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9. Sampling DistributionsPrerequisites• noneA. IntroductionB. Sampling Distribution of the MeanC. Sampling Distribution of Difference Between MeansD. Sampling Distribution of Pearson’s rE. Sampling Distribution of a ProportionF. ExercisesThe concept of a sampling distribution is perhaps the most basic concept ininferential statistics. It is also a difficult concept because a sampling distribution isa theoretical distribution rather than an empirical distribution.The introductory section defines the concept and gives an example for both adiscrete and a continuous distribution. It also discusses how sampling distributionsare used in inferential statistics.The remaining sections of the chapter concern the sampling distributions ofimportant statistics: the Sampling Distribution of the Mean, the SamplingDistribution of the Difference Between Means, the Sampling Distribution of r, andthe Sampling Distribution of a Proportion.300Introduction to Sampling Distributionsby David M. LanePrerequisites• Chapter 1: Distributions• Chapter 1: Inferential StatisticsLearning Objectives1. Define inferential statistics2. Graph a probability distribution for the mean of a discrete variable3. Describe a sampling distribution in terms of “all possible outcomes”4. Describe a sampling distribution in terms of repeated sampling5. Describe the role of sampling distributions in inferential statistics6. Define the standard error of the meanSuppose you randomly sampled 10 people from the population of women inHouston, Texas, between the ages of 21 and 35 years and computed the meanheight of your sample. You would not expect your sample mean to be equal to themean of all women in Houston. It might be somewhat lower or it might besomewhat higher, but it would not equal the population mean exactly. Similarly, ifyou took a second sample of 10 people from the same population, you would notexpect the mean of this second sample to equal the mean of the first sample.Recall that inferential statistics concern generalizing from a sample to apopulation. A critical part of inferential statistics involves determining how farsample statistics are likely to vary from each other and from the populationparameter. (In this example, the sample statistics are the sample means and thepopulation parameter is the population mean.) As the later portions of this chaptershow, these determinations are based on sampling distributions.Discrete DistributionsWe will illustrate the concept of sampling distributions with a simple example.Figure 1 shows three pool balls, each with a number on it. Suppose two of the ballsare selected randomly (with replacement) and the average of their numbers iscomputed. All possible outcomes are shown below in Table 1.3011 2 3Figure 1. The pool balls.Table 1. All possible outcomes when two balls are sampled with replacement. OutcomeBall 1Ball 2Mean1111.02121.53132.04211.55222.06232.57312.08322.59333.0 Notice that all the means are either 1.0, 1.5, 2.0, 2.5, or 3.0. The frequencies ofthese means are shown in Table 2. The relative frequencies are equal to thefrequencies divided by nine because there are nine possible outcomes.Table 2. Frequencies of means for N = 2. MeanFrequencyRelative Frequency1.010.1111.520.2222.030.3332.520.2223.010.111 302Figure 2 shows a relative frequency distribution of the means based on Table 2.This distribution is also a probability distribution since the Y-axis is the probabilityof obtaining a given mean from a sample of two balls in addition to being therelative frequency.00.050.10.150.20.250.30.351 1.5 2 2.5 3RelaƟve Frequency (Probability)MeanFigure 2. Distribution of means for N = 2.The distribution shown in Figure 2 is called the sampling distribution of the mean.Specifically, it is the sampling distribution of the mean for a sample size of 2 (N =2). For this simple example, the distribution of pool balls and the samplingdistribution are both discrete distributions. The pool balls have only the values 1, 2,and 3, and a sample mean can have one of only five values shown in Table 2.There is an alternative way of conceptualizing a sampling distribution thatwill be useful for more complex distributions. Imagine that two balls are sampled(with replacement) and the mean of the two balls is computed and recorded. Thenthis process is repeated for a second sample, a third sample, and eventuallythousands of samples. After thousands of samples are taken and the meancomputed for each, a relative frequency distribution is drawn. The more samples,the closer the relative frequency distribution will come to the sampling distributionshown in Figure 2. As the number of samples approaches infinity, the relativefrequency distribution will approach the sampling distribution. This means that you303can conceive of a sampling distribution as being a relative frequency distributionbased on a very large number of samples. To be strictly correct, the relativefrequency distribution approaches the sampling distribution as the number ofsamples approaches infinity.It is important to keep in mind that every statistic, not just the mean, has asampling distribution. For example, Table 3 shows all possible outcomes for therange of two numbers (larger number minus the smaller number). Table 4 showsthe frequencies for each of the possible ranges and Figure 3 shows the samplingdistribution of the range.Table 3. All possible outcomes when two balls are sampled with replacement. OutcomeBall 1Ball 2Range111021213132421152206231731283219330 304Table 4. Frequencies of ranges for N = 2. RangeFrequencyRelative Frequency030.333140.444220.222 00.050.10.150.20.250.30.350.40.450.50 1 2Relave&Frequency&(Probability)RangeFigure 3. Distribution of ranges for N = 2.It is also important to keep in mind that there is a sampling distribution for varioussample sizes. For simplicity, we have been using N = 2. The sampling distributionof the range for N = 3 is shown in Figure 4.30500.050.10.150.20.250.30.350.40.450.50 1 2Relative(Frequency((Probability)RangeFigure 4. Distribution of ranges for N = 3.Continuous DistributionsIn the previous section, the population consisted of three pool balls. Now we willconsider sampling distributions when the population distribution is continuous.What if we had a thousand pool balls with numbers ranging from 0.001 to 1.000 inequal steps? (Although this distribution is not really continuous, it is close enoughto be considered continuous for practical purposes.) As before, we are interested inthe distribution of means we would get if we sampled two balls and computed themean of these two balls. In the previous example, we started by computing themean for each of the nine possible outcomes. This would get a bit tedious for thisexample since there are 1,000,000 possible outcomes (1,000 for the first ball x1,000 for the second). Therefore, it is more convenient to use our secondconceptualization of sampling distributions which conceives of samplingdistributions in terms of relative frequency distributions. Specifically, the relativefrequency distribution that would occur if samples of two balls were repeatedlytaken and the mean of each sample computed.When we have a truly continuous distribution, it is not only impractical butactually impossible to enumerate all possible outcomes. Moreover, in continuous306distributions, the probability of obtaining any single value is zero. Therefore, asdiscussed in the section “Distributions” in Chapter 1, these values are calledprobability densities rather than probabilities.Sampling Distributions and Inferential StatisticsAs we stated in the beginning of this chapter, sampling distributions are importantfor inferential statistics. In the examples given so far, a population was specifiedand the sampling distribution of the mean and the range were determined. Inpractice, the process proceeds the other way: you collect sample data, and fromthese data you estimate parameters of the sampling distribution. This knowledge ofthe sampling distribution can be very useful. For example, knowing the degree towhich means from different samples would differ from each other and from thepopulation mean would give you a sense of how close your particular sample meanis likely to be to the population mean. Fortunately, this information is directlyavailable from a sampling distribution. The most common measure of how muchsample means differ from each other is the standard deviation of the samplingdistribution of the mean. This standard deviation is called the standard error of themean. If all the sample means were very close to the population mean, then thestandard error of the mean would be small. On the other hand, if the sample meansvaried considerably, then the standard error of the mean would be large.To be specific, assume your sample mean were 125 and you estimated thatthe standard error of the mean were 5 (using a method shown in a later section). Ifyou had a normal distribution, then it would be likely that your sample mean wouldbe within 10 units of the population mean since most of a normal distribution iswithin two standard deviations of the mean.Keep in mind that all statistics have sampling distributions, not just themean. In later sections we will be discussing the sampling distribution of thevariance, the sampling distribution of the difference between means, and thesampling distribution of Pearson’s correlation, among others.307Sampling Distribution of the Meanby David M. LanePrerequisites• Chapter 3: Variance Sum Law I• Chapter 9: Introduction to Sampling DistributionsLearning Objectives1. State the mean and variance of the sampling distribution of the mean2. Compute the standard error of the mean3. State the central limit theoremThe sampling distribution of the mean was defined in the section introducingsampling distributions. This section reviews some important properties of thesampling distribution of the mean.MeanThe mean of the sampling distribution of the mean is the mean of the populationfrom which the scores were sampled. Therefore, if a population has a mean μ, thenthe mean of the sampling distribution of the mean is also μ. The symbol μM is usedto refer to the mean of the sampling distribution of the mean. Therefore, theformula for the mean of the sampling distribution of the mean can be written as:µM = µVarianceng Distribution of the Mean The variance of the sampling distribution of the mean is computed as follows:ଶ௠ߪ=ଶߪܰ௠ߪ=ܰ ߪξng Distribution of difference between meansߤெభିெమ = ߤଵ െ ߤଶThat is, the variance of the sampling distribution of the mean is the populationvariance divided by N, the sample size (the number of scores used to compute amean). Thus, the larger the sample size, the smaller the variance of the samplingdistribution of the mean.(optional paragraph) This expression can be derived very easily from the variancesum law. Let’s begin by computing the variance of the sampling distribution of the308sum of three numbers sampled from a population with variance σ2. The variance ofthe sum would be σ2 + σ2 + σ2. For N numbers, the variance would be Nσ2. Sincethe mean is 1/N times the sum, the variance of the sampling distribution of themean would be 1/N2 times the variance of the sum, which equals σ2/N.The standard error of the mean is the standard deviation of the samplingdistribution of the mean. It is therefore the square root of the variance of thesampling distribution of the mean and can be written as:istribution of the Meanଶ௠ߪ=ଶߪܰ௠ߪ=ܰ ߪξistribution of difference between meansߤெభିெమ = ߤଵ െ ߤଶߪெଶభିெమ = ߪெଶభ െ ߪெଶమߪெଶ = ߪଶܰߪெభିெమ = ඨ݊ߪଵଵଶ + ݊ߪଶଶଶThe standard error is represented by a σ because it is a standard deviation. Thesubscript (M) indicates that the standard error in question is the standard error ofthe mean.Central Limit TheoremThe central limit theorem states that:Given a population with a finite mean µ and a finite nonzero variance σ2, the sampling distribution of the meanapproaches a normal distribution with a mean of µ and avariance of σ2/N as N, the sample size, increases.The expressions for the mean and variance of the sampling distribution of the meanare not new or remarkable. What is remarkable is that regardless of the shape of theparent population, the sampling distribution of the mean approaches a normaldistribution as N increases. If you have used the “Central Limit TheoremDemo,” (external link; requires Java) you have already seen this for yourself. As areminder, Figure 1 shows the results of the simulation for N = 2 and N = 10. Theparent population was a uniform distribution. You can see that the distribution forN = 2 is far from a normal distribution. Nonetheless, it does show that the scoresare denser in the middle than in the tails. For N = 10 the distribution is quite closeto a normal distribution. Notice that the means of the two distributions are thesame, but that the spread of the distribution for N = 10 is smaller.309ling Distribution of the MeanFigure 2 shows how closely the sampling distribution of the mean approximnormal distribution even when the parent population is very non-normal. If youclosely you can see that the sampling distributions do have a slight positive skelarger the sample size, the closer the sampling distribution of the mean would bnormal distribution.Figure 2. A simulation of a samplingdistribution. The parent population is verynon-normal.Figure 1. A simulation of a sampling distribution. The parent population isuniform. The blue line under “16” indicates that 16 is the mean. Thered line extends from the mean plus and minus one standard deviation.Figure 2 shows how closely the sampling distribution of the mean approximates anormal distribution even when the parent population is very non-normal. If youlook closely you can see that the sampling distributions do have a slight positiveskew. The larger the sample size, the closer the sampling distribution of the meanwould be to a normal distribution.310onlinestatbook.com/2/sampling_distributions/samp_dist_mean.htmlFigure 2. A simulation of a samplingdistribution. The parent population is verynon-normal.Check Answer Previous Question Next QuestionFigure 2. A simulation of a sampling distribution. The parent population isvery non-normal.311Sampling Distribution of Difference Between Meansby David M. LanePrerequisites• Chapter 3: Variance Sum Law I• Chapter 9: Sampling Distributions• Chapter 9: Sampling Distribution of the MeanLearning Objectives1. State the mean and variance of the sampling distribution of the differencebetween means2. Compute the standard error of the difference between means3. Compute the probability of a difference between means being above a specifiedvalueStatistical analyses are very often concerned with the difference between means. Atypical example is an experiment designed to compare the mean of a control groupwith the mean of an experimental group. Inferential statistics used in the analysisof this type of experiment depend on the sampling distribution of the differencebetween means.The sampling distribution of the difference between means can be thought ofas the distribution that would result if we repeated the following three steps overand over again: (1) sample n1 scores from Population 1 and n2 scores fromPopulation 2, (2) compute the means of the two samples (M1 and M2), and (3)compute the difference between means, M1 – M2. The distribution of thedifferences between means is the sampling distribution of the difference betweenmeans.As you might expect, the mean of the sampling distribution of the differencebetween means is:ng Distribution of the Mean = = ng Distribution of difference between means = = which says that the mean of the distribution of differences between sample meansis equal to the difference between population means. For example, say that themean test score of all 12-year-olds in a population is 34 and the mean of 10-yearolds is 25. If numerous samples were taken from each age group and the mean312difference computed each time, the mean of these numerous differences betweensample means would be 34 – 25 = 9.From the variance sum law, we know that:which says that the variance of the sampling distribution of the difference betweenmeans is equal to the variance of the sampling distribution of the mean forPopulation 1 plus the variance of the sampling distribution of the mean forPopulation 2. Recall the formula for the variance of the sampling distribution ofthe mean:bution of difference between means = = = = + Since we have two populations and two samples sizes, we need to distinguishbetween the two variances and sample sizes. We do this by using the subscripts 1and 2. Using this convention, we can write the formula for the variance of thesampling distribution of the difference between means as:Since the standard error of a sampling distribution is the standard deviation of thesampling distribution, the standard error of the difference between means is:Just to review the notation, the symbol on the left contains a sigma (σ), whichmeans it is a standard deviation. The subscripts M1 – M2 indicate that it is thestandard deviation of the sampling distribution of M1 – M2.Now let’s look at an application of this formula. Assume there are twospecies of green beings on Mars. The mean height of Species 1 is 32 while themean height of Species 2 is 22. The variances of the two species are 60 and 70,313respectively, and the heights of both species are normally distributed. Yourandomly sample 10 members of Species 1 and 14 members of Species 2. What isthe probability that the mean of the 10 members of Species 1 will exceed the meanof the 14 members of Species 2 by 5 or more? Without doing any calculations, youprobably know that the probability is pretty high since the difference in populationmeans is 10. But what exactly is the probability?First, let’s determine the sampling distribution of the difference betweenmeans. Using the formulas above, the mean is = 32 22 = 10 = 60 10 + 70 14 = 3.317 = + = + = 2 = 2 = (2)8(64) = 4pling Distribution of Pearson’s rThe standard error is: = 32 22 = 10 = 60 10 + 70 14 = 3.317 = + = + = 2 = 2 = (2)8(64) = 4ampling Distribution of Pearson’s rThe sampling distribution is shown in Figure 1. Notice that it is normallydistributed with a mean of 10 and a standard deviation of 3.317. The area above 5is shaded blue.ng Distribution of Difference Between MeansThe sampling distribution is shown in Figure 1. Notice that it is normally dia mean of 10 and a standard deviation of 3.317. The area above 5 is shadFigure 1. The sampling distributionof the difference between means.The last step is to determine the area that is shaded blue. Using either a Znormal calculator, the area can be determined to be 0.934. Thus the probthe mean of the sample from Species 2 will exceed the mean of the samplSpecies 1 by 5 or more is 0.934.As shown below, the formula for the standard error of the difference bmeans is much simpler if the sample sizes and the population variances arethe variances and samples sizes are the same, there is no need to use theand 2 to differentiate these terms.Figure 1. The sampling distribution of the difference between means.The last step is to determine the area that is shaded blue. Using either a Z table orthe normal calculator, the area can be determined to be 0.934. Thus the probabilitythat the mean of the sample from Species 1 will exceed the mean of the samplefrom Species 2 by 5 or more is 0.934.As shown below, the formula for the standard error of the differencebetween means is much simpler if the sample sizes and the population variances314are equal. When the variances and samples sizes are the same, there is no need touse the subscripts 1 and 2 to differentiate these terms. = 60 10 + 70 14 = 3.317 = + = + = 2 = 2 = (2)8(64) = 4Sampling Distribution of Pearson’s rThis simplified version of the formula can be used for the following problem: Themean height of 15-year-old boys (in cm) is 175 and the variance is 64. For girls,the mean is 165 and the variance is 64. If eight boys and eight girls were sampled,what is the probability that the mean height of the sample of girls would be higherthan the mean height of the sample of boys? In other words, what is the probabilitythat the mean height of girls minus the mean height of boys is greater than 0?As before, the problem can be solved in terms of the sampling distribution ofthe difference between means (girls – boys). The mean of the distribution is 165 –175 = -10. The standard deviation of the distribution is: = 32 22 = 10 = 60 10 + 70 14 = 3.317 = + = + = 2 = 2 = (2)8(64) = 4ing Distribution of Pearson’s rA graph of the distribution is shown in Figure 2. It is clear that it is unlikely thatthe mean height for girls would be higher than the mean height for boys since inthe population boys are quite a bit taller. Nonetheless it is not inconceivable thatthe girls’ mean could be higher than the boys’ mean.ng Distribution of Difference Between Meansmean height for girls would be higher than the mean height for boys since inpopulation boys are quite a bit taller. Nonetheless it is not inconceivable thamean could be higher than the boys’ mean.Figure 2. Sampling distribution ofthe difference between meanheights.A difference between means of 0 or higher is a difference of 10/4 = 2.5deviations above the mean of -10. The probability of a score 2.5 or more stdeviations above the mean is 0.0062.Figure 2. Sampling distribution of the difference between mean heights.315A difference between means of 0 or higher is a difference of 10/4 = 2.5 standarddeviations above the mean of -10. The probability of a score 2.5 or more standarddeviations above the mean is 0.0062.316Sampling Distribution of Pearson’s rby David M. LanePrerequisites• Chapter 4: Values of the Pearson Correlation• Chapter 9: Introduction to Sampling DistributionsLearning Objectives1. State how the shape of the sampling distribution of r deviates from normality2. Transform r to z’3. Compute the standard error of z’4. Calculate the probability of obtaining an r above a specified valueAssume that the correlation between quantitative and verbal SAT scores in a givenpopulation is 0.60. In other words, ρ = 0.60. If 12 students were sampled randomly,the sample correlation, r, would not be exactly equal to 0.60. Naturally differentsamples of 12 students would yield different values of r. The distribution of valuesof r after repeated samples of 12 students is the sampling distribution of r.The shape of the sampling distribution of r for the above example is shownin Figure 1. You can see that the sampling distribution is not symmetric: it isnegatively skewed. The reason for the skew is that r cannot take on values greaterthan 1.0 and therefore the distribution cannot extend as far in the positive directionas it can in the negative direction. The greater the value of ρ, the more pronouncedthe skew.317-0.2 0.1 0.4 0.7 1.0Figure 1. The sampling distribution of r for N = 12 and p = 0.60.Figure 2 shows the sampling distribution for ρ = 0.90. This distribution has a veryshort positive tail and a long negative tail.g Distribution of Pearson’s rFigure 2 shows the sampling distribution for ρ = 0.90. This distribution has a vpositive tail and a long negative tail.Figure 2. The sampling distributionof r for N = 12 and ρ = 0.90.Referring back to the SAT example, suppose you wanted to know the probthat in a sample of 19 students, the sample value of r would be 0.75 or highermight think that all you would need to know to compute this probability is theand standard error of the sampling distribution of r. However, since the samplindistribution is not normal, you would still not be able to solve the problem. Fortthe statistician Fisher developed a way to transform r to a variable that is normFigure 2. The sampling distribution of r for N = 12 and p = 0.90.Referring back to the SAT example, suppose you wanted to know the probabilitythat in a sample of 12 students, the sample value of r would be 0.75 or higher. Youmight think that all you would need to know to compute this probability is the318mean and standard error of the sampling distribution of r. However, since thesampling distribution is not normal, you would still not be able to solve theproblem. Fortunately, the statistician Fisher developed a way to transform r to avariable that is normally distributed with a known standard error. The variable iscalled z’ and the formula for the transformation is given below.z’ = 0.5 ln[(1+r)/(1-r)]The details of the formula are not important here since normally you will use eithera table or calculator (external link) to do the transformation. What is important isthat z’ is normally distributed and has a standard error of1 3Distribution of a Proportion(1 )=(1 )= (1 )= 0.60(110 .60) = 0.155where N is the number of pairs of scores.Let’s return to the question of determining the probability of getting a samplecorrelation of 0.75 or above in a sample of 12 from a population with a correlationof 0.60. The first step is to convert both 0.60 and 0.75 to their z’ values, which are0.693 and 0.973, respectively. The standard error of z’ for N = 12 is 0.333.Therefore, the question is reduced to the following: given a normal distributionwith a mean of 0.693 and a standard deviation of 0.333, what is the probability ofobtaining a value of 0.973 or higher? The answer can be found directly from thenormal calculator (external link) to be 0.20. Alternatively, you could use theformula:z = (X – µ)/σ = (0.973 – 0.693)/0.333 = 0.841and use a table to find that the area above 0.841 is 0.20.319Sampling Distribution of pby David M. LanePrerequisites• Chapter 5: Binomial Distribution• Chapter 7: Normal Approximation to the Binomial• Chapter 9: Introduction to Sampling DistributionsLearning Objectives1. Compute the mean and standard deviation of the sampling distribution of p2. State the relationship between the sampling distribution of p and the normaldistributionAssume that in an election race between Candidate A and Candidate B, 0.60of the voters prefer Candidate A. If a random sample of 10 voters were polled, it isunlikely that exactly 60% of them (6) would prefer Candidate A. By chance theproportion in the sample preferring Candidate A could easily be a little lower than0.60 or a little higher than 0.60. The sampling distribution of p is the distributionthat would result if you repeatedly sampled 10 voters and determined theproportion (p) that favored Candidate A.The sampling distribution of p is a special case of the sampling distributionof the mean. Table 1 shows a hypothetical random sample of 10 voters. Those whoprefer Candidate A are given scores of 1 and those who prefer Candidate B aregiven scores of 0. Note that seven of the voters prefer candidate A so the sampleproportion (p) isp = 7/10 = 0.70As you can see, p is the mean of the 10 preference scores.320Table 1. Sample of voters. VoterPreference112031415160718091101 The distribution of p is closely related to the binomial distribution. The binomialdistribution is the distribution of the total number of successes (favoring CandidateA, for example), whereas the distribution of p is the distribution of the meannumber of successes. The mean, of course, is the total divided by the sample size,N. Therefore, the sampling distribution of p and the binomial distribution differ inthat p is the mean of the scores (0.70) and the binomial distribution is dealing withthe total number of successes (7).The binomial distribution has a mean ofµ = NπDividing by N to adjust for the fact that the sampling distribution of p is dealingwith means instead of totals, we find that the mean of the sampling distribution ofp is:µp = πThe standard deviation of the binomial distribution is:1 3g Distribution of a Proportion(1 )(1 ) (1 )321Dividing by N because p is a mean not a total, we find the standard error of p:(1 )=(1 )= (1 )= 0.60(110 .60) = 0.155Returning to the voter example, π = 0.60 (Don’t confuse π = 0.60, the population proportion, with p = 0.70, the sample proportion) and N = 10. Therefore, the meanof the sampling distribution of p is 0.60. The standard error isroportion (1 )=(1 )= (1 )= 0.60(110 .60) = 0.155The sampling distribution of p is a discrete rather than a continuous distribution.For example, with an N of 10, it is possible to have a p of 0.50 or a p of 0.60, butnot a p of 0.55.The sampling distribution of p is approximately normally distributed if N isfairly large and π is not close to 0 or 1. A rule of thumb is that the approximation isgood if both Nπ and N(1 – π) are greater than 10. The sampling distribution for thevoter example is shown in Figure 1. Note that even though N(1 – π) is only 4, theapproximation is quite good.The sampling distribution of p is approximately normally distributed if N is farge and π is not close to 0 or 1. A rule of thumb is that the approximation isgood if both N π and N(1 – π) are both greater than 10. The sampling distributfor the voter example is shown in Figure 1. Note that even though N(1 – π) is o4, the approximation is quite good.Figure 1. The sampling distribution of p.Vertical bars are the probabilities; the smoothcurve is the normal approximation.Check Answer Previous Question Next QuestionQuestion 1 out of 4.The binomial distribution is the distribution of the total number ofFigure 1. The sampling distribution of p. Vertical bars are the probabilities;the smooth curve is the normal approximation.322Statistical Literacyby David M. LanePrerequisites• Chapter 9: Introduction• Chapter 9: Sampling Distribution of the MeanThe monthly jobs report always gets a lot of attention. Presidential candidates referto the report when it favors their position. Referring to the August 2012 report inwhich only 96,000 jobs were created, Republican presidential challenger MittRomney stated “the weak jobs report is devastating news for American workersand American families … a harsh indictment of the president’s handling of theeconomy.” When the September 2012 report was released showing 114,000 jobswere created (and the previous report was revised upwards), some supporters ofRomney claimed the data were tampered with for political reasons. The mostfamous statement, “Unbelievable jobs numbers…these Chicago guys will doanything..can’t debate so change numbers,” was made by former Chairman andCEO of General Electric.What do you think?The standard error of the monthly estimate is 100,000. Given that, what do youthink of the difference between the two job reports?The difference between the two reports is very small given thatthe standard error is 100,000. It is not sensible to take any singlejobs report too seriously.323ExercisesPrerequisitesAll material presented in the Sampling Distributions chapter1. A population has a mean of 50 and a standard deviation of 6. (a) What are themean and standard deviation of the sampling distribution of the mean for N =16? (b) What are the mean and standard deviation of the sampling distribution ofthe mean for N = 20?2. Given a test that is normally distributed with a mean of 100 and a standarddeviation of 12, find:a. the probability that a single score drawn at random will be greater than 110b. the probability that a sample of 25 scores will have a mean greater than 105c. the probability that a sample of 64 scores will have a mean greater than 105d. the probability that the mean of a sample of 16 scores will be either less than95 or greater than 1053. What term refers to the standard deviation of a sampling distribution?4. (a) If the standard error of the mean is 10 for N = 12, what is the standard errorof the mean for N = 22? (b) If the standard error of the mean is 50 for N = 25,what is it for N = 64?5. A questionnaire is developed to assess women’s and men’s attitudes towardusing animals in research. One question asks whether animal research is wrongand is answered on a 7-point scale. Assume that in the population, the mean forwomen is 5, the mean for men is 4, and the standard deviation for both groups is1.5. Assume the scores are normally distributed. If 12 women and 12 men areselected randomly, what is the probability that the mean of the women will bemore than 2 points higher than the mean of the men?6. If the correlation between reading achievement and math achievement in thepopulation of fifth graders were 0.60, what would be the probability that in asample of 28 students, the sample correlation coefficient would be greater than0.65?3247. If numerous samples of N = 15 are taken from a uniform distribution and arelative frequency distribution of the means is drawn, what would be the shape ofthe frequency distribution?8. A normal distribution has a mean of 20 and a standard deviation of 10. Twoscores are sampled randomly from the distribution and the second score issubtracted from the first. What is the probability that the difference score will begreater than 5? Hint: Read the Variance Sum Law section of Chapter 3.9. What is the shape of the sampling distribution of r? In what way does the shapedepend on the size of the population correlation?10. If you sample one number from a standard normal distribution, what is theprobability it will be 0.5?11. A variable is normally distributed with a mean of 120 and a standard deviationof 5. Four scores are randomly sampled. What is the probability that the meanof the four scores is above 127?12. The correlation between self-esteem and extraversion is .30. A sample of 84 istaken. a. What is the probability that the correlation will be less than 0.10? b.What is the probability that the correlation will be greater than 0.25?13. The mean GPA for students in School A is 3.0; the mean GPA for students inSchool B is 2.8. The standard deviation in both schools is 0.25. The GPAs ofboth schools are normally distributed. If 9 students are randomly sampled fromeach school, what is the probability that:a. the sample mean for School A will exceed that of School B by 0.5 or more?b. the sample mean for School B will be greater than the sample mean forSchool A?14. In a city, 70% of the people prefer Candidate A. Suppose 30 people from thiscity were sampled.a. What is the mean of the sampling distribution of p?b. What is the standard error of p?325c. What is the probability that 80% or more of this sample will preferCandidate A?15. When solving problems where you need the sampling distribution of r, what isthe reason for converting from r to z’?16. In the population, the mean SAT score is 1000. Would you be more likely (orequally likely) to get a sample mean of 1200 if you randomly sampled 10students or if you randomly sampled 30 students? Explain.17. True/false: The standard error of the mean is smaller when N = 20 than when N= 10.18. True/false: The sampling distribution of r = .8 becomes normal as N increases.19. True/false: You choose 20 students from the population and calculate the meanof their test scores. You repeat this process 100 times and plot the distributionof the means. In this case, the sample size is 100.20. True/false: In your school, 40% of students watch TV at night. You randomlyask 5 students every day if they watch TV at night. Every day, you would findthat 2 of the 5 do watch TV at night.21. True/false: The median has a sampling distribution.22. True/false: Refer to the figure below. The population distribution is shown inblack, and its corresponding sampling distribution of the mean for N = 10 islabeled “A.”326 12*,-*$ %$ ‘*$ %$$ ‘13*,- $ $+”‘14*,- “” (+”$ $*” # $ 10*”+” %”*# “” ), – “$ ‘,- $ ‘,- % %”$” #! ,.12-%” ‘ 14*, -$ %$ # ‘15*, – $ $ %’,*2*-Questions from Case StudiesAngry Moods (AM) case study23. (AM)a. How many men were sampled?b. How many women were sampled?24. (AM) What is the mean difference between men and women on the Anger-Outscores?25. (AM) Suppose in the population, the Anger-Out score for men is two pointshigher than it is for women. The population variances for men and women areboth 20. Assume the Anger- Out scores for both genders are normallydistributed. Given this information about the population parameters:(a) What is the mean of the sampling distribution of the difference betweenmeans?(b) What is the standard error of the difference between means?(c) What is the probability that you would have gotten this mean difference(see #24) or less in your sample?Animal Research (AR) case study32726. (AR) How many people were sampled to give their opinions on animalresearch?27. (AR) What is the correlation in this sample between the belief that animalresearch is wrong and belief that animal research is necessary?28. (AR) Suppose the correlation between the belief that animal research is wrongand the belief that animal research is necessary is -.68 in the population.(a) Convert -.68 to z’.(b) Find the standard error of this sampling distribution.(c) Assuming the data used in this study was randomly sampled, what is theprobability that you would get this correlation or stronger (closer to -1)?328