HP340: Statistical Methods, Spring 2015Due Date: 3:30PM, Tuesday Feb 24, 2015Homework 5Total Points: 55Mark the letter of the correct answer for multiple-choice questions. Show your work/explain your reasoning when calculations are required. Late assignments will not be accepted.1. (12 points total) Consider the following population of 7 scores:Subject 1 2 3 4 5 6 7Score 6 3 9 4 6 8 2(This is a toy example created to help you cement the concept of sampling distribution of the mean. Of course the population mean of its 7 scores can be easily computed exactly and without any need for sampling to estimate it, but we will pretend otherwise for the sake of learning. In real situations, sampling is required because it would be too costly/difficult to examine each member of the very large populations we are typically interested in)a. (3 points) List all possible different samples of 3 subjects (N=3) that can be obtained from this population (you should get 35 samples in total) and compute the sample mean for each sample. The distribution of the 35 sample means is the ‘sampling distribution of the mean’ with N=3 based on the ‘parent’ population given by the table above. Sample 1: subjects 1,2,3  6+3+9=18/3=6Sample 2: subjects 2,3,4  3+9+4=16/3= 5.33Sample 3: subjects 3,4,5  9+4+6=19/3=6.33Sample 4: subjects 4,5,6  4+6+8=18/3=6Sample 5: subjects 5,6,7  6+8+2=16/3=5.33Sample 6: subjects 7,1,2  2+6+3=11/3=3.66Sample 7: subjects 4,1,2  4+6+3=13/3=4.33Sample 8: subjects 5,1,2  6+6+3=15/3=5Sample 9: subjects 6,1,2  8+6+3=17/3=5.66Sample 10: subjects 4,1,3  4+6+9=19/3=6.33Sample 11: subjects 5,1,3  6+6+9=21/3=7Sample 12: subjects 6,1,3  8+6+9=23/3=7.66Sample 13: subjects 7,1,3  2+6+9=17/3=5.66Sample 14: subjects 5,1,4  6+6+4=16/3=5.33Sample 15: subjects 6,1,4  8+6+4=18/3=6Sample 16: subjects 7,1,4  2+6+4=12/3=4Sample 17: subjects 6,1,5  8+6+6=20/3=6.66Sample 18: subjects 7,1,5  2+6+6=14/3=4.66Sample 19: subjects 7,1,6  2+6+8=16/3=5.33Sample 20: subjects 5,3,2  6+9+3=18/3=6Sample 21: subjects 6,3,2  8+9+3= 20/3=6.66Sample 22: subjects 7,3,2  2+9+3=14/3=4.66Sample 23: subjects 5,4,2  6+4+3=13/3=4.33Sample 24: subjects 6,4,2  8+4+3=15/3=5Sample 25: subjects 7,4,2  2+4+3=9/3=3Sample 26: subjects 6,5,2  8+6+3=17/3=5.66Sample 27: subjects 7,5,2  2+6+3=11/3=3.66Sample 28: subjects 7,6,2  3+8+2=13/3=4.33Sample 29: subjects 6,4,3  9+4+8=21/3=7Sample 30: subjects 7,4,3  2+4+9=15/3=5Sample 31: subjects 6,5,3  8+6+9=23/3=7.66Sample 32: subjects 7,5,3  2+6+9=17/3=5.66Sample 33: subjects 7,5,4  2+6+4=12/3=4Sample 34: subjects 7,6,4  2+8+4=14/3=4.66Sample 35: subjects 7,6,3  8+9+2=19/3=6.33b. (3 points) Construct a grouped frequency distribution table of thesample distribution of the mean computed in a. i=8-3=5ClassIntervalReal limits(lower-upper)MidpointonintervalTally Frequencyrf %f cf crf c%f3.0-3.4 2.95-3.45 3.2 I 1 .03 3 1 .03 33.5-3.9 3.45-3.95 3.7 II 2 .06 6 3 .09 94.0-4.4 3.95-4.45 4.2 IIIII 5 .14 14 8 .23 234.5-4.9 4.45-4.95 4.7 III 3 .09 9 11 .31 315.0-5.4 4.95-5.45 5.2 IIIIIII 7 .20 20 18 .51 515.5-5.9 5.45-5.95 5.7 IIII 4 .11 11 22 .63 636.0-6.4 5.95-6.45 6.2 IIIIIII 7 .20 20 29 .83 836.5-6.9 6.45-6.95 6.7 II 2 .06 6 31 .89 897.0-7.4 6.95-7.45 7.2 II 2 .06 6 33 .94 947.5-7.9 7.45-7.95 7.7 II 2 .06 6 35 1 100c. (3 points) Compute the population mean of the ‘parent’ population (the scores in the table above) and the mean of the sampling distribution of the mean. Comparethe two and interpret. Mean of parent population table above values/7=5.42Mean of sampling distribution all sample means/35= 5.42Both of the means are the same, which means that the data could bebroken down into samples and the average mean could be calculated or the original subjects mean could be calculated and both ways would give the same mean value.d. (3 points) Graph a histogram of the sampling distribution of the mean. Commenton the shape of the distribution. Does it look normal? Would you expect it tolook normal? Explain. 3.0-3.4 3.5-3.9 4.0-4.4 4.5-4.9 5.0-5.4 5.5-5.9 6.0-6.4 6.5-6.9 7.0-7.4 7.5-7.9012345678Sampling Distribution of MeanIntervalsFrequencyThe graph does not look normal because it has a large gap where the peak should be and it its not symmetrical, resembling a bell-shaped curve. We would expect it to look normal because the samples means should be somewhere around 5.42 give or take 1 or 2 with a few outliers.2. (8 points total) a. (2 points) A statistical test with a rejection region comprised of both tails of thesampling distribution of the test statistic is called a(an) _____ test. (Chooseall that apply.) A. two-tailed B. one-tailedC. non-directional D. directional E. None of the above b. (2 points) When performing a two-sided Z-test you obtain Zobs = +2.10. If the significance α=0.05 levelthenisthse tdecisionat shouldbe: _____ the null hypothesis and _____ the alternative hypothesis____.A. reject ; do not reject B. reject ; accept C. fail to reject ; do not accept D. fail to reject ; accept E. reject ; reject c. (2 points) Suppose a non-directional Z-test was performed and the null hypothesis was rejected at a 5% significance level. Which of the following could have been the value of Zobs in the statistical test? (Choose all that apply) A. -1.98 B. -1.23 C. 0 ---D.1.69E. None of the above d. (2 points) Which of the following could be an appropriate null hypothesis for a statistical test concerning the mean cost of regular gas in LA county? (Choose all that apply.) A. H0 : = $2.70/Gallon B. H0 : ≠$ 2.70/Gallon C. H0 : µ = $2.70/Gallon D. H0 : ≠µ $2.70/Gallon E. H0 : ≥µ $2.70/Gallon 3. (10 points total) Given the following information calculate the appropriate test statistics (Zobs or tobs) for testing the equality of the population mean to the given value µ0, and determine whether the test statistic falls into the appropriate two-tailed rejection region at the 0.05 significance level.4. (25 points total) You are analyzing a dataset of current rental prices from N=97 randomly selected two-bedroom rental houses in a particular neighborhood of LA (neighborhood A). The average rent of the 97 houses is $1,832/month. You want to compare the mean rentalprice of two-bedroom houses in this neighborhood to the mean rental price of 2-bedroom houses in a comparable neighborhood (neighborhood B) for which the current mean rentalprice of two-bedroom houses is