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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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STAT 200 OL4 US2 Sections Final Exam Spring 2015 Complete Solution answersA+++++++++++
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Date Created: 11/06/15
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. c c b)  As there is no common region between A and A  so P(A and A ) = 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 t­value (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 1.0 ­ 1.9 4  4/25 = 0.16 2.0 ­ 2.9  25*0.4 = 10 0.4 3.0 ­ 3.9 25­(4+10+5)= 6  6/25 = 0.24 4.0 ­ 4.9 5  5/25 = 0.20 Total 25 1 3. The required percentage is = (0.24+0.20)*100% = 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. 2.0 ­ 2.9. 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 average/mean would also decrease. 6. Number of outcomes in the sample space = 6*6 = 36. 7. P(X2>4 |X1 = even) = (6/36)/(18/36) = 1/3 st nd 8. Here A is representing an event associated with 1  dice and B is the event related to 2   dice. As each dices are independent so as A and B. 9. Interquartile range for Quiz 1 = 95­40 = 55 and Interquartile range for Quiz 2 = 90­35 =  55 so both quiz has same Interquartile range. 10. As the Q3 is lower for Quiz 2 so Quiz 1 has the greater percentage of students with  grades 90 and over. 11. The median for Quiz 2 is lower than 60 so quiz 2 has a greater percentage of students  with grades less than 60. 12. P(Atleast one) = P(Stat or Psy) = P(Stat) + P(Psy) – P(Stat and Psy)  = 200/1000 + 100/1000 – 80/1000 = 0.22 13. P(Psy|Stat) = P(Stat and Psy)/ P(Stat) = (80/1000)/(200/1000) = 0.40 14. As here the positions matter so here we need to use the permutation thus the required  number of ways = 10P3 = 720 15. The calculated table is given below, Profit/Loss  Probability  X*P(X X^2*P(X (X) P(X) ) ) 100 0.1 10 1000 20 0.1 2 40 5 0.4 2 10 ­20 0.4 ­8 160         Total 1 6 1210 a) Mean = Expected winning = sum of X*P(X) = 6 b) Variance = sum of (X^2*P(X)) ­Mean  = 1210­6^2 = 1174   Std dev = sqrt(variance) = sqrt(1174) = 34.26 16.  The answers are given below, a) n = number of serves = 10; p = probability of return = 0.3; q = 1­p = 0.7 b) P(X ≥ 1) = 1­P(X=0) =1­0.7  = 0.9718 c) Expected value = np = 10*0.3 = 3 17. P(10< X < 12) = P((10­10)/2 < Z< (12­10)/2) ) = P(0<Z < 1) = P(Z<1) – P(Z<0)  = 0.8413 ­ 0.5000 = 0.3413 rd rd 18. The 3  quartile for a standard normal distribution is 0.6745 so the 3  quartile for the  given distribution is = 10+0.6745*2 = 11.349 feet 19. Required std dev = 2/sqrt(100) = 0.2 20. Here population standard deviation is known so a Z distribution can be used. Now as we  need a 95% confidence interval so Z multiplier = Z(0.025) = 1.96 hence, Required 95% confidence interval for population mean is, Sample¿¿ √¿ 95% CI =  Samplemean± Z 0.025 = 1500± 1.96∗300 ¿ ( √225 ) ¿ = (1460.80, 1539.20)  21. The required answers are given below. p−p 0 0.53−0.5 = 2 a) Test Statistic (Z) = 0∗(1−p0) 0.5  = 0.9 √ n √ 225 b) As the test is right tailed so P­value = P(Z> test statistic) = P(Z > 0.9)  = 1­ P(Z< 0.9) = 1­0.8159 = 0.1841 c) As the p­value is larger than the significance level of 0.01 so we are failing to reject  the null hypothesis implying no sufficient evidence to justify th0 rejection of H . 22. . a) The null and alternative hypotheses are,  H 0µ d0∧H :µ a0 d b) Here the data is paired data so we need to use the paired sample t­test. From the given data, Sample mean for difference(before­after) = ­1.4; sample sd of difference =  1.1402; sample size = 5. So the test statistic is, −1.4 1.1402 Test statistic =  = ­2.7456 √5 c) As sample size is 5 so df = 5­1 = 4 and the test is left tailed thus the p­value is, t <−2.7456 P­value = P (4 ) = 0.0258 d) The p­value is smaller than significance level of 0.1 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.  23. . S2 122 1= =¿ a) Test Statistic = S2 142 0.7347 b) As the test is left tail and numerator df = 1=31−1=30  and denominator df = n −1=30−1=29 2  so, p­value = P(F(30,29) < 0.7347) = 0.2029 c) As the p­value is larger than the significance level of 0.01 so we are failing to reject  the null hypothesis implying no sufficient evidence to justify the rejectio0 of H . 24. Consider the following table of calculation,   X Y X^2 Y^2 X*Y   0 1 0 1 0   1 2 1 4 2   3 3 9 9 9   5 8 25 64 40             Total 9 14 35 78 51 Averag 2.25 3.5 8.75 19.5 12.75 e a) So, xy− 1 x y 51− ∗9∗14 ∑ n∑ ∑ 4 b= = 2  =1.3220 ∑ x − 1 ∑ x)2 35− 9 n 4 1 1 a= ∑ y−b ∑ x=3.5−1.3220∗2.25  = 0.5255 n n Thus the required regression equation is, Y = 0.5255+1.3220*X b) The predicted value for X = 4 is, Y = 0.5255+1.3220*4 = 5.8135 25. The answers are given below, H : p =0.4; p =p =0.2; p =p =0.1 a) 0 brown ¿ orange ¿ tan ¿H aAtleast one proportiondiffers signific ntly 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, 2 2 2 2 2 (42−40 )+(21−20 )+(12−20 ) + 7−10 ) + 18−10 ) Chi­sq =  40 20 20 10 10  = 10.65 c) Df = number of groups­1 = 5­1=4. P­value = P(Chi­sq(4) > 10.65) = 0.0308 d) As the p­value is smaller than the considered significance level of 0.05 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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