solutions_to_third_midterm[1]

solutions_to_third_midterm[1] - Math 124 Section 13 Spring...

Info icon This preview shows pages 1–8. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 124 Section 13 Spring 2011 Helen Finkelstein Midterm #3 - May 10 1. In the town of Prolific Heights, the families are large. The average family size is p = 5.8 people and the standard deviation 0' = 1.5 people. The town has 1000 families, and the distribution of family sizes is approximately normal.. Nonna will get a simple random sample of n =200 families from the town, and calculate the average family size for her sample families, X. 3. Find the chance that Nonna’s value of X will be less than 6 people. Well—>2“ 9817“?“ "- 9.2- 5' ' ...§3‘ Spbfx'fi‘JfizJoéWle 2" Lye/94 /,?fi i It A: ,q’lolo Mw= ~77°6£@ b. Did you need to know that the distribution of family sizes was normal to find the chance in question a? Answer “yes” or “n o” and Oexplain why or why not. We 50mph was Is big 71“ CQMM Limi‘l’ W Says W‘l’luoLI‘shibufio—n 59X WW In appVlema/iltj Will 49 W SWple-siee is larj. 2. In the town of Castoff Valley the average and standard deviation of family size is the same as in Prolific Heights. And the distribution of sizes is also approximately normal. But Castoff Valley is 2 times bigger than Prolific Heights - 2,000 families. Nonna would like to get a sample in Castoff Valley that would give her an estimate of the mean with the same accuracy as her Prolific Heights sample. How big should her Castoff Valley sample be? WSW $7476. 3. Chatti claims that he spends less than 20 minutes per day, on the average, talking on the phone. In fact, on a simple random sample of n = 15 days, the average time he spent was 16.5 minutes. The null and alternate hypotheses for Chatti’s claim are: Ho: u = 20 minutes and Ha: u < 20 minutes.vYouccanvassumethatthetimesjollowsanormaLdistribrutionwand that the standard deviation of all of the times is c = 8 minutes. a. Find the value of P for this data. {4 Hersh/we. .. ‘ _. llus‘L’OA I mole’J/OW 22‘ TN“ 43'? -5E‘3—7 A ‘ SUV} X W'm—Alolm. A1 .0975 r(//' [90" ‘1‘8 20 211' —, , 0 L} 7 6 it b. State your conclusion. m M is shfioh’ufiflb Sikh/{1900M c. Complete the sentence: Pis the chance that... we W 8% a W 0‘4 32 M 5M4” M‘v 16-5 M‘M’J—w H” jA wow 10 WW 4. Here is a report of a study: “Our snake oil helps people lose weight effectively. In a recent study of a random sample of 150 people using our snake oil, they lost an average 0 f37 pounds (P = .025). 3. Identify the claim in this study. Smala. 01"!) W WM lose W7” b. identify the null hypothesis 5 I: o-LL 0L0“ M+ M P 0. Identify the sample size. [5 o d. True or false: The data is statistically significant. W- 5. Teddy works in a factory that produces toy bears. The length of the bears varies according to a normal curve. Every day, Teddy gets a simple random sample of 9 bears and finds the average of their lengths. He uses the sample average as an estimate of the average length of the bears produced that day. On Monday the sample bears had average length i = 23.5 cm. and standard deviation s = 8.4 cm. a. Give a 95% margin of error of the estimate. 5E O’fiyffsfi: Via—1: 2.30m- 7L*= 2301a ‘ Max?“ atrium = (2.30e)(2. 8’) 4: @ b. Give a 95% confidence interval for the population mean. .. I 117.0 . 7.5.? (,5 S l7+030wm 1%.; +(o'g': 30'0 c. If Teddy took a larger sample, what effect would you expect that to have on the margin of error? 1+ waM be. Similar- id. Say whether the following is true or false: 95% of all of the bears produced on Monday had lengths in the interval in question c. (lulu. e. Suppose Teddy gets a simple random sample and calculates a confidence interval every day for the month of May - 31 intervals. How many of the intervals would you expect to contain the average length in the population? own?“ 0/} ‘Htm 957% 9+ a! a 0%)sz 2.9 iml—waifi 6. Question 5, continued: Every day Teddy uses his sample to test the claim that the population average, u, is less than 25 cm. That is, he tests Ha: p < 25 cm against Ho: u = 25 cm. On most days the data is not significant. But about once a month - 1 day out of 30 or so — the datajs,significant. a. It would be wrong for Teddy to conclude on those days that l1 is less than 25 cm. Explain briefly why it would be wrong. 5% 0+ MM‘B «AW H—om‘l‘Vu-em Sljmhfiw JUAijWI-L SOKM‘SlgMfiM WWMMWWM - b. When the data is significant, Teddy suspects that u might be less than 25 cm. What might he do to see whether his suspicion is correct? Ireplrr/W' #321“ Ct 5W Sample mm” “at?” W W (vain‘ 7. One way to limit junk food consumption is to write down what we eat. Here are the number of candy bars eaten by three people before and after they started writing it down: Ann—¥-_ "-- _-—- a. What assumptions do you need in order to be able to make a 95% confidence interval for the average difference in the population? H:— Died}. Is {Va—ma motum sample 2 m oustvi but-om a‘F an ”WM ”pm ”PM“ is W- b. Make a 95% confidencejntervaLtoLtheAaverageditference. _ WMLOQFWw Mil-$450M?” ’5‘” 3’ i, ,_ _ Cchq ma Mimi/fl .s . 3," 1 , , 5E6}, X w fix; \Fb n [73 i W 9. OFF? {,1} Lil-5°} Mfiijvv MC yum ., _(q.7103)U-733%'l.'-l Math. 124 Section 13 Spring 2011 Helen Finkelstein. Midterm #3 - May 10 1. In the town of Prolific Heights, the families are large. The average family size is y. = 5.2 people and the standard deviation 0' = 1.6 people. The town has 2000 families, and the distribution of family sizes is approximately normal.. Nonna will get a simple random sample of n =100 fainilies from the town, and calculate the average family size for her sample families, X. 3. Find the chance that Nonna’s value of X will be more than 5 people. Wfliafiwizww H : O." = Lb— : 01b [e :1: : S“ 532‘ “-— ”LL; SD 0‘; X “73.4100 PM!) .Ib WA: .1051: ////// 7 Ww=l~.10§é‘ ,89q4 b. Did you need to know that the distribution of family sizes was normal to find the chance in question a? Answer “yes” or “no” and explain why or why not. No. Aces-1W +0 +1412 (LANA/Q Lil/raj Mam, MWWkM 0+ Y tux/U be Rppmximigij MM {Q‘HM ‘SanLa is lat/03%. 2. In the town of Castoff Valley the average and standard deviation of family size is the same as in Prolific Heights. And the distribution of sizes is also approximately normal. ' But Castoff Valley is 4 times bigger than Prolific Heights — 8,000 families. - Nonna would like to get a sample in Castoff Valley that would give her an estimate of the mean with the same accuracy as her Prolific Heights sample. How big should her Castoff Valley sample be? ‘Piu. 3&1!“— size- 3. Chatti claims that he spends less than 30 minutes per day, on the average, talking on the phone. In fact, on a simple random sample of n = 20 days, the average time he spent was 26.9 minutes. The null and alternate hypotheses for Chatti’s claim are: Ho: u = 30 minutes and Ha: u < 3.0 minutes.iYoucan.as_s.ume_that.thewtimesvtollowfiahogrmaLdistrjbutionwand that the standard deviation of all of the times is o = 10 minutes. a. Find the value of P for this data. l“; H's \‘s‘lYue MW tat—i @EOM “= L9.» ‘- 50¢y m~z¢4w~w : Zia-3.1320 cal.“ 1.52." WA} 00838 ”/4031 2,8 0 32 P: .0338 [email protected] b. State your conclusion. mm Ls Mt minnow“- c. Complete the sentence: Pis the chance that... we. W 84—0- x M W M2013 mwifflmao MW. ‘ 4. Here is a report of a study: “Our snake oil is proven to help most people feel. better faster than the other brand. in our study of a random sample of 200 people who took our snake oil, 85% of them felt better within 3 days (P = .01)....” a. ldentifythe claim in this study. Swab M W8 W [WM M bld t'fyth llh th ' lull-92v flow. . eni enu ypo eels Smluz the. OM M MP c. Identify the sample size. 1o 0 d. True or false: The data is statistically significant. Jnme 5. Teddy works in a factory that produces toy bears. The length of the bears varies according to a normal curve. Every day, Teddy gets a simple random sample of 16 bears and finds the average of their lengths. He uses the sample average as an estimate of the average length of the bears produced that day. On Monday‘thevsample hears had average length 2? = 45.5 cm. and standard deviation 3 = 9.6 cm. a. Give a 95% margin of error of the estimate. *— .3. z 33,9“- SEorx/Xj W m—zflm yt*:1.l'bl MW «P am = (2.l31)(Z-‘l) 2: @ b. Give a 95% confidence interval for the population mean. 95. e~€l = 40.9 } ugr. +5.! = 507 c. if Teddy took a larger sample, what effect would you expect that to have on the margin of error? it We! be smlLor d. Say whether the following is true or false: 95% of all of the bears produced on Monday had lengths in the interval in question c. Mex e. Suppose Teddy gets a simple random sample and calculates a confidence interval every day for a year - 365 intervals. How many of the intervals would you expect to contain the average length in the population? (“97% 4%- (15°70 ”TI; 32 y Mia/vats : Q95)(3_c5')_s3$7_,:flnzgwwaw 6. Question 5, continued: Every day Teddy uses his sample to test the claim that the population average, p, is less than 50 cm. That is, he tests Ha: p < 50 cm against Ho: u = 50 cm. On most days the data is not significant. But about once a month ~ 1 day out of 30 or so - the datajssignificant. a. it would be wrong for Teddy to conclude on those days that u is less than 50 cm. Explain briefly why it would be wrong. Embp H'O xtA +V'M-Q—I a W M w 6W 8;7n£/pcd¢hw “‘de @ WLQ. So at 4%,“; sfim‘fiwjj rig/“@1143 W055 WW moot..- b. When the data is significant, Teddy suspects that l4 might be less than 50 cm. What might he do to see whether his suspicion is correct? Gd: WW sample an We. plays, W W W. (Vepidaje’) 7. One way to limit junk food consumption is to write down what we eat. Here are the number of candy bars eaten by three people before and after they started writing it down: ‘ ---- —-n— —lllll- _aerge() I, n I, I - .—--- a. What assumptions do you need in order to be able-to make a 95% confidence interval for the average difference in the population? l. ‘Hv; dailioxcwma simple WSW/nix 1. V‘l’lw wal’vfbui’iém of oUHumex/l We populafiu‘m 1‘s Myrna—Z. b. Make a 95% confidencejntervaLtoratheaverageudifierence. SWAP/hdal‘He/MLM.‘ 9“ 001414 WQW‘L‘J‘M " ' 1- Ll.0!§= ~1fl§ 11%.? ST ‘ ‘93 g N . 3; sEdfiWMfiWW T3“ "’5 1:99? L£303 mag/we one we» : CV«303>CI'19>*‘+.9§ ...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern