5.18 the Table Below Shows the Values of P and (Which Depends on P) for a Given Value Of

5.18 The table below shows the values of p and s (which depends on p) for a given value of n. The maximum value ofs occurs when p = .5 .

p .01 .30 .50 .70 .90 s .99 4.58 5.00 4.58 3.00

5.26 Define x to be the number of readers 14 or older. Then p= P[success] = .6 and n = 12. You can use either the binomial formula or Table 1 to calculate the necessary probabilities. a P[ x�5] - 1 � P[ x - 4] = 1 .057 .943 b P[ x=9] = P[ x� 9] � P[ x - 8] = .917 .775 .142 c P[ x<3] = P[ x � 2] .003

5.34 Define x to be the number of Canadians who consume fewer vegetable servings on a typical winter day than on a typical summer day. Then, n = 15 and p = .5 . Using the binomial tables in Appendix I, a P( x= 8) = P ( x� 8) � P ( x - 7) = .696 .500 .196 b P( x � 4) .059 c P( x> 10) = 1 - P ( x � 10) - 1 = .941 .059

5.56 The random variable x has a hypergeometric distribution with N=10, M = 5 and n = 4. Then 5 5 Cx C4- x p( x )=10 for x = 0,1,2,3,4 C4 5 5 C4 C 0 5 1 a P( x =4) =10 = = C4 210 42 1 41 b P( x�3) - 1 = P( x = 4) - 1 = 42 42 C5 C 5+ C 5 C 5 10(10)+ 10(5) 150 5 ＃ 2 2 3 1 c P(2 x 3) = p (2) + p (3) =10 = = = C4 210 210 7 6.22 It is given that x is normally distributed with m = 15 ands = 2.5 . 骣 12- 15 a P( x�12) � P琪 z � = P( z 1.20) .1151 桫 2.5 骣 15- 15 P( x�15) � P琪 z � P( z 0) .5 桫 2.5 b In order to avoid a collision, you must brake within 18 metres or less. Hence, 骣 18- 15 P( x�18) � P琪 z � P( z 1.20) .8849 桫 2.5

6.34 The pulse rates are normally distributed with m = 78 ands = 12 . x -m 60 - 78 a The z-values corresponding to x = 60 and x = 100 are z = = = -1.5 . and s 12 x -m 100 - 78 z = = = 1.83 Then s 12 P(60< x < 100) = P( - 1.5 < z < 1.83) = .9664 - .0668 = .8996 b From Exercise 6.11b, we found that the 95th percentile of the standard normal (z) distribution x-m x - 78 is z = 1.645. Since z = = , solve for x to find the 95th percentile for the pulse s 12 rates: x - 78 1.645=� x + 78 = 1.645(12) 97.74 12 x -m 110 - 78 c The z-score for x = 110 is z = = = 2.67 and s 12 P( x> 110) = P ( z > 2.67) = 1 - .9962 = .0038. The z-score is between 2 and 3; the probability of observing a value this large or larger is quite small. This pulse rate would be somewhat unusual.

52.1 Define x to be the number of British Columbians who feel they will have to cut back on spending. Then the random variable x has a binomial distribution with n = 30 and p = .47 . a Since the Table 1 in Appendix I does not mention the case of n = 30, use Calculating Binomial Probabilities applet to compute P( x= 20; n = 30, p = .47) = 0.0145. b Calculate m=np =30(.47) = 14.1and s = 30(.47)(.53) = 7.473 = 2.7337 Using the normal approximation with correction for continuity, we get, 骣19.5- 14.1 20.5 - 14.1 P( x= 20) = P (19.5 < x < 20.5) = P琪 < z < 桫 2.7337 2.7337 =P(1.98 < z < 2.34) = .9904 - .9761 = .0143 This is approximately equal to the probably calculated in part a. c Using the normal approximation with correction for continuity, we find the area between x = 19.5 and x = 25.5 : 骣19.5- 14.1 25.5 - 14.1 P(19.5< x < 25.5) = P琪 < z < 桫 2.7337 2.7337 =P(1.98 < z < 4.17) = 1.0000 - .9761 = .0239 7.6 The questionnaires that were returned do not constitute a representative sample from the 1000 questionnaires that were randomly sent out. It may be that the voters who chose to return the questionnaire were particularly adamant about the Parks and Recreation surcharge, while the others had no strong feelings one way or the other. The nonresponse of half the voters in the sample will undoubtedly bias the resulting statistics.

7.24 a If the sampled population is normal, the sampling distribution of x will also be normal (regardless of the sample size) with mean m = 106 and standard deviation (or standard error) given as s n =12 25 = 2.4

x -m 110 - 106 b Calculate z = = = 1.67 , so that P( x>110) = P( z > 1.67) = 1 - .9525 = .0475 s n 2.4 c P(102< x < 110) = P( - 1.67 < z < 1.67) = .9525 - .0475 = .9050 7.34 The sampled population has a mean of 5.97 with a standard deviation of 1.95. x -m 6.5 - 5.97 a With n = 31, calculate z = = = 1.51, so that s n 1.95/ 31 P( x�6.5) � P( z 1.51) .9345 x -m 9.80 - 5.97 b Calculate z = = = 10.94 , so that s n 1.95/ 31 P( x�9.80) 郴 P( z 10.94) - 1 = 1 0 c The probability of observing an average diameter of 9.80 or higher is extremely unlikely, if indeed the average diameter in the population of affected tendons was no different from that of unaffected tendons (5.97mm). We would conclude that the average diameter in the population of patients affected with AT is higher than 5.97. 7.44 a The random variable pˆ , the sample proportion of students who used the Internet as a major resource in the past year, has a binomial distribution with n = 1000 and p = .66 . Since np = 660 and nq = 340 are both greater than 5, this binomial distribution can be approximated by a normal .66(.34) distribution with mean p = .66 and SE = = .0150 . 1000 骣 .68- .66 b P( pˆ >.68) = P琪 z > = P( z > 1.33) = 1 - .9082 = .0918 桫 .0150 c P(.64< pˆ < .68) = P( - 1.33 < z < 1.33) = .9082 - .0918 = .8164 pˆ - p .70 - .66 z = = = -2.67 d The value pˆ = .70 lies pq .0150 standard deviations from the mean. This n is a somewhat unlikely occurrence, assuming that p = .66 , and would tend to contradict the reported figure. 8.100 a The approximate 95% confidence interval for p is

pqˆ ˆ .67( .33) pˆ �1.96 � .67 1.96 .67 .020 or .65

c The approximate 98% confidence interval for p1- p 2 is

pˆ1 q ˆ 1p ˆ 2 q ˆ 2 ( pˆ1- p ˆ 2 ) �2.33 n1 n 2 .37( .63) .27( .73) (.37- .27) � 2.33 160 (2170- 160)

.10� .092 or .008-( p1 < p 2 ) .192 9.72 a The hypothesis to be tested is

H0 :m= 2.71 versus H a : m 2.71 . The test statistic is x -m 2.71 - 2.71 z � = 0 s n .54 100 and the p-value is p-value= 2 P( z > 0) = 2(0.5) = 1

H0 cannot be rejected. There is insufficient evidence to indicate that the average GPA score for all Black applicants is different from the overall average. b The hypothesis to be tested is

H0 :m= 142 versus H a : m 142 . The test statistic is x -m 150 - 142 z � = 2.67 s n 30 100 and the p-value is p-value= 2 P( z < - 1) = 2(.0038) = .0076

Since the p-value, .0076, is smaller than a = .05 , H0 is rejected and we can conclude that the average LSAT score for all Black applicants in the class of 2005 is different from the national average. c Since the same students are used to measure GPA and LSAT scores, there would not be two independent samples, and the two sample z-test would not be appropriate. 9.84 a In Example 9.8, the critical values of x that specify the acceptance region were found to be 874.18 and 885.82 and b = P(874.18＃ x 885.82) for various alternative values of m . In this exercise, the sample size has been reduced to n = 30 . Although the procedure remains the same, the

value for xC is changed because the standard deviation of the random variable is s n =21 30 = 3.834 . The critical values of x corresponding to the acceptance region -1.96＃z 1.96 are calculated as x-880 x - 880 1= -1.96 and 2 = 1.96 3.834 3.834

x1=872.485 and x 2 = 887.515 Remember that the corresponding z-value must be calculated in each case, so that normal curve areas may be obtained. When m = 870 , b = P(872.485＃ x 887.515) and the z-

values corresponding to x1 = 872.485 and x2 = 887.515 are 872.485- 870 887.515- 870 z = = .65 z = = 4.57 1 3.834 2 3.834 then b =P(.65＃ z 4.57) = 1 - .7422 = .2578 and the power is 1-b = .7422 b When n = 70 , the standard deviation of the random variable is s n =21 70 = 2.510 . The critical values of x corresponding to the acceptance region -1.96＃z 1.96 are calculated as x-880 x - 880 1= -1.96 and 2 = 1.96 2.51 2.51

x1=875.08 and x 2 = 884.92 When m = 870 , b =P(875.08＃ x 884.92) = P( 2.02 ＃ z 5.94) = 1 - .9783 = .0217 and the power is 1-b = .9783 c The applet shows power values between .70 and .78 for m = 870 (correct to the nearest integer) when n = 30 and a power of .98 when n = 70 . These are consistent with the hand calculations. d When the sample size is increased, so is the power of the test. 10.66 a The hypothesis of interest is 2 2 2 2 H0 :s 1= s 2 versus H a : s 1 s 2 and the test statistic is 2 s1 .273 F =2 = = 2.904 . s2 .094

The upper portion of the rejection region witha =2( .005) = .01 is F> F.005 = 6.54 (from

Table 6) and H0 is not rejected. There is insufficient evidence to indicate that the supplier’s shipments differ in variability. 2 b The 99% confidence interval for s 2 is

2 2 (n2-1) s2 ( n 2 - 1) s 2

~~9( .094) 9( .094)~~