.pg

Is the number of games won by a major league baseball team ina season related to the team batting average? The table belowshows the number of games won and the batting average of 8teams.

Team Games Won Batting Average1 100 0.2772 94 0.2753 111 0.2624 104 0.2735 64 0.2846 81 0.2717 79 0.268 112 0.286

Using games won as the independent variablex, do the follow-ing:(a) The correlation coefficient is

r =(b) The equation of the least squares line is

y = + x


/ur stt 6 6.pg

The amounts of 6 restaurant bills and the correspondingamounts of the tips are given in the below.

Bill 97.34 106.27 49.72 43.58 52.44 32.98Tip 16.00 16.00 5.28 5.50 7.00 4.50

Use a 0.05 confidence level to find the following:1

The test statisticr =Is there a significant correlation?

• A. Yes• B. No

The regression equation is ˆy = + x.If the amount of the bill is $80, the best prediction for the

amount of the tip is


/ur stt 6 6a.pg

The amounts of 6 restaurant bills and the correspondingamounts of the tips are given in the below.

Bill 88.01 52.44 106.27 32.98 70.29 97.34Tip 10.00 7.00 16.00 4.50 10.00 16.00

Use a 0.05 confidence level to find the following:The test statisticr =The test statistict =The critical valuet =Is there a significant correlation?

• A. No• B. Yes

The regression equation is ˆy = + x.If the amount of the bill is $55, the best prediction for the

amount of the tip is ,and a prediction interval estimate of the amount amount of thetip is < tip <


/ur stt 6 7.pg

Construct both a 95% and a 80% confidence interval forβ1.β1 = 34, s= 7.5, SSxx = 50, n = 1395% : ≤ β1 ≤80% : ≤ β1 ≤


/ur stt 6 8.pg

Find the multiple regression equation for the data given below.

x1 −2 −2 0 1 3x2 −4 1 0 −1 2y −7 −10 2 10 18

The equation is ˆy = + x1+ x2.


/ur stt 6 9.pg

Consider the data set below.

x 9 3 2 3 2 6y 5 6 9 9 6 4

For a hyopthesis test, whereH0 : β1 = 0 andH1 : β1 6= 0, andusingα = 0.05, give the following:(a) The test statistic

t =(b) The degree of freedom

d f =

(c) The rejection region|t|>


• A. We can reject the null hypothesis thatβ1 = 0 andaccept thatβ1 6= 0.

• B. There is not sufficient evidence to reject the null hy-pothesis thatβ1 = 0.


/ur stt 6 10.pg

In some cases, the best-fitting multiple regression equation isof the form y = b0 + b1x+ b2x2 + b3x3. The graph of such anequation is called a cubic. Using the data set given below, andlettingx1 = x, x2 = x2, andx3 = x3, find the multiple regressionequation for the cubic that best fits the given data.

x −10 −6 −4 −1 3 7 9y 1 9.9 8.2 3.2 −2.9 2 13.5

The equation is ˆy = + x+ x2+ x3.


/ur stt 6 11.pg

A study was conducted to detemine whether a the final grade ofa student in an introductory psychology course is linearly relatedto his or her performance on the verbal ability test administeredbefore college entrance. The verbal scores and final grades for10 students are shown in the table below.

Student Verbal Scorex Final Gradey1 39 912 27 763 53 804 78 825 60 666 55 697 30 638 75 799 56 6810 69 87

Find the following:(a) The correlation coefficient

r =(b) The least squares line

y = + x

2


/ur stt 6 12.pg

For each paired data set, construct a scatterplot and identify themathematical model that best fits the given data.

x 1 2 3 4 5 6 7y 2 4 6 8 10 12 14• A. Exponential• B. Power• C. Logistic

• D. Quadratic• E. Linear

x 1 2 3 4 5 6 7y 1 0 1 4 9 16 25

• A. Quadratic• B. Exponential• C. Linear• D. Logarithmic• E. Power


3

Arnie Pizer Rochester Problem Library Fall 2005WeBWorK assignment Statistics7MultinomialContingency due 07/07/2008 at 02:00am EDT.

1. (1 pt) rochesterLibrary/setStatistics7MultinomialContingency-

/ur stt 7 1.pg

A multinomial experiment withk = 3 cells andn = 440 pro-duced the data shown below.

Cell 1 Cell 2 Cell 3ni 104 118 218

If the null hypothesis isH0 : p1 = .25, p2 = .25, p3 = .5 andusingα = 0.05, then do the following:(a) Find the expected value of Cell 1.

E(Cell 1)=(b) Find the expected value of Cell 2.

E(Cell 2)=(c) Find the expected value of Cell 3.

E(Cell 3)=(d) Find the test statistic.

χ2 =(e) Find the rejection region.

χ2 >The final conclustion is

• A. There is not sufficient evidence to reject the null hy-pothesis thatp1 = .25, p2 = .25, p3 = .5.

• B. We can reject the null hypothesis thatp1 = .25, p2 =.25, p3 = .5 and accept that at least one of the multi-nomial probabilities does not equal its hypothesizedvalue.


/ur stt 7 2.pg

A computer random number generator was used to generate 850random digits (0,1,...,9). The observed frequences of the digitsare given in the table below.

0 1 2 3 4 5 6 7 8 982 70 84 71 76 74 86 75 73 159

Test the claim that all the outcomes are equally likely usingthe significance levelα = 0.05.

The expected frequency of each outcome isE =The test statistic isχ2 =The critical value isχ2 =Is there sufficient evidence to warrant the rejection of the

claim that all the outcomes are equally likely?

• A. No• B. Yes


/ur stt 7 3.pg

It has been suggusted that the highest priority of retirees istravel. Thus, a study was conducted to investigate the differ-ences in the length of stay of a trip for pre and postretirees. Asample of 702 travelers were asked how long they stayed on atypical trip. The observed results of the study are found below.

Number of Nights Pre-retirement Post-retirement Total4−7 250 174 4248−13 80 60 14014−21 30 51 81

22 or more 25 32 57Total 385 317 702

With this information, construct a table of estimated expectedvalues.

Number of Nights Pre-retirement Post-retirement4−78−1314−21

22 or more

Now, with that information, determine whether the length ofstay is independent of retirement usingα = 0.05

χ2 =rejection region isχ2 >The final conclustion is

• A. There is not sufficient evidence to reject the null hy-pothesis that the length of stay is independent of retire-ment.

• B. We can reject the null hypothesis that the length ofstay is independent of retirement and accept the alter-native hypothesis that the two are dependent.


/ur stt 7 4.pg

Among drivers who have had a car crash in the last year, 100were randomly selected and categorized by age, with the resultslisted in the table below.

Age Under 25 25-44 45-64 Over 64Drivers 39 24 15 22

If all ages have the same crash rate, we would expect (be-cause of the age distribution of licensed drivers) the given cat-egories to have 16%, 44%, 27%, 13% of the subjects, respec-tively. At the 0.05 significance level, test the claim that the dis-tribution of crashes conforms to the distribution of ages.

The test statistic isχ2 =The critical value isχ2 =The conclusion is

• A. There is sufficient evidence to warrant the rejectionof the claim that the distribution of crashes conforms tothe distibuion of ages.

• B. There is not sufficient evidence to warrant the rejec-tion of the claim that the distribution of crashes con-forms to the distibuion of ages.

1


/ur stt 7 5.pg

Test the null hypothesis of independence of the two classifica-tions, A and B, of the 3×3 contingency table shown below. Testusingα = 0.05

B1 B2 B3 TotalA1 68 40 60 168A2 52 47 56 155A3 55 66 66 187

Total 175 153 182 510

χ2 =rejection region isχ2 >The final conclustion is

• A. We can reject the null hypothesis that A and B areindependent and accept that A and B are dependent.

• B. There is not sufficient evidence to reject the null hy-pothesis that A and B are independent.


/ur stt 7 6.pg

The number of men and women among professors in Math,Physics, Chemistry, Linguistics, and English departments of acertain college were counted, and the results are shown in thetable below.

Dept. Math Physics Chemistry Linguistics EnglishMen 36 62 35 17 25

Women 3 2 3 2 16

Test the claim that the gender of a professor is independentof the department. Use the significance levelα = 0.01

The test statistic isχ2 =The critical value isχ2 =Is there sufficient evidence to warrant the rejection of the

claim that the gender of a professor is independent of the de-partment?

• A. No• B. Yes


2

Arnie Pizer Rochester Problem Library Fall 2005WeBWorK assignment Statistics8ANOVA due 07/08/2008 at 02:00am EDT.

1. (1 pt) rochesterLibrary/setStatistics8ANOVA/ur stt 8 1.pg

Complete the ANOVA table for a completely randomized designbelow.

Source df SS MS FTreatments 15 15.5

ErrorTotal 48 42.7


The table below lists the body temperatures of six randomly se-lected subjects from each of three different age groups. Use theα = 0.01 significance level to test the claim that the three age-group populations have different mean body temperatures.

16-20 21-29 30 and oldersubject 1 98.6 98.4 98subject 2 98.2 98.7 97.1subject 3 97.3 99 97.9subject 4 97.4 98.3 97.8subject 5 97.4 97.8 98.4subject 6 97.3 98.8 97.3

mean 97.7 98.5 97.75standard deviation 0.559 0.429 0.476

The variance between samples isns2x =

The variance within samples iss2p =

The test statistic isF =The critical value isF =Is there sufficient evidence to warrant the rejection of the

claim that the three age-group populations have the same meanbody temperature?

• A. Yes• B. No


An experiment is conducted to determine whether there is a dif-fernce among the mean increases in growth produced by fiveinculins (A, B, C, D and E) of growth hormones for plants. Theexperimental material consists of 20 cuttings of a shrub (all ofequal weight), with four cuttings randomly assigned to each ofthe five different inoculins. The increase in weight and the stan-dard deviation of the experiment are given in the table below.

A B C D EPlant 1 13 16 22 17 8Plant 2 8 15 22 19 9Plant 3 16 28 25 14 11Plant 4 11 25 25 15 9Mean 12 21 23.5 16.25 9.25

Standard Dev. 2.9155 5.6125 1.5000 1.9203 1.0897

Compute the following:(a) SST=(b) SSE=(c) MST=(d) MSE=(e) F=


Which of the following changes the analysis of variance results?

• A. the same constant is added to every one of the sam-ple values

• B. each value in one of the samples is multiplied by thesame constant

• C. the same constant is added to each value in one ofthe samples

• D. each of the sample values is multiplied by the sameconstant

• E. the order of the samples is changed• F. each of the sample values is converted to a different

scale


245 245 Suppose the Total Sum of Squares for a completelyrandomzied design withp = 4 treatments andn = 12 total mea-surements (SS(Total))is equal to 490. In each of the followingcases, conduct anF -test of the null hypothesis that the meanresponses for the 4 treatments are the same. Useα = 0.025.(a) Sum of Squares for Treatment (SST) is 50% of SS(Total)

F =Rejection regionF >


• A. There is not sufficient evidence to reject the null hy-pothesis that the mean responses for the treatments arethe same.

• B. We can reject the null hypothesis that the mean re-sponses for the treatments are the same and accept thealternative hypothesis that at least two treatment meansdiffer.

(b) Sum of Squares for Treatment (SST) is 70% of SS(Total)F =Rejection regionF >


• A. We can reject the null hypothesis that the mean re-sponses for the treatments are the same and accept thealternative hypothesis that at least two treatment meansdiffer.

• B. There is not sufficient evidence to reject the null hy-pothesis that the mean responses for the treatments arethe same.

1

(c) Sum of Squares for Treatment (SST) is 90% of SS(Total)F =Rejection regionF >


• A. There is not sufficient evidence to reject the null hy-pothesis that the mean responses for the treatments arethe same.

• B. We can reject the null hypothesis that the mean re-sponses for the treatments are the same and accept thealternative hypothesis that at least two treatment meansdiffer.


Use the Minitab display to test the claims. Use theα = 0.05 sig-nificance level. The sample data are SAT scores on the verbaland math portions of SAT-I.

Analysis of Variance for SATSource DF SS MS F PGender 1 52506 52506 4.99 0.032Ver/Math 1 6011 6011 0.57 0.455Interaction 1 31413 31413 2.98 0.093Error 37 376156 10524Total 40 466086

Test the claim that SAT scores are not affected by an interac-tion between gender and test (verbal/math).

TheF−test statistic isThe P-value isDoes there appear to be a significant effect from the interac-

tion between gender and test?

• A. Yes• B. No

Test the claim that gender has an effect on SAT scores.TheF−test statistic isThe P-value isIs there sufficient evidence to support the claim that gender

has an effect on SAT scores?

• A. Yes• B. No

Test the claim that the type of test (math/verbal) has an effecton SAT scores.

TheF−test statistic isThe P-value isIs there sufficient evidence to support the claim that the type

of test has an effect on SAT scores?

• A. No• B. Yes


A study was conducted to see how people reacted to certain fa-cial expressions. A sample group ofn = 36 was randomly di-vided into six groups. Each group was assigned to to view onepicture of a person making a facial expression. Each group saw

a different picture, and the different expressions were (1) Sur-prised (2) Nervous (3) Scared (4) Sad (5) Excited (6) Angry.After viewing the pictures, the subjects were asked to rank thedegree of dominance they inferred from the facial expressionthey saw. (The scale ranged from -10 to 10) The data collectedis summarized in the table below.

Surprised Nervous Scared Sad Excited Angry−0.8 −0.7 −0.7 −0.0999999999999999 1.8 1.20.3 1.2 1.3 −2 −0.8 1.10.5 1.6 0.9 −1.8 −0.9 −1.9−0.5 1.3 −0.9 0 0.7 −0.6

0 −1.2 1.9 2 −1.8 0.9−1.1 −0.2 0.8 0.8 −1.9 1.7

Complete the following ANOVA table

Source df SS MS FExpressions

ErrorTotal


Use the Minitab display to test the claims using the significancelevel of α = 0.05. The sample data are the numbers of supportbeams manufactured by 5 different operators using 4 differentmachines. Assume that there is no interaction effect from oper-ator and machine.

Analysis of Variance for BeamsSource DF SS MS F POperator 4 59.88 19.76 2.37 0.111Machine 3 93.69 46.12 5.52 0.009Error 12 47.73 8.35Total 19 201.3

Test the claim that the four operators have the same meanproduction output.

TheF−test statistic isThe P-value isIs there sufficient evidence to warrant the rejection of the

claim that the four machine operators have the same mean pro-duction output?

• A. No• B. Yes

Test the claim that the choice of machine has no effect on theproduction output.

TheF−test statistic isThe P-value isIs there sufficient evidence to warrant the rejection of the

claim that the choice of machine has no effect on the produc-tion output?

• A. Yes• B. No

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Arnie Pizer Rochester Problem Library Fall 2005 WeBWorK ...webwork.maa.org/viewvc/rochester/trunk/rochester_problib/set... · WeBWorK assignment Statistics1Data due 07/01/2008 at