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## STAT 200 OL4 US2 Sections Final Exam Spring 2015 Complete Solution answersA+++++++++++

by: kimwood Notetaker

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# STAT 200 OL4 US2 Sections Final Exam Spring 2015 Complete Solution answersA+++++++++++

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1. . a) Variance represents the average of the squared difference from the mean. Variance 0 means all values are same as mean which indicates identical values. Thus the statement is true. b) As...
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Date Created: 11/05/15
a Variance represents the average of the squared difference from the mean Variance 0 means all values are same as mean which indicates identical values Thus the statement is true b As there is no common region between A and Ac so PA and Aquot 0 and thus the given statement is false c As the normal distribution is symmetric about the mean so the mean and median is the same value and thus the given statement is true d For 99 confidence interval we need a larger Z or t value or any other value as appropriate to multiply with standard error to find out the confidence interval Thus the statement is correct e Smaller significance level gives larger critical value and hence rejecting the null hypothesis becomes more difficult so the given statement is false 2 The obtained table is given below Checkout Time in minutes Frequency Relative Frequency 10 19 4 425016 20 29 2504 10 30 39 25 4105 6 625 024 40 49 5 525 020 Total 25 3 The required percentage is 024020100 44 4 The median is the middle most value in the ordered dataset Here the values are already ordered in terms of class intervals So the class interval containing the middle value would be the median class Here total frequency is 25 so middle most value is 13 And from the frequencies we can see 13th observation falls in class interval 20 29 5 Median is middle most value in the ordered dataset as here the maximum value is changing but number of observation is same so median would remain same However as mean is average of all observations and here the maximum value is decreasing after correction so the averagemean would also decrease 6 Number of outcomes in the sample space 66 36 10 11 12 13 14 15 a b 16 PX2gt4 X1 even 6361836 13 Here A is representing an event associated with 1st dice and B is the event related to 2rld dice As each dices are independent so as A and B Interquartile range for Quiz 1 9540 55 and Interquartile range for Quiz 2 9035 55 so both quiz has same Interquartile range As the Q3 is lower for Quiz 2 so Quiz 1 has the greater percentage of students with grades 90 and over The median for Quiz 2 is lower than 60 so quiz 2 has a greater percentage of students with grades less than 60 PAtleast one PStat or Psy PStat PPsy PStat and Psy 2001000 1001000 801000 022 PPsylStat PStat and Psy PStat 8010002001000 040 As here the positions matter so here we need to use the permutation thus the required number of ways 10P3 720 The calculated table is given below ProfitLoss Probability XPX Xquot2PX X PX 100 01 10 1000 20 01 2 40 5 04 2 10 20 04 8 160 Total 1 6 1210 Mean Expected winning sum of XPX 6 Variance sum of XA2PX Mean2 12106A2 1174 Std dev sqrtvariance sqrt1174 3426 The answers are given below a n number of serves 10 p probability of return 03 q 1p 07 b PX 2 1 1PX0 10710 09718 c Expected value np 1003 3 17 18 19 20 21 P10lt X lt 12 P10102 lt Zlt 12 102 P0ltZ lt 1 PZlt1 PZlt0 08413 05000 03413 The 3rd quartile for a standard normal distribution is 06745 so the 3rd quartile for the given distribution is 10067452 11349 feet Required std dev 2sqrt100 02 Here population standard deviation is known so a Z distribution can be used Now as we need a 95 confidence interval so Z multiplier Z0025 196 hence Required 95 confidence interval for population mean is Sample Z Z xZ Z SD 196gtIlt300 95CI s l iL21500i amp 6 mean 6 m L 146080 153920 The required answers are given below 3 190 053 05 a Test Statistic Z 1901 P0 09 n 225 b As the test is right tailed so Pvalue PZgt test statistic PZ gt 09 1 PZlt 09108159 01841 c As the pvalue is larger than the significance level of 001 so we are failing to reject the null hypothesis implying no sufficient evidence to justify the rejection of H0 22 b C d 23 a b The null and alternative hypotheses are HozdeOAHazpdlt0 Here the data is paired data so we need to use the paired sample ttest From the given data Sample mean for differencebeforeafter 14 sample sd of difference 11402 sample size 5 So the test statistic is 14 11402 15 Test statistic 27456 As sample size is 5 so df 51 4 and the test is left tailed thus the pvalue is Pvalue P ff Z7456 00258 The pvalue is smaller than significance level of 01 so we are rejecting the null hypothesis and concluding that there is sufficient evidence to support the claim that the consumption of 2 ounces of alcohol increases mean reaction time Si Si 122 Test Statistic n1 131 130 As the test is left tail and numerator df and denominator df n2 130 129 so pvalue PF3029 lt 07347 02029 As the pvalue is larger than the significance level of 001 so we are failing to reject the null hypothesis implying no sufficient evidence to justify the rejection of H0 24 Consider the following table of calculation X Y XA2 YA2 XY Total 9 14 35 78 51 Averag 225 35 875 195 1275 a So ny leZy 51 914 n b 2l 2 92 13220 2x 35 1 1 6122 y bZZ x35 13220225 05255 Thus the required regression equation is Y O525513220X b The predicted value for X 4 is Y 05255132204 58135 25 The answers are given below 3 H0 pbrown p zpomnge pl39JZPtan391 3 H a Atleast one proportion di ers signi cantly b Here total frequency is 100 so the expected frequency for each color is the total frequency multiplied With the corresponding expected probability Thus test statistic 42 402 21 202 12 202 7 102 18 102 Chi39SCF 40 20 20 10 10 1065 c Df number of groupsl 5l4 PValue PChisq4 gt 1065 00308 d As the pValue is smaller than the considered significance level of 005 thus we are rejecting the null hypothesis and concluding there is sufficient eVidence to reject the claim that the published color distribution is correct

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