homework 6&7 key1 - T 1 of 3 ID Msncrcmnenmca...

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Unformatted text preview: T 1 of 3 ID: Msncrcmnenmca Internet company Gurgie Is carrvlng out testing on the effldencv of Its search engine. a sample of 40 searches have been can1ed out and the time takien to displaliI the results has been recorded for each search. The mean search time for the sample was calculated as 0.2572 1 _ seconds. The standard deviation oi the search times ior the sample was calculated as 0.0153 seconds. The population standard devlatIon Ll r 8 e of search times ls unknown. at} Select all the techniques that are commonly used to construd a confidence interval for the mean when the population standard deviation (a) is unknown: Approximate the population standard deviation {0} with the sample standard deviation is) Replace the sample size in} with n-l J Decrease the confidence level to compensate for the Increased margin of error J Approximate the standard normal dlsb'lbutiori with the Student's t distribution It} Calculate the uppe‘ and lower bounds cl'the 95% confidence interval for the mean searcl'i time for the Gurgle search alpine. You mallr lind this film W useful. lee your answers in seconds to 4 decimal places. Upper bound = 0.3231' seconds Lower bound - class seconds —' Feedback [3 out at a] a) You are partly correct. - Wale the maiden standard deviation (0’) with the sample standard Man (I): this option should have been selected. - Wale the standard norntai distribution with the Student's t distribution: you are correct. - Decreale the confidence level to compensate for the lnu'euad martin of error: this Is not correct. It) A 95% confidence interval for the mean is constructed by taking the sample mean as the center cfthe interval and defining the width cfu-ie interval using the margin of error. The margin of error is calmlated using the camping distribution ofthe mean. According to the central lelt Theorem (CLTJ, the sampling distribution of the mean follows a normal distribution with mean u and standard deviation aN'n. In most stuatlons the population standard deviation (cl) ls unknown and thus the fiandard deviation of the sampling distribution of the mean is unknown. To solve this issue. the populatlon standard deviation Is aoproxlmated using the sample standard deviation (5]. When this approximation Is made. the normal distribution cannot be used to construct: the confidence interval because (it - ii).l'(ei'v'n) is not norrr'ialilr distributed. This W500 actualllir Follows the Student‘s t distribution whfl'i has a smllar shape in the normal dstrhudon, exmpt It B thldtler a: the tells. That is, It has a larger standard deviation. Therefore, to construct a 95% confidence interval when Ute population standard deviation is unknown, or is approxi'riated by s and the correspondhg t value obnlned from the Student‘s t dlstnbutlon. b) Tito a-iticai trope correspondng to a 95% confidence interval can be obtahed ushg the Wage. The upper bound can be oaioriatod ushg the Following formula: W I i = sample mean search time = 0.25?2 seconds s = sample standard deviation of the search times = 0.0163 seconds n = sample size = 40 ' t‘ = upper critical value corresponding to 95% confidence = 2.022? U = upper bound = unknown ' auuu..u.mau"unnumuumumo — s U =x-i- ‘x— ‘ t/n 0.0163 - 0.2572 + 2.022? K 74? 026241301. . 0.2624 seconds Rounded as last step — 3. of 3 TD: M5T.c_I_CIr-Lr.m.0(120b A team of software englneers are tesllng the tlme taken for a partlmlar type of modem computer to exewte a oomplleated algorltnm lbr factorlng large numbers. They would Ila: to astlmaha the mean tlma taken for a mmputarbn mm the algnrthm. A random sample of 31 tlmes are collected. The mean time In fills samfle ls 855.9 seconds and the sample standard Man Is found to be 5?.5. Calculate the 90% confidence Interval fur the mean time taken to emanate the algarlahm. You may and this W uaelhI. lee your answers to 2 fledl'llal m -sus- Feedback [I] out of 1] True False a) The t procedure is robust, so that as long as the population does not depart too much from normal your results will be reasonable b) The t distribution procedures are insensitive to changes in the sample size c) When using a sample size of less than 15, the t distribution procedures should not be used for data that is highly skewed andror contain extreme outliers d) If the sample size is less than 15. you should use the t distribution only If sure that the population Is close to normal -- :| Feedback [3 out of 4] Legend vi You are correct. Thls option should have been selected. I This is not correct. "lieu-sh" .. .. .. .. .. .. .. .. 11:2 Student's t tistrlbutlon is used Instead of the standard non'nal distribution In Instanoes what the population :aodard deviation is not known (which is usually the case]. There Is not a single t dlstrlaution but a family ofdstrlhutlons, each oorrupondlng to a particular number of deprea of freedom. There are several I'nportant assumptions and chat-acterlstles «the t dlstrlbutlons: - uslng the t distribution assumes that the populatlon from which the sample Is drawn Is normally dlstrllwted o Mwever, the t distribution procedure ls robust - meaning that as long as the population does not differ too much from normality then your results will not be affected boo greatly - uslng a larger sample size helps to Improve aocuracy of results - you should not use a t distribution If your data ls hlghly skewed or contalns very extreme outliers Sorne guidellns for Ltdng the t clstrihutlon Iwith a sample she n: o n is less than 15: use a t distribution only If you are confident that the population is close to normal - 15 5 I1 < 40: t distributions can be used except for highly skewed data or ls extents outllers are present o rt ls greater than 4-0: you can use t dlstrltulions even with skewed data I 1 off. ID: MST.HT.TM.06.[ID2[Ia A psyrhologist has developed an aptlwde test which oonslsts of a series of mathematical and vocabulary problems. They want to test the hypothesis that the mean test score Is Bfl. A random sample of 40 people have spleen the test and thelr results recorded: You may find thls figment; t dlfinmtmn tafle useful throughout thls question. at) (him late the test statistic (t) for the hypothesis test. Give your answer to 4 decimal places. i: = 1.639s I.) At a level ofsimiflcanoe ofo = 0.05. the result ofthistest lsthat the null hypothesis is —. - Feedback [1 out of 2] a) This is not: correct. 1 - 4.82?3 II) You areoorrect. ‘3th” Calculation a) Using a statistical sdftware package, the following resJIts can be obtained: Sample Stat s sample mean 68.325 sample standard deviation 15.29603183 test statistic 4.3213 F-Value < 0.001 Alternatively, the test statistic ran be mluilated using the following furrnula: show variablE X ' "(l i #11 68.325 - 80 : 1529603183 J40 = 432734242. .. 4.5273 Rounded as last step b) This test statistic follows the t distflbutlon with n , 1 = 39 degreesoffreedom. This isa tweetaiied test, because the alternative hypothesis dalms that the population mean is not equal to 80.111e hypotheses for the test are therefore Hg: p = 80 and H3: u s 80. The null hypothesis rejected ifthe Pavalue ofthe test statistic is less than the level of signifiahcie ofthe test. The Fi-value for thB test statistic is equal to 2 X Probability“ > 4.8273) < 0.001. Therefore, since the P—value is less than the level of signifimnoe , the null hypothesis is rejected. This means that, based upon the sample data, there is enough evidence to conclude thatthe mean test score Is not equal to 80. That Is, the hypothesized claim the mean tat score Is 80 is not supported by the sample data. 0f3 ID: MST.HT.TM.05-°020h A team of software engineers are testing the time taken fora particular type cfmodern computertn execute a complicated algorithm for factoring large numbers. may know that the mean time to execute the algorithm was 830 seconds. However, the algorithm has recently been upgraded and it is suspected that the mean time has now been decreased because of the improvements made. They would like to test this hypothesis. A random sample of 40 times are collected: 595 l 584 951 l 977 You may find this Student's t dlstrlhutlon table useful throughout this quatlon. a) Calmlate the test statistic (t) for the hypothesis test. Give your answer to 4 decimal places. I = 1.3062 b) At a level of significance ufu = 0.1. the result cifthis test is that the null hypothesis is —. :| Feedback a) This is not correct. t = 0.4862 b) You are correct. Calculation a) Using a fiatistical soflware package. the following results can be obtained: Sample Statistics sample mean [1 nutol 2] sample standard deviation 19235877489... test statistic 0.4852 P—value Alternatively, the test stat'stic an be alculated using the following formula: shuw variables ; - no i V’n 344.825 ' B30 0.4861665... 0.4861 Rounded as last step II) This test statistic follows that distribution with n - 1 = 39 degrees nffreednm. This is a one-tailed test, because the alternative hypothesis claims thatthe population mean is leg than 530. The hypotheses for the test are therefore Ho: p = 830 and Ha: u < 330. The null hypothesis is rejected ifthe P-value ufthe test statistic is less than the level ofsignificanoe of the test. The P-value for thistest statistic is equal to Probability(t > 0.4362) = 0.6852. Therefore, since the P-value E greater than the level of significance , the null hypothesis is not rejected. This means that, based upon the sample data, there is not enough evidence to conclude that the mean time taken to execute the algodthm Is not equal to 830. That is, the hypothesized claim the mean time taken to execute the algorithm is 830 seconds is not contradicted by the sample data. — 3 of3 ID: MST.CI.CIM.O?.0010b 111e following sample data has been drawn from a population with unknown variance. Sample data 166-5 mum-“simm- mm- ommmmmmmmm 0102030405060108090 commonsense» a) It would be appropriate to use the t procedure forthis data when the histogram for the data most closely resembles graph @. b) Calculate the 95% confidence Interval for the population mean. Give your answers to 1 decimal place. 166.! S |.I 5 170.6 3 Feedback [2 outnl3] a) This Is not correct. It would be appropriate to use the t procedure for this data when the histogram for the mta most closely resembls graph 5. b) You are correct. 9189mm, , , ,, ,, ,, ,, , ,, , ,, ,, ,, , a) The use of t procedures relies on the underlying population being normally dbtrlbuted. However, the t procedurfi are robust which means that even lfthat population Ls not quite normal the resultswlll still be applicable. So before you use some t procedure to make an Inference you fliould check that your data seem reasonably normal. One way you can do this Is to create a histogram. The resulting histogram should be approximater symmetric and bell-shaped, in which case It Is appropriate to use a t procedure. Therefore, It Is appropriate to uset procedure forthls data when the histogram for the date most closely raembles graph B. b) Using a statisti-l software package, the following mullscan be obtained: Sa m pie Statistics sample mean 16632666667... sample standard deviation 8.6661208... 95% confidence interval “6608796511.... 17056536823...) Altematlvely, the 95% confidence Interval can be found In the following way.- M — a 5 8.6661208... = . i: . x— 168 32666667 2 001 1/60 X :i: t “V." = 16832666667... J: 2.001 K 1.11879138... = 16832666667... J: 223870156... 16638796511... 5 u S 170.56536823... 1.66.1. S u 5 170.6 Rounded as Iaststep Dr Carl Is a mathematics and statistics lecturer. He Is Interested In studying a new approach at teaching mathematlis and statistlls In high school. Recently this new approach has been adopted. It was previously the case that the mean score for all high school students doing their end of year exam was 73, however, the population standard deviation is unknown. The scores of all fludenu are approximately normally distributed. He thinks that, with this new teaching approach, the mean score may have Increased. A random sample of 46 scores Is gathered and the sample mean and standard deviation calculated. Dr Carl will conduct a hypothesis test using a level of significance of 0.05. Select the correct null and alternate hypotheses for this test: Ha: p: 73 Ha: p< 73 3 Feedback [1 outol' 1] You are correct. Piacusala n In this hypothesis test, the mearcher Is ainnng to test a single population mean. Since the population standard devratlon Is unknown, this will be a one-sample t test for the mean. Since the researcher suspects that the mean has changed in a specific direction, i.e. that the mean has inn-eased, this will be a one-tailed tat. The alternate hypothesis will reflect the change in the mean, Le. it Will state that the mean is greater than Previously known. The status duo is that the mean has not increased, i.e. it has remained unchanged from its previous value of73. Therefore, the hypotheses forthis test are: "0:“ =73 H.:p>7a :I Z of 3 ID: MST.HT.TM.05.0030 The printout below shcwsthe results of a one-tailed hypothesis test that has been carried out by a first year Statistics student. 111e purpose of the hypothesis test was to determine whether there was evidence to suggest that the population mean (u) ls greaterthan 55. The test was carried out at a 5% level of significance. Select all the lina on the printout that contain errors. You may find WSW Or this W useful. Hypotheses: {Hg:p> 55 {Ha:p=55 Dascrlpflva stifled“: Sample mean: E = 56.700 , Sample standard deviation: 5 = 7.530 I Sample size: n = 30 InferenceI: Inferenoaa: i: Significance level: 0.05 ._ Critical value: 1.96 w/ Test statisb'c: 1.23 . Result: Accept the null "I Conclusion: The sample mean ls greater than 55 [1 outof2] You are partly ourrect. . H“: p > 55: you are oorrect. - H.: u = 55: you are correct. I (mural value: 1.56: this option should have been selected. - Result: Anespt‘flle null: this option should have been selected. - conclusion: The lamaie mean II greater than 55: you are correct. - Tait statistic: 1.23: this is not correct. Hypotheses: A hypothesis tat Involve two hypothess: the null hypothesis and the altematlve hypothsls. The null hypothesis represents. the status quo. That Is, what Is assumed to be true. The altematlve hypothesis represents what Is trying to be shown. It was given that the Statistics student was attempting to show that p > 55. Therefore, the curred; alternative hypothesis Is H.: p > 55. Accordingly, the correct null hypothesi is HD: p = 55. It would also be acceptable to have a null hypothesis of HD: p S 55. unsure-es: A hypothesis“: Iivolves two hypotheses: the null hypothesls and the ahematlve hypothesls. The null hypothesis WIS the anus qua. That Is, what is assumed to be true. The doematlve hypothesis represents whfi Is trying to be shown. it was Wm thfi the warm audent wee mfllng to Show that II > 55. Them the eon-red: alternative “was Is H.: II > 55. mm, the m null hypothus ls H“: p - 55. ltwould also beaooeptabletohayea null hypodeeelsol'l-lo: p s 55. Ill-admin em The sample mean sen be oelouiateil using the fulcrum fermuh: W i - E n I 563'“) The ”mole standard deviation can he ealorieteu nesrq me Mowing Ibrlnule: W _ 2m - £32 n - 1 - 15301-1643... - 7.560 Rounded as iastsbep The sample else can be doubted by oounttig the number ol’oells in the tflle ofdata. The table arm oontelns 3 rows, each oomahlng 10 data points. flierelbne, the simple sine ls emal to n = [3 a: 10) = 30. Therefore, by comparing three calculations “Rh the results provided by the Sunnis Rodent, It can be seen that there have been no errus made In the desulptlve sta‘tlfilu. lute-renew: It was glval that the hypotheds we would be condom at a 5% level dignll'loance. Thetefore. It Is correct to state that the level ddpllllcauw Is 0.05. The hypdilesls best can-led out by the statistlu stumt is a one-tailed test because the almatlve MM is WHO!" m with whether ll is greater film 55. Forthls modeled test. the full hmesls Is moo fthe \fch Ii the test medal: is greater than the a'ldall value correspond“ to the level ofslglillunoe ofthe tea. This is because the level of significant dell-lee the sandal-d by which mittens dedoe that there Is enough evldenoe to armed: that the nul hypothesis ls false. The test mlltlc that Is minced follows a Studs-it‘s t distribution with 3D - 1 - 29 dares offleedom. This is because the valve offl'le population mandala deviation is unlmom and has been amimated using the semis standard deviatlm. The critical value that corresponds some level ofslanlieenee ofthe test can be obtained using the W. Moo-dog heme table. the mlvalue that modern level efaimlheenaearons and 29 mum 151.6991. Theatre. the 9.“thth n stating that the amoelvaluewae 1.96. Aaltleel valueot1.96 would beoorlett [the hypothalstut uasatwu-talled Hat eta level ofslglillmoeofons. The tea mtlatlc an be mlwlaoed use-lg the lbllmvhg lbrmula: W X | I lie fine 56.700 - 55 - 7.580 :30 - 112340151... - 1.23 Romdedasiaatsteo Moortlilp to the tan statistic for the hypethals tat. the null hypodeeels should not be rejected at a 5% levd ofslpnll'loalloe since 1.23 rt 1.6991. The nudent was Incorrect III noting that the nuI hypomaala ShouIfl have been muted. Ill fact. the null hypothesis VIII llmr be am because this lswtsloe the m «hypothesis MM. mug to the arm d m Whefls Why. it Is truly possible he Iejea: the null hypothesis (fa-we Ia eneum evidence be m that the null Is false) or not rejec the I'ILII hypothesis [If there Is not alough evidence team that the null Is Pelee). The option ol’aooqxlng the nul hypothesis II not available because It is not pliable to plum that the null hypothesis is true. Sineethenullhypoflleds was not Iejettedflt lspoeslnletoeondude that fllelelsnot enmph evldenoetoshdw that u lspleeterfllm 55. The maria-Items Incorrect hoondudlng that sunple mean leg-eater than 55. Bid: a oonehlslon ls error-laws since the purpose ofa hymhuls tn furthe mean is In maloe lrlhrenoel about the popullflell men, not the sample mean. The CEO ofthe Jen and Benny's ice cream company is oonoemed about the net weight of ioe cream in their 50 nunoe ice cream tubs. He decides that he wants to be fairly sure that the mean weight of these ice cream tubs (U) is greaterthan 55 ounces. A hypothesis test is conducted with the following ”YDUUWESES: Hail-l = 55 Haw > 55 The level of slgnlfirance used in this test Isa = 0.05. A random sample of 26 Ice cream tubs are collected and weighed. The sample mean weight ls mlmlated to be i = 57.71 and the sample standard deviation L s = s53. You may find this Student's t distribution table useful throughout this question. 3) Calculate the tst statistic (t) for this test. Give your answer to 4 decimal plats. t = 1.6012 b) The result ofthls test is that the null hypothesis rejected. Ti Feedback [2 out of 2] I) You are correct. b) You are correct. Falcu latlon 1.60119843... Feed a) You are correct. b) You are mnect. Calculation a) The tat statistic can be calculated us‘ng the following formula: show variables x - no i \/n 57.7]. - 55 8.53 V25 1.60119848... 1.6012 Rounded as last step .1- H II) This test statistic follows the t distribution with n - 1 = 25 degres nffreedorn. This is a one-tail test, berause the alternative hypothesis claims that the population mean is greatnr than 55. Therefore with a level Msionificanoeofa - 0.05. the region of rejection is the set ofvaluos greater than the 0.05 critical value oftlle t distrimtion with 25 dedreesoffreedom. That is, the region ofrejection is the set ofvalues larger than 17051, Since the test statlstlc (1.6012) ls loss than 1‘7081. the null hypothesis to not rejected. J of 3 ID'. H5T.HT.TM.05.DDZD Rick the reoeptionlg claims that he types an average oi" ?0 words per mlnuoe. In oi'der to tea this. M's boss was him 15 dillerent one minute typing diallenpee diroufl'iout the day. On each challenge. Riolt'e boas records the nu mherol «was wood. The mandates the «male mean to be 5‘1 «was per...
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