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**Unformatted text preview: **Fahd! a) This is not correct.
t = 2.76
h) This is not correct. According to the above data and at a level of signiﬁcance of 0.1. the claim that Macrohard accessories are more expensive is justified given the results of the hypothesis test. 9?'.F!‘.'€'t.i‘?.'!...... .. The two samples are related by the type of accessory in question, and the population standard deviation is not known. Therefore, the appropriate test to use is the paired t-test. The claim states that
Macrohard products are more expensive which will be treated as the alternative hypothesis. If we treat the differences equal to the Macrohard price minus the Fear Computer price, then: H”: ud = 0 and H5: pd > 0. The following table lists the amount by which Macrohard is more expensive than Pear Computer: Computer Accessory Price Data
Computer Accessory Macrohard Price ($) Pear Computer Price ($) Difference ($)
keyboard 71.21 52.24 18.97 speaker
modem 107.95 29.99
433.03 19.38 monitor a) Using a statistical software package, the following results can be obtained: a) Using a statistical software package. the followmg results can be obtained: sample statistics sample mean of 15.34
the differences sample standard deviation 1143957443...
of the differences test statistic 2.76
P—va lue 0.0254 Alternatively, these results can be calculated as follows: Before the sample standard deviation of the differences can be calculated It Is necessary to calculate the sample mean. This can be calculated using the following formula: show variables a_&
n 18.97 + -3.75 + 12.11 + 29.99 + 19.38 = 15.34 The sample variance of the differences in weight can be calculated using the following formula: show variables 55, = EM 7 W
n - 1
: (3.53;;2 + (-19.09),2 + (-3.23)2 + (14.6.5),2 + (4.04): 4
5‘: = 12.43967443... Therefore, the test statistic for the mean difference can be calculated using the following formula: show variables Therefore, the test statistic forthe mean difference can he calmlated using the following formula: show variables d - pa
5:1
W
15.34 - 0
1143967443 . \ls : 2.75741001...
: 2.76 Rounded as last step r?
H h) This Is a one-tailed test since the null hypothesis states that Macmhard products are not more expensive (which impliesthat the mean difference should he not be signiﬁcantly positive). The test statistic
follows the t distribution With 4 degrees of freedom. A software package can he used to ﬁnd the Fevalue for this test statistic to be 0.0254 (to 4 decimal places) which IS less than or = 0.1. Alternatively,
you can consult a table of t distribution critical values to see that: Probabllityﬁ > 2.76) < Prohabllity(T> 2.132) : 0.05 < o = 0.1 Therefore, the null hypothesis is rejected and the periodicals claim lsjustiﬁed at a level of signiﬁcance of0.1. [2 polnlsl A study compared the individual pro-tax yeady income earned by residents from two states. The following table lists the statistics resulting from this study: Yearly Income
Location I sample size I sample Mean (s‘uoos) Sample Standard Deviation (s'oous)
112 16 Calculate the upper and lower bound of the 95% confidence interval of the mean difference (State A a State B) between the Income earned by Individuals from the two states. Give your answers to 2 decimal
places. You may ﬁnd this Student‘s t distribution table useful. a) Lower bound = -204 41 b) Upper bound = 326.41 a) This Is not correct. Lower hound : 50.99
b) This is not correct.
Upper bound = 71.01 Calelilatieri. . . . . . The two-sample t conﬁdence interval can he aldulated using the following formula: show variables [a outof a) This is not correct.
Lower bound = 50.9!
I!) This Is not correct. Upper bound : 71.01 Falcula '0" . .. The tworsample t conﬁdence nterval can be calculated us 9 the following formula: show variables —+ ,( 152 3—32 )1/2
: (112 - 51) 2t 51 55 6!. :l: 4.93192313...t* The degrees offreedom can be calculated as the smaller of nl-l and n1-1.Therefore, the degrees of freedom here is 50. Since one is constructing a 95% conﬁdence interval, the corresponding upper tail area
[since it Is two tailed) is 0.025. Therefore, t‘ ls equal to 2.009. 61 i 438192313... x 2.009 = 61 : 10.nuassas...
: 503913064... or 7113036936... Therefore, In a 95% twoasample t conﬁdence Interval, the lower bound ls equal to 50.99 and the upper bound Is equal to 71.01, rounded to 4 decimal places. [1 point] A biologist wants to test whether chemically modifying a particular plant family Will reﬂqu their ability t0 retain waterand moistUre. 1" order to test this. a ﬁxed volume Of water is given to plants that have been
chemically modiﬁed, and plants in a control group that have not been modiﬁed. The volume of moisture retained by each plant some time after is then recorded. The statistlm are given in the table below: Volume of moisture retained
Group Sample Size Sample Mean (ml) Sample Standard Deviation (ml)
Control Group 28 176 45
Modified Group 23 1.22 93 At level a : 0.05, the null hypothesis that chemical modiﬁcation does not reduce the plant's ability to retain water and moisture is . You may ﬁnd this Student's t distribution table useful. This is not correct. The null hypothesis that chemical modiﬁcation does not reduce the plant‘s ability to retain water and moisture is rejected. The null hypothesis and alternative hypothesis are given as follows: show variables Ho: H1 = l-Iz
Ha: U1 < U2 The twosample t statistic where the population standard deviation is not known can be calculated using the following formula: show variables The null I‘lYDOtl‘leSiS and aitemative hypothesis are given as follows: maples
Ho: P1 = U2
Ha: |-|1 < I-Iz The twcrsarnpie t statistic where the population standard deviation is not known ran be calculated using the following formula: show variables §1-§2 t — (5—12+§)1/2
"1 "2 122 - 176 = i
21.1746233 = -2.55D2218... This test statistic follows the t distribution with k degrees of freedom, where k ls equal to the smaller of n1-1 and hZ-l, which is 22. From the t distribution table with 22 degrees of Freedom, the P-value of
this test statistic can been found to be: Probabilityfr < 72.5502218...) = Probabilityfl > 2.5502218...) < Probability(T > 2.508) = 0.01 < a = 0.05
Alternatively. a software package can be used to ﬁnd a more accurate value (to 4 decimal places) of FfODﬁDiIil-YU— < '25501213...) = 0.0091.
Therefore, the null hypothesis is rejected. Students from two schools take an identical mathematics exam. Although It is expected that students from both schools are equally proﬁcient in mathematim, an examiner suspects that the students in school A
are actually.. more proﬁcient in mathematics than those in school B. and decides to test this by looking at a sample of their esanl scores and comparing the means orthose soora [roan both schools. Select the most aporopnate hypotheses for the above scenario: Ho: Mo = 0 Ho: Up = 1:3
Ha:yo>D Hammad-Ia
-§D=n ’H0:E“=EB
Hamobﬂ H,:xn>x5
3 Feedback [a out at 1] This is not correct.
The most appropriate hyipdhems are: "D: IJA = Ha
Ha: IJA ’ l-la Discussion The null hvpothﬁis (Hp) reprwents Hie relationship that is assumed to be true by delault. The aloemative hypothals (Ha) WE the relationship that the timer suspects to be true instead and is
attempting to ﬁnd evident: liar. “HEM. the rlull h‘iDDKl’IﬁlS should be that the mean score lrorn school a {LIA} la equal he the mean snore lrorn school B {up}, since it Is expected that students from both sdiools are eouallv proﬁcient in
mathema‘dist The alternative hypothesis diwld be that the mean score [mm school A (Ha) ll pramr than the mean ll'om school 3 (pa). This Is what the examiner modem; and Is the reason whyI the test Is being
concluded. The mat mean to use when nonstruan hypotheses Is the population mean {u} and not the sample mean G). The point of hypothesis testing Is to draw an inference from sample statistics about the
population statistiis. The gxamina’ is inberﬁted In knowing whether the students in general are better at madiematlcs in School 1‘. not whether the students in his chosen sample are better at mathemath
In School R. Lastly, It is not IoaaI to use the mean oftne aimnoa {Hg} in the hymesls. sinoa the data Involved In the maria is not matched pairs data. The sample taken [rum both smoois do not necessarin have to
be equalI and there is no mention of any basis upon which the indent: are paired. 3 3 at a I s1xco.c~m.os.nnzn A biologist wants to has: whether diemically modifying a particular plant family will reduce their ability to retain water and moisture. In order to mt thisf a ﬁxed
volume of water I5 given to plants that have been chemically modiﬁed. and plants in a control group that have not been modiﬁed. The volume of moisture retained
by each plant some time after is then recorded. The statistics are given in the table below: Volume of moisture retained Group Sample Size Sample Mean (ml) Sample Standard Deviation (ml)
Control Group 23 152 | 43 |
Modified Group 23 115 | 115 | At level o - 0.05. the null hypothesis that mammal modiﬁcation does not heduoe the plant's ability to retain water and moistune is . You may ﬁnd
this Student's t distribution table useful. 3 Feedback [1 out of 1] Yul are ﬂowed. Discussion Discussion The null hypothesis and alternative hypothesis are given as follows: Shﬂ variables Ha:
Ha: The two-sample 1: statistic where the population standard deviation is not known can be calculated using the following formula: Shﬂ variagjes P1: “2
“1‘: N2 il-iz -36
2531868311... -1.4218?49... This test statistic follows the t distribution with k degrees offi'eedom. where k is equal to the smaller of nl-l and n2-1. which is 22. From the t distribution table with 22 degrees offi'eedom, the P-value of this test statistic can been found to be:
Probabilitva< -1.4213749...)= Probabilityfl’ > 1.4213741"): ProbabilitvfI'b 1.717) = 0.05 2 o = 0.05
nltemativelv. a soﬁwai'e padtage can be used to ﬁnd a more accurate value (to 4 decimal plaoes) of Probabilitvfl' < 4.4218749...) = 0.0845. Therefore. the null hypothesis is not rejected. :l 3 «3 ID: MST.CP.CTM.DB.001[I A study was to be conducted to determine whether the time by which trains run late is different in the morning compared to that In the evening. A random sample of 30 times for morning trains was gathered
(from a list of times ofall morning trains) and a separate random sample of EU times for evening trains was also formed (from a list oftirnes of all evening trains). The following table of observations lists how
many minutes each morning or evening train was late by: Number of Minutes Tﬁins were Running Late by in the ﬂaming 16 15 18 19 20 6
17 21 7 13 14 1
6 5 5 14 17 8
3 4 11 12 5 9 Download the data Using the data prowded above, at o = 0.01. the null hypothesis that trains run late equally during moming and evening is . You may ﬁnd this Student's t distribution table useful. 3 Feedback You are correct. [1 out of 1] Feedback [1 out of 1] You are oorreet. DEC“ IOI‘I Us‘ng a soﬁcware package, the following statistics can be obtained: Train Delay Statistics
Period of the day Sample Size Sample Mean (minutes) Sample Standard Deviation (minutes) | Morning | 30 1133333333... 5.67713705... |
| Evening | 30 10.13333333... 516442295... | The null hypothesis and alternative hypothesis are given as follows: show variables Ho: u: = I-I2
Ha! “I at “2 The two-sample t statistic where the population standard deviation is not known can be calculated using a software package to give t = 035640651... The degrees of freedom are equal to the smaller of n1-1 and n2-1. Since the sample sizes are equal. the degrees of freedom are equal to 29. The alternative hypothesis claims that the mean of one group is merely not equal to the mean ofthe other group, so therefore the test is a two-tailed test. The
p-value that oonesponds to this t-statistic at 29 degrees of freedom is equal to 0.3987944]. Since this p-value is greater than the signiﬁcance level o (0.05),
the null hypothesis is not rejected at a = 0.01 This means that, based upon the sample data, there is not enough evidence to conclude that the mean delay time for morning trains is not equal to the mean
delay time for evening trains. That is, the hypothesized claim that the mean delay time is equal for morning trains and evening trains is not oohtradioted by the
sample data.. Calculating the t Statistic 3 of! ID: MST.CP.CI'M.D§.0050 Driving Force is a golf ball manufacturer. Their R & D deuartrnent have been developing a hive of golf ball with a new dimuling pattern that is designed to increase the flight distance ofthe ball.
They are at the testing stage and run an experiment to test the mean ﬂight distance forthe new type of ball compared to a standard one. A special device is used to fire the balls at a set force V 7'
and angle. A sample of 40 ofthe new balls (sample A) and 40 standard balls (sample B) are ﬁred using the device and the distance travelled (in feet) is recorded for each ball. wnload he data
Sample 3 761 797 747 329 514
739 775 861 791 311
764 769 734 787 354
E40 745 595 see 761
sou 734 779 771 797
E34 831 326 303 313
m
757 762 BDE 772 BEE Conduct a one-tailed hypothesis test for equality of the population means. Assume that the population variances are not equal.
You may ﬁnd this Student's t distribution table useful.
a) From the following options, select the curled null and alternate hypotheses furthis test: 1‘: Ho PA = Ha. Ha: “A < lie 3: “01 HA : Ha: Ha: i-IA * Ho '3: “0‘ HA = H57 Ha: HA > He D: Ha: PA ‘ Ha» Ha‘ PA > He The correct null and alternate hypotheses for this test are b) Calculate the test statistic. Give your answer to 3 decimal places. t = 2.019 c) At a signiﬁcance level ofD.OS,the null hypothesis is .
That is, you can state that there Is to oonclude that the prulatlon mean flight disrance : of the new type of golf ball is that of the standard golf ball. Feedback [5 out of 5] a) You are correct.
b) You are correct. c) You are correct. DISCUSSION 3) The null hypothesis (Ha) represents the relationship that is assumed to be true by default. The alternative hypothesis (Ha) represents the relationship that the tester suspects to be true instead and is
attempting to ﬁnd evidence for. Therefore] the null hypothesis should he that the mean ﬂight distanoe of the new type of golf ball (pm) is equal to the mean ﬂight distance of the standard ball (Hg). The alternative hypothesis should be that the mean ﬂight distance of the new type of golf ball (pA) Is greater than the mean ﬂight dlstanm of the standard ball (pa). This Is what the tester suspects to
he the case and isthe reason why the testis being conducted. Therefore, the null hypothesis and alternative hypothesis forthls test are those given in option C: show variables Ha: MA = Ha
Ha: HA > lie b) Using a statistical software package, the following results can he obtained: b) Using a statistical software padrage. the loilowlng results can be obtained: sample A mean 803.175 sample A standard deviation 293353244... sample B mean 785.475 sample B standard deviation 42.907?6482... test statistic 1.019 degrees of freedom 39 F-yalue 032521102...
Calculating the I Statistic 5) The degreﬁ of lreedom of the t distribution that this test statistic follows can be awouimated by i-ia - 1 I 39. This is a One-tailed test, bemuse the alternative hypothesis claims that the population mean
ﬂight distance oi the new type ofooll ball {us} Is weaker than the population mean ﬁloht dstance of the standard ooll' ball ill 5}. Since the ﬂvalue ofthe rm statistic is Iﬁs than the level ofsigniﬁcanoe, the null hypothals is rejected. This means that there Is enough evidenoe (as deﬁned by the level ofsigniﬂcanoe} to oondude the
alternate hypothesis. That is, you can state that there is dgniﬂunl avid-hm to oondude that the population mean ﬂight dial-non of the new type of golf ball is urn-liar than that of the standard
golf oall. when responding to Ale email, you don't have all of the data for the annual lﬂCOi'l'IES of the private owners of Perdisoo Percolator outlets. However. you do recall that 15 of the 400 owners were randomlyi selected to provide their Incomes for a newspaper article that was reoently published. You decide to send this data to Alex. An important point to consider here Is that, while you only have data for 15 of the Perdisco Percolator outlet owners, Alex would like to gather information about all 400. In other words, Alex Will need to draw
conclusions about the population using sample data. These conclusions Will need to be reached alterseveral steps ofanaiysis. You don‘t have time to do all of this for Alex, and so you decide to give Alex an idea of
where to start analyzing the data. Avaricious Alex From: Sam Statistic Sent: 3 Octoherlﬂli" 9'01AM
Tn: Avarioiuus Alex
Subject: Re: Money in Colfoc? Hi Alex. Small businesses can offer a good income, but you do have to be careful. Some
suooeed, some fail. There are 400 Perdisoo Percolator outlets across the country. Here are the
incomes] for the last ﬁnancial year, of the owners of 15 randomly selected
SIGFE. These are the only income ﬁgures that are publicly available and the
only ones that I can provide you with. Income (thousands $) Unfortunately, I don‘t have the time to do any analysis for you. You should
measure the central tendency and variation In this data. In particular you should calculate a income and a sample standard dlvialiﬂh t . 111ese statistii: are important, because they represent all of the data in the
sample equally and will best allow you to draw oonciusions about the average
income ofall Pendisco Peroolator franchisees. marl“. Dhc union Inhodllcﬂnll Your adrioe boNex is to analyze the data by measuring its Denter and variation. These two statistical measures farm the basis oranv subsequent data analysis. An important point here is that Ala: only has
sample data [15 moomes) from which conduslons are to be drawn about the entire population (400 insane-s]. In particular. Neat would liliie to linow about the average insane ofall Perdisoo Permlator
rrandtlsees. There are several measurﬁ ol‘ oenter and variation for sample data but some afthese are more appropriate than others for the purposes of drawing the sort afoondu dons that Alex needs to
draw. Measures of Center Alex’s goal should be to use the sample data to draw conclusions about the average income at all 400 dullness. not Just the 15 for whid'i data is available. To do this. the measure or center [at the sample
data) mat is calculated should allow Ale: to know more about the average efﬂie population data. The sample mean is the m statin measure ofcerioer for this purpose. The sample mean i'ndudﬁ in its caiariation ail arthe data in the sample. as such, it is an apprmrimation to the population mean or the larger set that the sample came from and can be used to draw
condusions regarding that population mean. On the other hand, while the median and mode are aeo measures of neuter for a set or sample data, they are not represematlire oftlie entire sample data set. For elrarniller altering a particular data Dolrit
mayI not change the meaan and mode but it iiirlll change the mean. Since the median and mode do not represent the entire data sample. it Is much more difﬁcult to use these statistics bod-aw conclusions
about the larger population that the sample was drawn from. A sesame disadvantage pfuslng the mean as a measure or center for a set afsample data Is its smsitlirltv to data points that diller greatly from the “619m afthe rest ofthe sample (nailed outliers). While
there are some outliers present. there Is no reason on believe that they do not represent the population as well as the rest or the sample data. its a result at these considerations. the sample mean is the
most appropriate measure orderioer to use. Wll'l' la the mean the beat measure «center? m Measures of earl-Man in order tip make Mislemm about the population mean. ...

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- Fall '13
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