problem set 4 answers - HW4 Answers Exercise 9.10(21 H0 p =...

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Unformatted text preview: HW4 Answers Exercise 9.10 (21) H0 : p = 9.5 hours H1 : p 75 9.5 hours, two—tailed test (b) Ho : p = 105 dollars H1 : ,u < 105 dollars, left-tailed test (c) Ho : p. = 39, 000 dollars H1 : p > 39, 000 dollars, right—tailed test ((1) H0 : p = 10 minutes H1 : u aé 10 minutes, two—tailed test (8) Ho : p = 30 hours H 1 : p < 30 hours, left-tailed test Exercise 9.23 If H0 is not rejected, the difference between the hypothesized value of p and the observed value of 5: is ”statistically not significant". Exercise 9.29 (a) Exercise 9.37 2=0 353 (a) Step 1: Ho : p = 2320 square feet, H1 : p > 2320 square feet Step 2: Since 11 > 30, use the normal distribution Step 3: a 312 = .— = -——— = 15.6 uare feet ”2 «E 400 sq Then, 5 — [1 2365 — 2320 z a; 15.6 Piom the normal distribution table, area to the right of z = 2.88 is 1 —— .9980 = .0020, Thus, p—value: .0020. Step 4: Reject Ho since .0020 < .02. 0)) Step 1: Ho : p. = 2320 square feet, H1 : p. > 2320 square feet Step 2: Since n > 30, use the normal distribution. Step 3: For a = .02, the critical value of z is 2.05. Step 4: a 312 (r— : -——- = —— = 15.6 uare feet ”‘ x/fi 400 sq Then, z = 22—11 = 2365 —2320 x 2.88. 0': 15.6 Step 5: Reject Ho since 2.88 > 2.05. For both parts (a) and (b), conclude the mean size of new homes in the United States exceeds 2320 square feet. Exercise 9.44 (a) The first two steps and the calculations of the observed values of the test statistics are the same for both the p—value approach and critical value approach. So, I do not write twice. Step 1: Ho : p Z 30 hours, H1 2;; < 30 hours Step 2: Since the population is normally distributed, use the normal distribution. a 7 0'- = —- = — =1 1.4 hours ’ x/5 V23 and z: 521‘ =27_—a=_2.14 a5: 1.4 (i) p-value approach Step 3: From above, 2 = —2.14. From the normal distribution table, area to the left of z = —2.14 is .0162. Thus, p—value: .0162. Step 4: Rejecting Ho, since .0162 < .025. (ii) Critical value approach Step 3: For a = .025, the critical value of z is -1.96. Step 4: Horn above, 2 = —-2.14. Step 5: Reject Ho since —2.14 < —1.96. Conclude that the time per month on such reading by adults in this city is less than 30 minutes. (1)) For a = .01, do not reject Ho since .0162 > .01. The critical value of z = —-2.33, do not reject H0 since —2.14 > 2.33. The decision in part (a) and (b) are different. The result of this sample are not very conclusive, since lowering the significance level from a = .025 to a = .01 reverses the decision. Exercise 9.95 (a) Step 1: Ho :p. _<_ .05 1112p. > .05. Step 2: Since up = (200) (.05) = 10 > 5 and 'nq = (200) (.95) = 190 > 5, we can use the normal distribution. Step 3: For a = .025, the critical value of z is 1.96. Step 4: _ [—m _ /(.05) (.95) N 0,, _ n _ 200 N .0154. 17 z; — p .085 — .05 = = ————— m 2.2 l (7,; .01541 7 Step 5: Reject Ho since 2.27 > 1.96. Conclude that the percentage of defective DVDs is statistically greater than 5%, and the production process should be stopped to make necessary adjustment. 0)) If a = .01, the critical value of z is 2.33.Since the value of the test statistic is 2.27, we do not reject Ho. Thus, our decision differs from that of part (a); we conclude that the process should not be stopped. The results of this sample are not very conclusive, since lowering the significance level from 2.5% to 1% reverses the decision. Exercise 9“? Fl To make a hypothesis test about the percentage of bank customers who are satisfied with the services provided: (1) Take a large sample of the bank’s customers such that up > 5 and n? > 5. Exercise 9.108 (a) Step 1: H0 : p .—.—_ 55,693 dollars H- : p > 33.693 dollars Step 2: Since the population standard deviation is unknown and the sample size is large (n > 30), we use the 15 distribution. Step 3: For a = .025 with df = n — 1 — 65 ~ 1 = 64, the critical value of t is 1.998. Step 4: s 6300 35 = —-—- = -—- 3 781.4188 dollars W 1/55 _ :1— ,1 “ 58,940—55,693 .0 r t “ si ’" 781.4188 “4'15”” Step 5: Reject Ho since 4.155 > 1.998. Conclude that the current mean salary of all California school teachers exceeds 55,693 dollars. (b) l-Yom the t distribution table, 3.223 < 4.144 < oo : hence. the range of pvalue is 0 < p—value< .001. For a = .025, reject Ho since p—value < .025. Exercise 9.115 (a) Step 1: Ho : p = .156, H1 :1) > .157. Step 2: Since np = (1200) (.156) = 187.2 > 5 and 1112 = (1200) (.844) = 1012.8 > 5, we can use the normal distribution. Step 3: For a = .02, the critical value of z is 2.05. - Step 4: . _ Ja- (11132122.. ”P ‘ n — 1200 ~.0105. 216 = — = .1 p 1200 8 13 - p .18 — .156 = z _ z 2.29. 7' 0,; .0105 Step 5: Reject Ho since 2.29 > 2.05. Conclude that the current percentage of Americans who have no health insurance coverage exceeds 15.6%. (b) The Type I error would occur if the current percentage of Americans who have no health insurance coverage does not exceed 15.6%, but we concluded this percentage does not exceed 15.6%. The probability of error is a = .02. (C) From the normal distribution table, area to the right of z = 2.29 is 1 — .9890 — .0110. pvalue = .0110. For a = .02, reject Ho since .0110 < .02. Exercise 9.127 The following are two possible experiments we might conduct to investigate the effectiveness of middle taillight. Method I: Let p) = proportion of all collisions involving cars build since 1984 that were rear—end collisions. Let p; = proportion of all collisions involving cars build before 1984 that were rear-end collisions. (172 would be known) We would test Ho : p] =p2 versus H1 :1); < p2. We would take a random sample of collisions involving cars build since 1984 and determine the number that were rear—ended collisions. We would have to assume the following: (1) They only change in cars build since 1984 that would reduced rear-end collisions is the middle taillight. (2) None of the cars built before 1984 had middle taillight. (3) People’s driving habits, traffic volume, and other variables that might affect rear-end collisions have not changed appreciably since 1984. Method 11: Let [1.1 = mean number of rear-end collisions per 1000 cars built since 1984. Let #2 = mean number of rear-end collisions per 1000 cars built before 1984. (112 would be known) We would test Ho : #1 = #2, H1411 < #2. We could take several samples of 1000 cars built since 1984. We would determine the number of rear-end collisions in each sample of 1000 cars. We would find the mean and standard deviation of these numbers and would then to form the test statistic. We would require the same assumptions as these listed for the test in Method 1. Also, if we took less than 30 samples, we would have to same that the number of rear—end collisions per 1000 has a normal distribution. 10.13 3. am, = 3+9; =‘l(800)'+(1000)' =$183.9295119 .1 " n1 ,1) 45 51 The 99% confidence interval for #1 — ,uz is (351—372): :afx-Yz = (3300 — 3850) i 7580839295119) = —550 i 474.54 = —$1024.54 to —$75.46 b. Step 1: Ho: p1—,113 = 0. H1: 1‘11 —/13 :5 0 Step 2: Since 01 and 0-3 are known, the samples are independent. and the sample sizes are large (n1 3 30 and n] 2 30). use the normal distribution. Step 3: For a = .01. the critical values of: are —2.58 and 2.58. =(i1~fg)—(/q-#2)= —550—0 0— — 1839295119 3‘ I “X; Step 4: 3 = —2.99 Step 5: Reject Ho since —2.99 < —2.58. C onclude that such mean repair costs are different for these two types of can. c. If a = 0. there can be no rejection region. and we cannot reject Ho. Therefore. the decision would be "do not reject H0." (nI —1)s,2 +012 —1)s§ 111 + 113 — 2 l . 57% =51, —l—+—— = 1011599394 —1—+——1—— =2.957l6900 mmutes '1 ‘ nl 712 J25 22 The 99% confidence interval for 111 — ,uz is (25—1)(1 1)2 +(22—1)(9)2 25+22—2 = 1011599394 minutes (:71 ~f2)i:rsf1_— = (44 — 49) i 2.690(295716900) = —5 i- 7.95 = —12.95 to 2.95 minutes 1‘1 b. Step1: Ho: ,ul—y] =0.Hl: [UP/’2 <0 Step 2: Step 3: Step 4: Step 5: Since 01 and 03 are unknoxm but assumed to be equal. the samples are independent. the sample sizes are small. and the populations are normally distributed. use the r distribution. For a = .01 with df= 45. the critical value of r is —2.412. = (571 ‘le‘bui ”#2) = —5 —0 5} _.- 295716900 1‘! t =—1.691 Do not reject Ho since ——1 .691 > —2.412. C onclude that the mean relief time for Brand A is not less than that for Brand B. 10.32 2 , _ 2 _____________(700—1)(1.9) ’(740 ”(15) =1.7S225466 hours 700+740—2 1 s _ (nl—l)512+(n3—l)533 = i P n1+772—2 1 1 1 1 57—17. =3}, ———+— = 1.75225466 —+— =.09238758 hour " . n1 n 700 740 Since "1 = 700 and r13 = 740 are very large. we use the normal distribution to approximate the r 14 distribution The 95% confidence interval for ,ul — y: is (Y1 —.‘T'3)i 33,734: = (10 — 7.5) i l.96(.09338758) = 2.5 i .18 = 2.32 to 3.68 hours b. Step 1: Ho: #1 :11: = 0. H1: Jul —,ul > 0 Step 2: Since 01 and 0'2 are unknovm but assumed to be equal. the samples are independent. and both samples are large (n, 3 30 and n2 2 30). use the r distn'bution. However. since nl = 700 and "3 = 740 are very large. we use the normal distfibution to approximate the r distribution. Step 3: For a = .025. the critical value of: is 1.96. - . 5 Step 4: :=;——— = ——'— = 27.06 . .09238758 Step 5: Reject Ho since 27.06 > 1.96. Conclude that the mean time spent per week with friends by all men is greater than that for all women . a Sfi—Yq “ —+" = *791921534 minutes 711 nz 25 2.2 3 2 1 i 5: 1i 9_' n1 2:) 25 22 df = . .‘ , = , =44.78 44 2\- 2 -‘ 2y 2\ li l‘l lL [9“) n n 25 22 1 + 2 ' J + "1—1 —1 24 21 The 99% confidence interval for ,u, — y: is (\‘-—1 x,)+rs—:_ - =(44— 49)+ 3.692(2.9l921534)=—5 : 7.86=—12.86 to 2.86 minutes -X_ b. Step 1: Ho: #1 —;1: =0. H1: #1 —,u2 < 0 Step 2: Since (71 and (73 are milmown but assumed to be unequal. the samples are independent. the sample sizes are small. and the populations are normally distn'buted. use the r distn'butiou. Step 3: For a = .01 with df= 44. the critical value of r is —2.4l4. _(x1—-"—2) (/11" l’2)_ "5—0 5m» 2 9.2921534 Step 5: Do not tejectHo since —1.713 > —2.414. C onclude that the mean relief time for Brand A is not less than that for Brand B. 10.46 2 2 2 2 a. SW. = 1+3?— =1l09l +06) =.09282566 hours " 3 n1 "3 700 740 Since 271 = 700 and n; = 740 are very large. we use the nonnal distribution to approximate the t distribution. The 95% confidence interval for :“1 #13 is (El ——.\_'3)i ZS— — = (10 — 7.5) i 1.96(.09282566) = 2.5 i .18 = 2.32 to 2681101113 x1 —x2 Step 1; Ho: pl --,11; = 0.H1:;11—;13> 0 Step 2: Since 01 and a! are tmknown but assumed to be equal. the samples are independent. and both samples are large (:11 3 30 and :12 Z 30). use the t distribution. However. since HI = 700 and n2 = 740 are very large. we use the normal distribution to approximate the r distribution. Step 3: For a = .025. the critical value of .: is 1.96. ‘ —“., — — 2. — Step4; my. 24L =26” SWZ .09282566 Step 5: Reject Ho since 26.93 > 1.96. C onclude that the mean time spent per week with fiiends by all men is greater than that for all W 0111511. ...
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