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**Unformatted text preview: **3 Feedback [2 out I112] a) You are correct. b) You are correct. Discussion The hypothesis test carried out by Laid-back Larry is a Mmemar test for the dlﬁerenoe between two proportions because the proportionsof two related
DODUlations are being compared. The populations are related because a person's evaluation of the taste of the chocolate froth a new machine would be related to
that same person's evaluation of the taste of a chocolate from the current machine. a) Let B represent the number of people who thought that the chocolate from the current machine tasted good (but also thought that the chocolate from the new machine tasted bad). In other words. the cunent machine made better tasting chocolates than the new machine (the opposite of what is being tested
for). Let W represent the number of people who thought that the chocolate from the current machine tasted bad {but also thought that the chocolate from
the new machine tasted good). In other words. the wrrent machine made worse tasting chocolates than the new machine. Having B - W < 0 would indicate that. overall there is more evidence supporting the hypothesis that the current machine mam worse tasting chocolates. The value of (B 4- WW
represents the variation in the responses concerning whether the current machine made better or worse tasting chocolates. The value of B - W is standardized by dividing it by (B + W)”.
The test statistic can be calculated using the following formula: show variables B - w
(e + w)”
3 - a
(3 + a)” = -1.oon b) The hypothesis test carried out by Laid-back Larry is a one-tailed test because the alternative hypothesis is speciﬁcally concerned with whether pc is fees than on, as opposed to being generally concerned with whether p: is merely cﬁﬂierent to p". The lower the value Of the test statistic. the more evidence there is that more people thought the current machine created worse tasting chocolates.
Therefore, for this one-tailed test. the null hypothwis is rejected if the value of the test statistic is less than the critical value corresponding to the level of
significance oi the test. This is because the level of signiﬁcance deﬁnes the aandard by which statisticians decide that there is enough evidence to suggest
that the hull hvmthesis is false. A 95% conﬁdence level corresponds to a level of significance of 0.05. The critical value is obtained from the W
gape. According to this table. the critical value that corresponds to a signiﬁcance of 0.05 is -l .645. According to the test statistic for Larry's hypothesis test. the null hypothesis should not be rejected at a 95% conﬁdence level since -1.000 2 4.645. 3| 1 of3 m: nsr.c9.c9v.o1.ooao Consider the following two soenancs: Seen arln 1 It is known that 30% of registered small buslneﬁ owners across the nation currently apply for a particular deduction when paying their annual tax. A new
scheme is being proposed that may increase the proportion oi s’nall business owners that apply for the deduction. A survey is conduixed in which a random
sample of 1 00 small business owners are asked if they would apply for the deduction, should the new scheme come into Effect. A total of 39 respondents said
bhat they would. A hypothesis test is conducted to debermine if the proportion of small business owners that apply for the deduction would increase with the
new scheme. Soon arlo 2 It is believed that 40% of small businesses across the nation currently employee more than ten people. A new scheme ls being proposed that may Increase ohe
likelihood that a small busineﬁ will employ more people. Asurvey is cmducted in which a random sample of 100 small busineﬁ owners are asked ii they
currently employ more than 10 staff. A total of 41 respondents said that they do. Another survey is conducted in which a random sample of 100 Small business
owners are asked if they would employ more than 10 staff, should the new scheme come Into effect. A total of 48 respondents said that they would. A
hypothesis tat ls conducted to determine if the proportion of small businessa that would employ more than 10 staff under the new scheme is greater than the
propiortion of small businﬁses that currently employ more than 1|] staff. . Consider the twovparameter altematlve hypothﬁls: Ha: no > p: This would be the stated alternative hypothﬁis for: the hypothesis test in Scenario 1 only the hypothesis tests in both Scenario 1 and Scenario 2 the hypothesis test in Scenario 2 only neither of the hypothesis tests presented in the two scenarios This would be the stated alternative hypothesis for: the hypothesis test in Scenario 1 only
- the hypothesis tests in both Scenario 1 and Scenario 2
the hypothesis test in Scenario 2 only
neither of the hypothesis tests presented in the two scenarios [0 outof 1] This is not correct.
This would be the stated alternative hypothesis for the hypothesis test In Scenarlo 2 only. Discussion Only one of the two scenarios presented in this question involves actually comparing two unknown population parameters. This is scenario 2. In soenarlo 2f the
proportion of small businesses that currently employ more than 10 staﬁ is not known. thougi it is believed to be 40% (or 0.4}. In that scenario, it is also not
known what proportion of small businases would employ more than 10 staff if the proposed scheme comes into effect, though it is believed that t is greater
than the current proportion. The key feature to note is: In Scenario 2, there are two population propomons that are unknown. It Is suspected that one population proportion Is larger than the
other. The hypothesis tact relates to this auerﬂon. So in scenario 2, the null hypotheﬁs would state that there Is no difference between the proportion of Small businesses that would employ more than ltl staﬁ
under the new scheme (p2) and the proportion of Snail businesses that currently employ more than 10 staﬁ (p1). That is, the null hypothesis would state that
the two population proportions. 91 and p2. are equal. The alternative hypothesis would state that the proportion of snail husinases that would employ more
than 10 staff under the new scheme is greater than the proportion of small businesses that currently employ more than 10 staff. That is, the alternative
hypothesis would state that p: 3 p1. In contrast. Scenario 1 is a one-parameter best, because there is only one unknown pomlation parameter. The unknown population parameter is the proportion
of small business owners that would apply for the tax deduction, should the new scheme take eﬁect. It is believed that this proportion is greater than the
proportion of small business owners that currently apply for the deductionr which is known to be 30% (or 0.3). So in this scenario, the null hypothesis would
state that the unknown population proportion was equal to 0.3. and the alternative hypothesis would state that this proportion is greater than 0.3. :I 3 of 3 ID: MST.cH5T.cTP.oz.oolo The chief of quality control at Badbury chocolatesI Metimldus Martin, wants to test whether
Implementing new machlnes will adversely affect the ﬂavor of its chocolates. A random sample of 52 people have been selected and they were asked to taste a chocolate that had been produced using a cum.“ New machine cun‘ent machine. 111ese people then noted whether they thought the chocolate tasted good. That same machine Good Bad Tat” sample was then asked to taste a block of chowlate that had been produced using a prototype of the new machine and. once again. they noted whether they thought the chocolate tasted good. The results Good of this taste tsting are displayed in the table. Bad
You may find this Willie useful throughout the following questions. Total 52 a) The chief accountant at Badbury chocolates. Laid-back Larry. does not want to incur the expense of purchasing new
machines unless It can be shown that there would be a signiﬁcant Improvement In the ﬂavor of the chocolates. In
oﬂier words. Larry wants to know whether the proportion of good tasting chocolates created by the current machine (pc) is less than the proportion of good
tasting chmolates created by the new machine to"). Laid-back Larry has construded a hypothesis test with Hg: pc = p" and H3: pc < p". Calculate the test
statlstic (2) that corresponds to this hypothesis test. Give your answer as a decimal to 3 declmal places. 2 I 1.414 b) using the test statistic for Larry's hypotnﬁis test and a 95% conﬁdence level, Larry should the null hypotnﬁis. [1 auto! 2] a) This is not correct.
2: 4.414 Ia) You are correct. Discussion : Feedback [1 out of z] a) This is not correct.
2- -1.414 b) You are correct. Discussion The hypothesis test carried out by Laid-back Larry Is a MCNemar test for the difference between two proportions because the proportions of two related
populations are being compared. ‘Ihe populations are related because a person's evaluation of the taste of the chocolate from a new machine would be related to
that same person's evaluation ofthe taste ofa chocolate from the current machine. a) Let 5 represent the numberof people who thought that the chocolate from the current machine tasted good (but also thought that the chocolate from the
new machine tasted bad). In other words. the current machine made better tasting chocolates than the new machine (the opposite of what is being tested
for). Let W represent the number of people who thought that the chocolate from the current machine tasted bad (but also thought that the chocolate from
the new machine tasted good). In other words. the current machine made worse tasting chocolates than the new machine. Having B - w < 0 would
indicate that, overall there is more evidence supporting the hypothesis that the cunent machine makes worse tasting chocolates. The value of(El + W)"1
represents the variation in the rﬁponses concerning whether the current machine made better or worse tasting chocolates. The value of B - W is standardized by dividing it by [B + w)“.
The test statistic can be calculated using the following formula: hide variaﬂa B = the number of pe0ple who thought that the current machine made better tasting chocolates = 2
w = the number of people who thought that the current machine made worse tasting chocolates =
z = test statistic = unknown 1<>>>I<<<>>><<<>>>I<<<>>><<<>>>I1 o - w
(a + W)” A (2 + 6)“ = 4.4142355... = -1.414 Rounded as last step The test statistic can be calculated using the following formula: mules m = the number of people who thought that the current machine made better tasting chocolates = 2 W = the number of people who thought that the current machine made worse tasting chocolates = 6
= test statistlc = unknown N B - W
(e + W)”
2 — 6
(2 + 5)“2
4.414213%...
4.414 Rounded as last step b) The hypothesis test carried out by Laid-back Larry is a one-tailed test because the alternative hypothesis is speciﬁcally concerned with whether pc is less
than on. as opposed to being generally concerned with whether pc is merely dIHErEnt to p". The lower the value of the test statistic. the more evidence there is that more people thought the cunent machine created worse tasting chocolates.
Therefore. for this one-tailed test. the null hypothesis is rejected if the value of the test statistic is less than the critical value corresponding to the level of
signiﬁcance of the test. This is because the level of signiﬁcance deﬁnes the standard by which statisticians decide that there is enough evidence to suggest
that the null hypothesis is false. A 95% conﬁdence level corresponds to a level of signiﬁcance of 0.05. The critical value is obtained from the Mm
tame. According to this tabler the critical value that corresponds to a signiﬁcance of 0.05 is -1.645. According to the test statistic for Larry's hypothesis test, the null hypothesis should not he rejected at a 95% conﬁdence level since -1.414 2 -1.645. 3| 1 of 3 ID: M5T.CP.crv.o1.oo1o The proportion of people from Crighton that go to the movies at least once a week l5 denoted p1. while the proportion of those from Lindille that go to the movies at
least onoe a week is denoted p2. If samples of size n; are to be drawn from the population of Crightorr, the sample proportion, p1, of people in these samples that go
to the movies at least onoe a week approximately follows the normal distribution with mean: p1 and variance: [31(1 v own 1. Slmiladv, if samples of size n; are to be drawn from the population of Undille. the Sample proportion. ﬁz, of people in that: samples that go to the moviﬁ at least onoe a week approximately follows the
normal distribution with mean: p2 and variance: p2(1 — pzlr'nz. The difference 61 — 62 approximater follows the normal distribution with: mean: p; + p2 and variance: 91(1 - Pﬁfni + p20. - p2)fn2
mean: p1 + p; and variance: p1(1 - plynl - pzll - pzlfnz
mean: p1 - p2 and variance: pﬂl - pllfnl - p2(1 - paynz - mean: p1 — p2 and variance: p1(1 — pljfnl + p2(.1. — pzjfnz :| Feedback [1 out of 1] YOU are CDH‘EII. Discussion Soho-anong random ea rlahles If two random-i variables are nonnallv distributed then their cil’ferenoe is also normaihir distributed: Suppose X and Y are two random variables following normal dlstrlbutions and that:
500 = H): and mm = ax’
Em = NY “d VAR“) = Orz-
111en the dil’ferenoe of the two randOm variables (X-Y) is also non11aily distributed with mean and varianoe: E(x-Y} = ux - [JV and va-Y} = ox2 + of. [1 outofl] You are correct.
Discussion Subtracting random variables
If two random variables are normally distributed then tnelr dl’ﬁerenoe is also normally distributed: Suppose X and Y are two random variables followlng normal distrlbutions and that:
:00 = llx and imam = ax2
E") = NV and mm = of.
ThEn the differeme of the two random variabla DH!) is also normally distributed with mean and variance: six-v) = ox - NY and want-v; = 0x2 + of. So while you do ﬁnd the difference of the means (that is, the diﬁ‘ereme betweEn ox and uy) to calculate the mean of the ciﬁerenoe between the random variables (that is, the mean of X-Y), you actually add the varianoes to ﬁnd the varianoe of the diﬁerenoe between the two random variables. This fact is used quite often in hypotth testing because the null hypothesis is usually that there is no differenoe between the two population parameters (and so the difference
between the sample statistics is olten investigated). 3 1 or: m: MST.CHST.6FOT.I)2.0030 A survey.r was conducted at four separate towns on whether the participant was a smoker (smokes at least once a week) or a non-smoker. The table below lists the
results oi this survey: Town A Town B Town C Town D
Smoker Non—smoker
Total 401 ?00 1,996
802 1, 129 3,634 3) Calculate the proportion ofparticipanu from all towns who are non—smokers. Give your answer as a decimal to 3 decimal planes.
Proportlon = 0.549 b) Calculate the proportion ofparticipanls from Town B who are smokers. Glue your answer as a decimal to 3 decimal plats.
Proportion = 0.4 50 :) Calculate the proportion ofnon-smokers that are from Town A. Give your answer as a decimal to 3 decimal places. Proportion = 0.4 90 [2 out of 3] a) You are correct.
b) You are correct.
:1) This is not oonept. Proportion = 0.169 Calculation
c) This is not correct.
Proportion = 0.169 Calculation The two-way table is shown below: TOWI'IA TOWI'IB TOWI'IC TOWI'ID Total
457 429
337 | 553 | 401 | 700 |1,996
633 | 1,015 | 302 | 1,129 |3,634 a) The total number of people in the study is 3.634 and the number of people who are non-smokers from all towns is equal to 1.996. Therefore, the proportion
of people from all towns who are non-smokers is equal to: 1,995
3,634 = 0.549 Rounded as last step 054925702. . . b) The total number of people from Town B is equal to 1.015 and the number of people from Town B who are smokers is equal to 457.1'herefore. the required
proportion of people from Town B who are smokers is equal to: 457
1,015 0.45024631... 0.450 Rounded as last step c) The total number of people who are non-smokers ls 1,996. The number of people who are nonvsmokers and are also from Town A Is equal to 337. Therefore,
the required proportion of nonvsmokers that were from Town A is: 337
1,995 0.16333765... (I. 169 Rounded as last step 3| 2 of 3 ID: nsr.cnsr.cro'r.os.omo Select whether a chi—square test for association is the most appropriate for the following scenarios: Most Not appropriate appropriate 3) Gabrielle wishes to know whether gender has any relation with whether one prefers wine, beer, or liqueur.
She surveys a group of men and women and records the number of participants who prefer wine, the number -
who prefer beer and the number who prefer liqueur. b) Gabrielle surveys a group of people on how often they eat dinner at restaurants and diners. The age and
annual income of each participant are recorded. Gabrielle is interested in knowing whether these two factors
are important in determining how often one goes out for dinner. c) A study intends to determine whether the average annual interest rate has any relation with the number of
households who go overseas on a holiday at least once a year. Danielle collects data on the average annual '
interest rate and the number of households that travelled overseas at least once in that year over }' years. d) It has been proposed that one's education level has a strong relation with one's income level. Fiona collects
data from a large group of 31 year-old adults about their education level (primary and lower, secondary,
tertiary and higher} and their income level (to the nearest $100). Fiona plans to test this data to see whether
the mean income in each education level is the same across all three levels. :| Feedback [3 out or 4] a) You are correct.
b) You are correct.
c] You are correct-
d) This Is not correct.
Using the chi-square test for association In the soenario in part d) is not appropriate. Discussion a) The chi-square test Is the most appropriate here because the two factors, gender and alcohol preference. are both categorical variables. The frequency of
each combination of the two factors is recorded on a two-way table as shown below: Discussion is) The chi-square tea: Is the most appropriate here because the two factors, gender and alcohol preference. are both cabegorlcal variables. The frequency of
each combination of the two factors is reoorded on a two-way table as shown below: IWine Beer Liqueur Total Male Ell—n rem-e nun-:-
ma- Ilnll Gabrielle plans to teat whether gender has any effect on alcohol preference. This can be interpreted as being a test of whether the factors - gender and
alcohol preference, are assoo'ated (null) or not (alternate). b) The chi-square test Is no: the most appropriate here because the factors - frequency or dining outsde. age, income level - are all numerical variables. The
inoome level, for example, has an almost inﬁnite number of levels (e.g. $1,000, $1,001 $100,000). Chi-square tests are used for categorical variables
that have only a few levels {e.g. Male. Female). Even if the levels are grouped Into categories of ranges, there will still be the problem of having more than
two factors each with dlﬁerent levels. The data would be tabulated as follows: Frequency of Dining Out Age Income Level a1 a2 a3
| b1 | b2 | ha |
| c1 | c2 | :3
| GI. | az | d3
| 21 | e2 | e3 | 'lhe most appropriate test here is the multiple regression test for oorrelation, due to there being two variables which (are hypothesised) to affect another
variable. c) The chi-square test is not the most appropriate here because the two factors are both numerical. Interest rates and frequency of overseas travelling can
both take a large number of values. while it Is possible to split numerical ranges into categories. it would be mum easier and more effective to use a simple
linear regression model, The most appropriate test here is a simple regressirm test for correlation. ﬁle data would be tabu...

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