Statistics Note Sheet Essay Sample

Practice Question # 1: A gambler claims he can foretell the axial rotation of a dice more frequently than opportunity would foretell. To prove this. a dice is rolled 100 times and the gambler guesses the figure that comes up 20 times and conjectures falsely the other 80 times. Is this strong grounds that the gambler’s claim is true? a. Stipulate the nothing and alternate hypothesis for this job. P = chance gambler conjecture right for an single axial rotation nothing ( Ho ) : P = 1/6 option ( Ha ) : P & gt ; 1/6

B. Find the trial statistic and cipher the p-value. What do you reason? p? = 0. 2 so z = ( 0. 2-0. 1667 ) / . 1667 ( . 8333 ) /100 = 0. 894. From the tabular array the p-value is between 18 % and 19 % . The void hypothesis is a sensible account of this informations so we do non hold strong grounds that the gambler can foretell the result of the dice. c. If the gambler had guessed right on 40 of the axial rotations. would the p-value go up or down? Since this would be further off from what is wexpected under the nothing. the p-value would acquire smaller.

Practice Question # 2: Does the individual paying the measure order more expensive repasts. or less expensive repasts. at a eating house? The measures from 100 parties of two are examined and it turns out that the individual paying the measure ordered a repast bing an norm of $ 0. 50 more than their comrade with a standard divergence of $ 2. 00 a. Clearly define the parametric quantity of involvement in this state of affairs and so province the nothing and alternate hypotheses as statements about this parametric quantity. ? = the mean sum the bill-payer orders more than their comrade for the whole population…null ( Ho ) : ? = 0 alternate ( Ha ) : ? ? 0

B. Find the trial statistic and cipher the p-value. What do you reason? ten = 0. 5 and z = ( 0. 5-0 ) / ( 2. 00/100 ) =2. 5. From the tabular arraies the p-value ? 2 ( 100-99. 38 ) = 1. 24 % . It is improbable that mean differences of this size would happen for 100 braces of clients. We have grounds that the individual paying the measure and their comrade make non order the same priced repasts on norm. c. In talk we learned that statistical significance does non connote practical significance. Explain how this is illustrated by this illustration. Even though the bill-payer spent significantly more. the existent difference of 50 cents is really little and non likely to be of any practical importance. Practice Question # 3: Which of these statements are true and which are false. Explain. A ) If the p-value is 100 % so the void hypothesis must be true. False – for a reversible trial this merely says the information came out precisely as expected under the void hypothesis. But this might besides go on under the surrogate ( though the opportunities would be lower ) . B ) If the p-value is 1 % so the alternate hypothesis has a 99 % opportunity of being true.

False – the p-value is calculated presuming the void hypothesis is true so it doesn’t tell you about the opportunities under the option. C ) If the p-value is 50 % this says that 50 % of the clip the trial statistic would be this far from what is expected when the void hypothesis is true. True D ) For the same information. a one-tailed significance trial will give a lower p-value than a two-tailed significance trial. True – since the p-value for the reversible trial includes the opportunity of acquiring both high and low values under the nothing. Tocopherol ) It is of import to look at the consequences before make up one’s minding on whether to utilize a one-tailed or two tailed trial. False – the nothing and alternate hypothesis are created before the informations are collected. F ) If you conduct 100 hypothesis trials you are likely to happen some important consequences even when the informations are all merely due to opportunity. True – for illustration about 5 of the trials will give p-values less than 5 % .

Sampling Distributions and Confidence Interval Practice Problems. ( note: you will necessitate to utilize Table 21. 1 and the tabular array on page 435 of the text ) Practice Question # 1: A research worker wants to cognize the per centum of Columbus occupants who would prefer a two cent addition in the gasolene revenue enhancement to fund route fixs. A random sample of 900 occupants is chosen and 278 favour the addition. a. Stipulate the parametric quantity and statistic for this job.

Parameter = per centum of Columbus occupants who would prefer a two cent addition in the gasolene revenue enhancement to fund route fixs = p Statistic = 278/900 ( 30. 9 % ) =p? B. Find an 80 % assurance interval for the parametric quantity

Estimated st. dev. Of p? = 100 % . 309 ( . 691 ) /900 ? 1. 54 %
Assurance interval is 30. 9 % ±1. 28 ( 1. 54 % ) or 30. 9 % ± 1. 97 % Practice Question # 2: Suppose 50 % of all OSU pupils drive to school every twenty-four hours. We randomly choice 36 OSU pupils. a. What is the standard divergence of the trying distribution of the sample proportion who drive to school in this job? The standard divergence is. 5 ( . 5 ) /36= 0. 083 B. Describe the trying distribution of the sample proportion from a sample like this. Specifically explain what form it would hold and where it will be centered. The trying distribution will look like the normal curve centered at 0. 5 with a standard divergence of 0. 083. c. What is the chance that in a random sample of 36 OSU pupils. more than 52 % thrust to school?

Standard mark = omega = ( . 52- . 50 ) /0. 083 = 0. 24
Using Table B the reply is about 100 % – 59 % = 41 % ( all right if pupils don’t bother to extrapolate reply from tabular array ) Practice Question # 3: A random sample of 400 OSU pupils are asked how much money they spent on amusement in the past hebdomad. These 400 spent an norm of $ 28 with a standard divergence of $ 18. a. Find a 90 % assurance interval for the parametric quantity of involvement in this job. Estimated st. dev. Of ten is $ 18/400 = $ 0. 9 so the assurance interval is $ 28 ± $ 1. 64 ( 0. 9 ) or $ 28 ± $ 1. 48 B. How would the interval alteration if you made a 70 % assurance statement alternatively? With 70 % assurance we would hold a shorter interval ( smaller border or mistake ) . In fact the new interval would be $ 28 ± $ 1. 04 ( 0. 9 ) or $ 28 ± $ 0. 94 Practice Question # 4: The books in the OSU library contain an norm of 425 pages with a standard divergence of 100 pages. a. Is it more likely that a sample of 20 books will hold an norm of more than 500 pages or is this more likely in a sample of 200 books? Explain. It is more likely for a sample of 20 books.

The Law of Averages says that being off from what you expected for an norm is more likely with a smaller sample ( here you expect 425 and are being asked about over 500 ) B. What is the chance that the mean figure of pages in the following 200 books checked out of the library will be between 410 and 440 pages? Show your computations and depict any premises you are doing in transporting out these computations. You must presume that the books being checked out have an independent figure of pages so this is like a random sample of 200 books doing the trying distribution of this mean about follow the normal curve. Here. the standard divergence of the trying distribution is 100 / 200 ?7. 07. The standard units are ( 440-425 ) /7. 07?2. 1 and ( 410-425 ) /7. 07 ? -2. 1 From the tabular arraies the reply is 98 % -2 % = 96 %

1 ) A trial statistic is used to mensurate the difference between the ascertained sample informations and what is expected when the void hypothesis is true. True 2 ) If a value has a standard mark of –1. 8 so it must be below the mean. True 3 ) A boxplot will rapidly demo if a distribution is bimodal. False 4 ) Most of the pupils in last quarter’s statistics 135 category. that responded to a study. did non smoke. although one pupil smoked three battalions a twenty-four hours. True or False: For this group of pupils the average figure of coffin nails smoked per twenty-four hours was larger than the average. True 5 ) Two newspapers report the consequences of the same Gallup Poll sing the per centum of people who favor term bounds for Senators. The first newspaper uses the canvass to show a 95 % assurance interval for this per centum. while the 2nd newspaper presents an 80 % assurance interval. True or False: The assurance interval presented by the first newspaper will be wider than the interval reported by the 2nd paper. True 6 ) If everyone who works at a eating house is given a $ 500 vacation fillip so the correlativity between their single incomes and hours worked would non alter. True 7 )

If the P-value is 99. 999 % so the void hypothesis does non supply a plausible account for the information. False8 ) If a list of Numberss has a mean of 0. so the SD of the list will besides be zero. False 9 ) You are more likely to acquire caputs on between 40 % and60 % of the flips when you toss a coin 800 times than when you toss the coin 8 times. True 10 ) One 1000 pupils are indiscriminately selected from the list of those presently registered at Ohio State and they each study how many stat mis they rode in a COTA coach during the past hebdomad. Since many pupils did non sit a coach at all. a histogram of the values does non look like the normal curve.

True or False: Even though the histogram did non follow the normal curve. it is still possible to utilize the normal curve to do a assurance interval for the mean figure of stat mis that all OSU pupils rode on COTA coachs last hebdomad. True B 11 ) One forenoon a pet shop weighed each coney it has for sale. The SD of these weights is 2 lbs. A neckband that weighs 0. 1 lbs is so put on each of these coneies. The SD of their weights at this point ( including the neckbands ) is so: A ) 2. 1 lbs B ) 2 lbs C ) 1. 9 lbs D ) changed by a factor matching to the standard units of 0. 1 old ages.

C 12 ) Below is a histogram of the size ( length of longest dimension in millimeter ) of optic tumours seen at the OSU ophthalmology clinic over the past three old ages. One patient had a tumour that was 1mm in size. Its standard units in this information set

A ) would follow the normal curve. B ) would be bigger than the average. C ) would be a negative figure. D ) would be a positive figure. 13 ) Boxs of birthday tapers in a cargo weigh an norm of 4 ounces with a standard divergence of 0. 2 ounces. A histogram ofthese weights followed the Normal distribution rather closely. Approximately what per centum weighed between 3. 78 ounces and 4. 22 ounces? A ) 4. 7 % B ) 9. 3 % C ) 54. 7 % D ) 72. 9 % Answer – D

omega = ( 3. 78-4 ) /0. 2 = -1. 1 ; omega = ( 4. 22-4 ) /0. 2 = 1. 1
86. 43 % – 13. 57 % = 72. 86 %
14 ) A spinster can set down in eight possible places so that all eight have the same chance of coming up. Each place must hold chance A ) between 0 and 1. but can’t say more. B ) between -1 and 1. but can’t say more. C ) 1/2. D ) 1/8. Answer: D they must all add to 1 so if they are the same so they each must hold chance 1/8 15 ) A sample is about to be taken from the 900 pupils taking Statisticss 135 this one-fourth. Which of the following would most closely follow the normal distribution? A ) The histogram of the weights of 2 indiscriminately selected pupils B ) The histogram of the weights of 20 indiscriminately selected pupils. C ) The trying distribution of the norm of the weights of 2 indiscriminately selected pupils D ) The trying distribution of the norm of the weights of 20 indiscriminately selected pupils Answer – D. The normal estimate works better for larger samples 16 ) Match the drumhead statistics with the histograms

I ) mean = 6. 6. average = 6. 8. standard divergence = 1. 3
two ) mean = 6. 6. average = 6. 0. standard divergence = 8. 65
three ) mean = 6. 6. average = 3. 75. standard divergence = 7. 4
17 ) Royals and Connor studied the relationship between the pH of the H2O in 53 lakes in Florida and the degrees of quicksilver ( in parts per million ) found in the lakes’ fish. ( note: pH measures the sourness of the H2O ) . One lake had a pH degree of 5. 3. You would anticipate the fish in this lake to hold a quicksilver content of __________ . About __________ per centum of the variableness in the quicksilver content of the fish is explained by the pH of the lake. Answers:

estimated quicksilver content = 1. 53092 – 0. 152301 ( 5. 3 ) ? . 72 ppm R2 = 33. 1 % = % of variableness in Y ( mercury content ) explained by X ( pH ) 18 ) The deepness ( in pess below sea degree ) of the deepest trench in the Pacific Ocean is measured 100 times independently utilizing the same procedure and the consequences are shown in the histogram below. If one more measuring is taken utilizing this same procedure. we would anticipate it to come out around __________________ give or take a standard divergence of about __________________ . Fill-in the spaces from the picks 20 ) A random sample of 1000 people who signed a card stating they intended to discontinue smoking on November 20. 1995 ( the twenty-four hours of the “Great American Smoke-out” ) were contacted in June. 1996. It turned out that 210 ( 21 % ) of the sampled persons had non smoked over the old 6 months. a ) Specify the population of involvement. the parametric quantity of involvement. the sample and the sample statistic in this job. • Population: everyone who signed the card • Parameter: % of card signers who didn’t fume for 6 months after that • Sample: the 1000 people contacted in June 1996 • Statistic: 21 % B ) Make a 90 % assurance interval for the per centum of all people who had stopped smoking for at least six months after subscribing the nonsmoking pledge.

p? ± z * p? ( 1? p? ) /n

0. 21±1. 64 0. 21 ( 1? 0. 21 ) /1000
0. 21 ± 1. 64 ( 0. 0129 )
0. 21 ± 0. 0211
21 ) A familial theory says that a cross between two pinkflowering workss will bring forth ruddy blossoming workss 25 % of the clip. To prove the theory. 100 crosses are made and 31 of them produce a ruddy blossoming works. Is this strong grounds that the theory is incorrect? Carry out the appropriate hypothesis trial. Be certain to compose down the nothing and alternate hypotheses. happen the trial statistic and the Pvalue. and province your decisions. Null hypothesis: p=0. 25 ( where P = the expected proportion of red-flowering workss ) Alternative hypothesis: p?0. 25 Test statistic: P-value ?16 % ( Table – be certain to include the part in both dress suits ) The void hypothesis is a sensible account of the informations. Such a big P-value indicates that we don’t have strong grounds that the familial theoretical account is incorrect. 22 ) Customers utilizing a self-service sodium carbonate dispenser take an norm of 12 ounces of sodium carbonate with an SD of 4 ounces. What is the opportunity that the following 100 clients will take an norm of less than 12. 24 ounces? The trying distribution of x-bar with n=100 is usually distributed with a mean of ?=12 and a standard divergence of = 0. 4. Standard mark = ( 12. 24 – 12 ) /0. 4 = 0. 6 From Table B. the reply is 72. 6 % 23 )

A research worker at The Ohio State University believes that a certain constituent of ant venom can be used to decrease the sum of swelling in the brass knuckss of people enduring from arthritis. The ant venom intervention has been made into a capsule signifier that can be swallowed. Explain how you would plan an experiment to look into whether this new intervention when taken orally each twenty-four hours for one hebdomad causes a lower grade of swelling in arthritis sick persons. ( You may say that 200 people enduring from arthritis have already volunteered to be experimental topics ) . Identify the explanatory variable and the response variable in your experiment. Randomly assign people to ant venom capsule or placebo capsule. At the terminal of one hebdomad cheque for an addition or lessening in swelling ( be certainly the individual look intoing doesn’t know which group the topic is in. This would be an illustration of a randomized. doubleblind. comparative experiment. Explanatory variable: whether the topic got ant venom or placebo Response variable: the grade of alteration in swelling. 24 ) Historical Polling illustration. Perot electors were less likely to be watching the argument since their campaigner wasn’t involved.

Therefore the polls that merely questioned people who had watched the arguments would supply colored estimations of the penchants of all electors. The polar thought: Always check whether what brings people into the sample might be related to the response being measured. 25 ) A pigment maker fills tins of pigment utilizing a machine that has been calibrated to make full the tins to incorporate an norm of 1 gallon ( 128 ounces ) each. To prove whether their machine has come out of standardization. and will be given to overfill the tins. the maker takes a random sample of 100 tins and finds that they average 128. 2 ounces with an standard divergence of 4 ounces. Is this strongevidence that the can-filling machine is set excessively high? Carry out the appropriate hypothesis trial. Be certain to compose down the nothing and alternate hypotheses. happen the trial statistic. the P-value. and province your decisions. Null hypothesis: ? = 128 ( where ? = long tally norm put in can ) Alternate hypothesis: ? & gt ; 128 Test statistic: = 0. 5 P-value = 30. 85 % The void hypothesis provides a sensible account of the informations. With such a high P-value. we don’t have strong grounds that the machine is out of standardization. omega = ( 128. 2 ?128 )

4 / 100 0. 5
26 ) Customers at a food market shop wage with a recognition card. with hard currency. or with a cheque. Sixty per centum of the clients pay with hard currency. Eighty percent wage don’t use a recognition card. What is the chance that a randomly picked client wages with a cheque? Explain. Since 80 % don’t use a recognition card. 20 % do 100 % ( all clients ) – 60 % ( opportunity of hard currency usage ) – 20 % ( opportunity of recognition card usage ) = 20 % ( opportunity of cheque usage ) 27 ) Trucks are weighed at a Truck Scale to set up the sum owed in route revenue enhancements. Person complains that the deliberation process has three jobs. Problem I: Sometimes the driver is sitting in the truckwhen it is weighed. Problem II: When the same truck is weighed independently more than one time. the truck graduated table will give different values. Problem III: When the legislative assembly established the route revenue enhancement. they intended to revenue enhancement harmonizing to the value of the goods being shipped. non harmonizing to the weight.

Which of the above indicates ( Explain each briefly )
a ) a job with prejudice? B ) a job with dependability? degree Celsius ) a job with cogency? Answers: a ) I b ) II degree Celsius ) III Lurking variable-a variable that has an of import consequence on the response variable but is non included as an explanatory variable. Confounded variables- two variables are confounded if their effects on the response variable can’t be distinguished from each other. Variables that are confounded with each other might be explanatory variables or they might be skulking variables. “explanatory – explains – response” Facts about Standard Deviation

1. the SD of a population is denoted by “sigma” ? . The SD of a sample is
denoted by s. 2. The SD is NEVER NEGATIVE. It is positive if there is any fluctuation at all in the information. 3. The SD is zero merely if all values in our informations are precisely the same. The bigger the SD. the flatter the curve. Small SD- pointy curve. Standard mark “z” = observed value-mean/SD

Shifting-adding or deducting the same figure to all of the values in a distribution Scaling- multiplying or spliting all of the values in a distribution by the same figure Measurements:
1. A measuring is RELIABLE if repeated measurings on the same single give the same consequences. 2. A measuring is BIASED if it systematically deviates from the true value in the same way. ( a measuring is UNBIASED if it doesn’t pervert in one way more than the other ) 3. A measuring is Valid if it is relevant or appropriate as a representation of the belongings it is meant to meas. Bettering measurings?

* To better RELIABILITY. we can take several measurings and average them. * Averages tend to be less variable than single measurings. * To better BIAS. we typically need to happen a new method of measuring. * Repeating a colored process does non repair prejudice.

* *To better VALIDITY. we need to happen a more valid step. * Repetition doesn’t aid with cogency either.
Simpson’s Paradox- an ascertained association b/ 2 variables can be deceptive or even change by reversal way when there’s another variable that interacts strongly with both variables. *look out for skulking variables-bc of SP they can wholly alter Scatterplots:

1. Describe the signifier ; linear or nonlinear
2. Describe the way ; positive ( up to the R ) . negative ( down to the R ) 3. Describe the strength ; strong ( bunched up around the line ) or weak ( can’t distinguished a line. scattered all over ) 4. Outliers ; divergence from the overall form. can fall on or off line Regression Output: Y=A ( m ) X+B

Beers: Ten
BAC: Yttrium
Coefficient: -0. 0127006 = B
Coefficient: 0. 0179638 = M

BAC= -0. 0127006 + 0. 0179638

Probability:
60 % of the clients at a gas station make full up their takes. which is more likely? A ) Between 58 % and 62 % of the following 100 clients fill up their armored combat vehicles B ) Between 58 % and 62 % of the following 1000 clients fill up their armored combat vehicles Fliping a just coin. which is more likely?

A ) Between 40 & A ; 60 of the following 100 flips are caputs. B ) Between 90 & A ; 110 of the following 200 flips are caputs. Thirty per centum of the autos go throughing through a turnpike toll country wage with a province issued “Fast Pass” . Which is more likely? A ) That between 25 % & A ; 35 % of the following 100 autos pay with fast base on balls. B ) That between 25 % & A ; 35 % of the following 800 autos pay with fast base on balls. Thirty per centum of the autos go throughing through a turnpike toll country wage with a province issued Fast Pass. Which is more likely? A ) That precisely 30 of the following 100 attention wage with fast base on balls B ) That precisely 240 of the following 800 autos pay with fast base on balls. The Law of Averages State that:

1. Averages or proportions are likely to be more stable with more tests. 2. Sums or counts are likely to be more variable when there are more tests. *This does non go on by compensation for a bad tally of fortune since independent tests have no memory.

Which is more likely?
A ) The avg amnt spent by the following 10 twosomes is between 40 & A ; $ 60. B ) The avg amnt spent by the following 40 twosomes is between $ 40 & A ; $ 60. more tests Which is more likely
A ) The avg amnt spent by the following 10 twosomes is over $ 60 less tests B ) The avg amnt spent by the following 40 twosomes is over $ 60

A sampling distribution tells the distribution of a statistic. It tells us what values the statistic can take on if different samples are chosen. and how frequently the statistic would take on each value if many samples were taken.

Determining the hypothesis:
* The void hypothesis is ever the “equals” statement. Holmium: µ=___ or Holmium: p=___ * What would we wish to reason? Ha: µ & gt ; ___ or Ha: µ ? ___ Always the same Numberss for both hypotheses.
Alternate hypothesis is what we want to cognize based on inquiry.

* Researchers studied the behaviour of drivers on a rural interstate main road in Maryland where the velocity bound was 55 stat mis per hr. They found that 5690 out of 12. 931 vehicles were transcending the velocity bound. Is this good grounds that ( in this location ) fewer than half of all drivers are rushing?

We have adequate grounds to reason that fewer than half of all drivers are rushing. * A baccy company claims that the sum of nicotine in its coffin nails is usually distributed with an norm of 2. 2 milligram with a standard divergence of 0. 3 mg. A random sample of 100 coffin nails had a average nicotine content of 3. 1 milligram. Perform an appropriate hypothesis trial.

We have adequate grounds to reason that the mean nicotine content is greater than 2. 2 milligram.

A recent Gallup canvass found that 55 % of Americans worry a great trade or a just sum about planetary heating. Consequences for this Gallup canvass are based on telephone interviews conducted March 8-11. 2012. with a random sample of 1. 024 grownups. aged 18 and older. life in all 50 U. S. provinces and the District of Columbia. One can state with 95 % assurance that the maximal border of trying mistake is ±4 per centum points. •Population: American grownups •Parameter: proportion of all Americans that worry a great trade or a just sum about planetary warming •Sample: 1024 American grownups •Statistic: 55 % •Confidence Degree: 95 % assurance •MoE: +/- 4 %

•What would go on to the border of mistake if we decided to use… –90 %
assurance? lessening
–99 % assurance? addition
•What would go on to the border of mistake if we sampled… –500 American grownups? addition
–2500 American grownups? lessening

* The Bureau of Labor Statistics uses 90 % assurance intervals in showing unemployment consequences from the monthly Current Population Survey. The January 2008 study interviewed an SRS 134. 163 people. Of these. 65. 481 were employed and 3293 were unemployed. Give a 90 % assurance interval for the proportion of those surveyed who were unemployed. ( 0. 0238. 0. 0252 )

* Suppose that the return of the S & A ; P 500 each twelvemonth. measured in per centum gained/lost is usually distributed with a standard divergence of 5 % . Suppose that a random sample of 36 old ages averaged 8 % . Trial to see if the mean return of the S & A ; P 500 is more than 7 % . on norm.

We do non hold adequate grounds to reason that the mean return is more than 7 % . •Give a 95 % assurance interval.
( 6. 4. 9. 6 )

* The eighteenth century Gallic naturalist Count Buffon tossed a coin 4040 times. He got 2048 caputs. Give a 95 % assurance interval for the chance that Buffon’s coin lands heads up. ( . 492. . 522 )

•Are you confident that this chance is non 1/2? No. the assurance interval contains 1/2