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HW6_Solutions-1 - Offline HW#6 C.50 summed i The diameter...

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Unformatted text preview: Offline HW #6 C .50 summed i, The diameter of 3.5 inch diskettes is normally distributed. Period‘ u rol in nectors randoml select a FEE sample of 23 diskettes. If the mean diameter of the diskettes i- too lar-e or too sm.|lthe diskette punch is shu down for adjustment; othenNise, the punching process continues. ' e ast sample showed a mean and standard 8 'W deviation of 3.47 and 0.08 inches, respectively. Using 0 = 0.01, the appropriate decision is : . do not reject the null hypothesis and do not shut down the punch u g 11‘ ; 3‘s- #0 H 1 ,‘ 1 3 5' o 0 ~ 0 reject the null hypothesis and shut down the punch I z 3}”, II: : 0,08 reject the null hypothesis and do not shut down the punch * : “K " , 33"" :5 s _ ' .1“ I f .03 ,r" ' do not reject the null hypothesis and shut down the punch Vs.” I" I ‘3 % m H “RES!“ ““Ft<':fiof «171‘: 3., l”, -1.%fi ‘~‘ .1 4i 3.:QH ' ’11s .0‘ ,3). 1 . The diameter of 3.5 inch diskettes 'is normally distributed Periodically, quality control inspectors randomly select a sample of 18 diskettes. If the mean diameter of the diskettes is too large or too small the diskette punch is shut 996.! down for adjustment; othenNise, the punching process continues. The last sample showed a mean and standard fl ”l3 deviation of 3.55 and 0.08 inches, respectively. Using a = 0.05, the appropriate decision is: NW 74: 3.§S,m=.°3, M»: m ., . ‘ t g x .R0 35:“ 3‘50 reject the null hypothesis and do not shut down the punch W! W 3 3" C53. do not reject the null hypothesis and shut down the punch ' 2 lid do not reject the null hypothesis and do not shut down the punch MIR“: ['1‘ \F t."# 08 1.71; U, reject the null hypothesis and shut down the punch Jug”. . 015,11 C_____———————— .2..____.__.«_*.r\ 5 tum '3 mic 3‘“ 2" 3. A market researcher is testing consumer opinions about a new product. Consumers are asked to rate the product on a scale of from O to 10 as to whether or not they would purchase the product if it were available (where O = 9% absolutely would not purchase, 10 = absolutely would purchase). Past studies have shown that this scale is 4‘. W normally distributed in the -o-ulation. The market researcher will not give the go ahead for further development of the DVOdUCt UNESSW is greater than 6.00. A sample of eight scores are obtained as follows: 7, 8, S, . - ", 8. The observed t statistic is: (F we sum «in: 14,, we 1.81 ' _ “5: K : (“0 W. A”. K>QOQ(HM£ “IHOIIXTQ'TIO I". >40 . , 4.83 . . ~\" ~ ~ FEQx;*X 2 i. ‘)t(-# A; : {.452 H. A company believes that it controls more than 40% of the total market share for one of its products. To prove this belief, a random sample of 16S purchases of this product are contacted. It is found that 70 of the 16S purchased this company's brand of the product. If a researcher wants to conduct a statistical test for this problem, the alternative hypothesis would be: ,: Noun-nut MTG. 0F Hakim SHfiRE h I h CM‘TKORLE“ 31 Conn!" t e popu ation proportion is greatert an 0.40. o H: ”w "m“: ,qo ir vans-smug the population mean is equal to 0.40. ‘ ~ g k ’ u: MINE DEfloflm 1;. the population proportion is not equal to 0.40. the population proportion is less than 0.40. «an»: (outset: MRI. mm 40']. 0V NRRKCT 3“” S. A company believes that it controls more than 40% of the total market share for one of its products. To prove this belief, a random sample of 165 purchases of this product is taken. It is found that 70 of the 165 purchased this company's brand of the product. If a researcher wants to conduct a statistical test for this problem, the observed RSV STR'USTK valueis: 76 A _ . . 0.63 (“~- ‘ “~3- , Q ‘ ”.5 ’ 0.933 f'. 1.03 _, A ,, -LL- .4»! mm: ' 0'35 2 ' - = 01.19 C (M (.c A . A bank inspector monitors the default rate on personal loans at the AFB member banks. One standard that she examines is that no more than 5% of personal loans should be in default. On each Friday, the default rate is calculated for a sample of 500 personal loans. Last Friday's sample contained 30 defaulted loans. The bank inspector's alternate hypothesis is: “005 Q: i’ol‘ULn'rioiI N16. or Manual. Loni-Is N 057mm- p=.05 H”: ’Y . .05 vs. H“. «>135- 1P Wt Rtxéd' H“ p¢.05 Wt HRVE DEHOI‘STRR‘WO THIlT no“ THM 5'7. «F Locus nut in OEFRJLT, WHICH VIOLMIS ”fl-it STRNDMD " Ho none THE“ ‘70. A bank inspector monitors the default rate on personal loans at the AFB member banks. One standard that she examines is that no more than 5% of personal loans should be in default. On each Friday, the default rate is calculated for a sample of 500 personal loans. Last Friday's sample contained 30 defaulted loans. If the bank inspector uses hypothesis testing techniques to test this and if alpha is .10, then the critical value is: 1-645 Rt‘sm Ho l? 2 9-1‘b5l.1,%1, (Tam afi -1.28 -1.645 t. 1.28 3. A bank inspector monitors the default rate on personal loans at the AFB member banks. One standard that she examines is that no more than 5% of personal loans should be in default. On each Friday, the default rate is y. 3:, calculated for a sample of 500 personal loans. Last Friday’s sample contained 30 defaulted loans. If the bank At”? inspector uses hypothesis testing techniques to test this and if alpha is .10, then the observed value is: 0° ,. A 0.046 .. _ b ‘0‘. ‘_ .05, 3.06 —1.03 in o .03 (.95) g 1.03 . 500 it; '0.046 = LOU. q. Management at a large company has recently implemented a more participatory management style. Management believes that at least half of the company's employees feel that they are more a part of the decision making process than before. To test this, the company's business researcher randomly polls 180 company employees. One hundred and seven of those polled say that they feel they are more a part of the decision making process than before. Suppose the researcher is using a 1% level of significance. The critical value is: 1-645 'Q : Minn-non MTG. or wheat: Fé'tLNG Till? w: non: A mgr 0F Tl-II mm“ Mun“ r8003: 1H»: strong 1.96 a 2.33 “0:,“ 3'“ "" ‘1‘”?th A 3 (i ‘ ,wi ~.50 ¢ 3‘ x "1 . _ w . ) .. 2575 2 . (Mu-.3 'S ., 1.3L ,RESECY HglF 2 '2.“ J i550 lo . Management at a large company has recently implemented a more participatory management style. Management believes that at least half of the company's employees feel that they are more a part of the decision making process than before. To test this, the company's business researcher randomly polls 180 company employees. One hundred and seven of those polled say that they feel they are more a part of the decision making process than before. Suppose the researcher is using a 1% level of significance. The observed value is: . 2.53 565 RBWS 1.84 ' 2.57 1.56 H. A graphical curve constructed by plotting beta values against various values of the alternative hypothesis is called: Pareto chart BtACK , VHCES 33o - 33\ power curve p operating characteristic curve ogive curve H.- The following data (in pounds), which were selected randomly from a normally distributed population of values, represent measurements of a machine part that is supposed to weigh, on average, 8.3 pounds. 8.1 8.4 8.3 8.2 8.5 8.6 8.4 8.3 8.4 8.2 8.8 8.2 8.2 8.3 8.1 8.3 8.4 8.5 8.5 8.7 Use these data and a = .01 to test the hypothesis that the parts average 8.3 pounds. Test of mu 8.3 vs not = 8.3 II Variable N Mean StDev SE Mean 99% CI T P Weights 20 8.3700 0.1895 0.0424 (8.2488, 8.4912) 1.65 0.115 . Us; THE Tm s-ra-nx-ric. ‘70 REACH a «Humor! Mm 14. it: K 4 4.00:," ”1.“! on 1: ’t.oo:’13’lcfi“3ffl"~1° “n‘fi Ho . “a 1n: Q Nam: To «Emu A concwtiwz 5mg ?:,n$ 7'°“>""‘ To Rfltc‘r H. . us: “m :1 'To arm! A codcwxiw: SNCE $3 i: M Tu: cIJFniL'm Rm "0- B.One survey conducted by RHI Management Resources determined that the Lexus is the favorite luxury car for 25% of CF05. Suppose a financial management association conducts its own survey of CFOs in an effort to determine whether this figure is correct. They use an alpha of .05. Following is the MINITAB output with the results of the survey. Discuss the findings, including the hypotheses, one- or two-tailed tests, sample statistics, and the conclusion. Explain from the data why you reached the conclusion you did. K0 {4 ‘l‘csmndi lint i is)? Pmimrlitm ”("1‘0“ '3 89¢ ‘f 7"” HSSOC ' B‘UEVCS VCR . “““‘““7——£"“‘“‘~‘ ‘3 “Win an Lawn. Test of p : 025 vs )9 not : 0.35 Exact- $1516 ’SIDEO H“. Sample X .‘I Sample p 95?; CI Pry-$111.19 1 79 384 0 206.729 (0.166390, 0.249653) 0.045 a use “a ,y nwé To REHCH a (cactusml 3 Since the p - value rounded to 3 decimal places is O. 045, then we reject the null hypothesis. ‘ USE THE “ST 379757“- TO REM“ a<ou<w$mk 17:5 3 :1, “ML-,2 m M —_““_= r 1 003 I, on” ~15“ Mk \lL—J “5‘2: 3 :2.) 85356 “a: Q3 .1: o 031 W: (I 76 “RR R Comma“ 3 glVKE gas is “a "1 WI: ‘11, RETECT THE H ‘ Fame» T -T$ST W; A researcher wants to conduct a before/after study on 15 subjects to determine if a treatment results in any difference in scores. The null hypothesis is that the average difference is zero while the altemative hypothesis is that the average difference is not zero. Scores are obtained on the subjects both before and after the treatment. After subtracting the after scores from the before scores, the average difference is computed to be ‘~2.13 with a sample standard deviation of 1.21. The degrees of freedom for this test are 1“- .r'l 2 = » 8 Hozbm vs 3:0160 ‘1 "‘3 g, 15 ‘-' Au: mu . 14 is suaSz<15=b4(:,~—i:|q / 13 is. A researcher wants to conduct a before/after study on 15 subjects to determine if a treatment results in any difference in scores. The null hypothesis is that the average difference is zero while the alternative hypothesis is that the average difference is not zero. Scores are obtained on the subjects both before and after the treatment. After subtracting the after scores from the before scores, the average difference is computed to be —2.13 with a sample standard deviation of 2.51. The observed t value for this test is: " ‘329 HEM A. 21.3w r", -1.76 .— ‘b J\ " \ ~1J3 '0 1:; ‘2.87 t = : ..___.__________ 2 . 3.187 .S\ f -3.04 All 1"; X I I: l H. . A researcher wants to conduct a before/after study on 15 subjects to determine if a treatment results in any difference in scores. The null hypothesis is that the average difference is zero while the alternative hypothesis is that the average difference is not zero. Stores‘a‘re obtained on the subjects both before and after the treatment. After subtracting the after scores from the before scores, the average difference is computed to be —2.13 with a sample standard deviation of 1.21. Using an alpha of .05, the table t value for this test is: (‘1, 2.131 2-145 :3 'r H < - = f 2252 u c ‘ W 1‘ t-oxs‘m °“ "Foam ”is g. 1.761 ...
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