Math 152. Rumbos Fall Solutions to Review Problems for Exam #2. Number of Heads Frequency

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1 Math 152. Rumbos Fall Solutios to Review Problems for Exam #2 1. I the book Experimetatio ad Measuremet, by W. J. Youde ad published by the by the Natioal Sciece Teachers Associatio i 1962, the author reported a experimet, performed by a high school studet ad a youger brother, which cosisted of tossig five cois ad recordig the frequecies for the umber of heads i the five cois. The data collected are show i Table 1. Number of Heads Frequecy Table 1: Frequecy Distributio for a Five Coi Tossig Experimet a) Are the data i Table 1 cosistet with the hypothesis that all the cois were fair? Justify your aswer. Solutio: If we let X deote the umber of heads observed i the five coi toss, ad all the cois are fair, the X biomial5, 0.5). Thus, the probability that we will see k cois out of the 5 showig heads is ) ) k ) 5 k p X k) =, for k = 0, 1, 2, 3, 4, 5. k 2 2 Thus, out of of the = 3590 tosses of the five cois, o average, we expect to see p X k) of them showig k heads. These expected values are show i Table 2. The table also shows the expected couts. We ca therefore compute the value of the Pearso Chi Square statistic to be ˆQ = I this case, the Pearso Chi-Square statistic has a approximate χ 2 5) distributio sice there are 6 categories. The p value of the goodess of fit test is the, approximately, p value = PQ > ˆQ) ,

2 Math 152. Rumbos Fall Category p k Predicted Observed k) Couts Couts Table 2: Couts Predicted by the Biomial Model which is very small. Thus, we may reject the ull hypothesis that the data i Table 1 follows a biomial distributio at the 1% sigificace level. Therefore, we ca say that the data do ot support the assumptio that the five cois are fair. b) Assume ow that the cois have the same probability, p, of turig up heads. Estimate p ad perform a goodess of fit test of the model you used to do your estimatio. What do you coclude? Solutio: Suppose ow that the cois are ot fair but they all have the same probability, p, of turig up head. We ca estimate p from the data as follows: 5 ˆp = , 3590 from which we get that We ow test the ull hypothesis ˆp H o : X biomial5, ˆp). I this case we get the expected couts show i Table 3 o page 3. The Pearso Chi Square statistic, Q, has the value ˆQ 8.75, ad the approximate p value is p value = PQ > ˆQ) 0.068, sice Q has a approximate χ 2 4) statistic i this case because we estimated p from the data. Thus, we caot reject the ull

3 Math 152. Rumbos Fall Category p k Predicted Observed k) Couts Couts Table 3: Couts Predicted by the biomial5, ˆp) Model hypothesis at the 5% sigificace level, but we could reject at the 10% level of sigificace. Hece, the data gives moderate support to the hypothesis that the are slightly loaded towards yieldig more heads o average. 2. I 1, 000 tosses of a coi, 560 yield heads ad 440 tur up tails. Is it reasoable to assume that the coi if fair? Justify your aswer. Solutio: Test the hypothesis H o : p = 1 2 versus the alterative H 1 : p > 1 2. We model the tosses by a sequece of = 1000 idepedet Beroullip) trials, X 1, X 2,..., X ad form the test statistic Y = X j. j=1 We reject the ull hypothesis if Y > c, for certai critical value c, determied by the level of sigificace, α, of the test. I this case, α = PY > c) for Y biomial1000, 0.5).

4 Math 152. Rumbos Fall Usig the Cetral Limit Theorem, we have that ) c α P Z >, )1 0.5) where Z ormal0, 1). Thus, if we let z α deote a value such that PZ > z α ) = α, we have that we ca reject H o at the α sigificace level if Y > z α 1000/4. if α = 0.05, z α is the value of z which yields F Z z) = 1 α. Thus, z α = F ) We will the reject the ull hypothesis Z if Y > 526. I this case, the observed value of Y is ˆY = 560. Hece, we may reject the ull hypothesis at the 5% level of sigificace ad coclude that the data led evidece to hypothesis that the coi is biased towards more heads. 3. I a radom sample, X 1, X 2,..., X, of Beroullip) radom variables, it is desired to test the hypotheses H o : p = 0.49 versus H 1 : p = 0.51 Use the Cetral Limit Theorem to determie, approximately, the sample size,, eeded to have the probabilities of Type I error ad Type II error to be both about Explai your reasoig. Solutio: We use Y = i=1 X i p 1 = 0.51, ad defie the rejectio regio as a test statistic. Put p o = 0.49 ad R: Y > c, where c is some critical. We the have that the probability of a Type I error is α = PY > c), give that Y biomial, p o ).

5 Math 152. Rumbos Fall Similarly, the probability of a Type II error is β = PY c), give that Y biomial, p 1 ). We approximate these errors usig the Cetral Limit Theorem as follows: ) Y p o α = P po 1 p o ) > c p o po 1 p o ) P Z > c p o po 1 p o ) ), where Z ormal0, 1). Thus, we set c p o po 1 p o ) = z α, 1) where z α is the real value with the property that PZ > z α ) = α. For the probability of a Type II error we get ) Y p 1 β = P p1 1 p 1 ) c p 1 p1 1 p 1 ) Thus, we may set P Z c p 1 p1 1 p 1 ) ). c p 1 p1 1 p 1 ) = z β, 2) where z β is the real value with the property that F Z z β ) = β. For the case i which α = β = 0.01, we have z α 2.33 ad z β Equatios 1) ad 2) the become ad c = 2.33 p o 1 p o ) + p o 3) c p 1 = 2.33 p 1 1 p 1 ). 4) Subtractig 4) from 3) leads to p 1 p o ) = 2.33 po 1 p o ) + ) p 1 1 p 1 ),

6 Math 152. Rumbos Fall which leads to 2.33 = po 1 p o ) + ) p 1 1 p 1 ). 5) p 1 p o Substitutig the values for p o ad p 1 i 5) we obtai 116.5, so that we wat to be at least 13, Let X 1, X 2,..., X be a radom sample from a ormalθ, 1) distributio. Suppose you wat to test H o : θ = θ o versus H 1 : θ = θ o, with the rejectio regio defied by X θ o > c, for some critical value c. a) Fid ad expressio i terms of stadard ormal probabilities for the power fuctio of this test. Solutio: The power fuctio of this test, γθ) is the probability that the the test will reject the ull hypothesis whe θ = θ o ; that is, γθ) = P X θ o > c ) give that X ormalθ, 1/), for θ = θ o. Thus, we ca write γθ) as γθ) = 1 P X θ o c ) = 1 P θ o = 1 P θ o θ c < X θ o + c ) c < X θ θ o θ + c ) θo = 1 P θ) c < X θ 1/ ) θ o θ) + c = 1 P θ o θ) c < Z θ o θ) + c ), where Z ormal0, 1). We therefore have that γθ) = 1 F Z θ o θ) + c) F Z θ o θ) c) ), 6) where F Z deotes the cdf of the stadard ormal distributio.

7 Math 152. Rumbos Fall b) A experimeter desires a Type I error probability of 0.04 ad a maximum Type II error probability of 0.25 at θ = θ o + 1. Fid the values of ad c for which these coditios ca be achieved. Solutio: The probability of a Type I error is γθ o ) where γθ) is give i Equatio 6). Thus, α = γθ o ) = 1 F Z c) F Z c)) = 2 2F Z c). Thus, if α = 0.04, we eed to set c so that which yields F Z c) = 0.98, c The probability of a Type II error for θ = θ o + 1 is β = 1 γθ o + 1) = 1 1 F Z + c) F Z c))) = F Z + c) F Z c) = P c < Z + c) P < Z + c) = F Z + c). Thus, i order to make β 0.25, we require that F Z + c) = This yields + c Thus, c ) Thus, we may take to be at least Let X 1, X 2,..., X be a radom sample from a ormalθ, σ 2 ) distributio. Suppose you wat to test H o : θ θ o

8 Math 152. Rumbos Fall versus with the rejectio regio defied by T θ) > H 1 : θ > θ 1 θ o θ) + c, for some critical value c. Here, T θ) is the statistic T θ) = X θ), where X ad S 2 are the sample mea ad variace, respectively. a) If the sigificace level for the test is to be set at α, what should c be? Solutio: The power fuctio of this test is ) γθ) = P θ T θ) > θ o θ) + c, where T θ) t 1); that is, T θ) has a t distributio with 1 degrees of freedom. Observe that, if the ull hypothesis is true, the θ θ o ad therefore θ o θ) + c c for all θ θ o. It the follows that ) P θ T θ) > θ o θ) + c PT θ) > c). Thus, α = sup θ θ o γθ) = PT θ) > c). where T θ) t 1). Thus, to choose c, we fid a real value, t, such that PT > t) = α, where T t 1). Deotig that value by t α, 1, we get that c = t α, 1.

9 Math 152. Rumbos Fall b) Express the rejectio regio i terms of the value c foud i part a), ad the statistics X ad S 2. Solutio: The rejectio regio is X θ) > θ o θ) + t α, 1, which ca be re-writte as X > θ o + t α, 1. c) Compute the power fuctio, γθ), for the test. Solutio: From part a) of this problem we have that ) γθ) = P θ T θ) > θ o θ) + t α, 1 = 1 P θ T ) θ o θ) + t α, 1, where T t 1). Hece, the power fuctio of the test is ) γθ) = 1 F T θ o θ) + t α, 1 for θ > θ A sample of ouce cereal boxes has a mea weight of 10.4 oz ad a stadard deviatio of 0.85 oz. Perform a appropriate test to determie whether, o average, the 10 ouce cereal boxes weigh somethig other tha 10 ouces at the α = 0.05 sigificace level. Explai your reasoig. Solutio: We assume that the weight i each 10 ouce cereal box follows a ormalμ, σ 2 ) distributio with mea μ ad variace σ 2. We would like test the hypothesis H o : μ = 10 oz agaist the alterative hypothesis H 1 : μ = 10 oz.

10 Math 152. Rumbos Fall We cosider the rejectio regio R : X μ o > t α/2, 1 where μ o = 10 oz, ad t α/2, 1 is chose so that P T > t α/2, 1 ) = α, for T t 1). The, if H o is true, the statistic T = X μ o /, where = 16, has a t 1) distributio, sice we are assumig the the sample, X 1, X 2,..., X, comes from a ormalμ o, σ 2 ) distributio. Cosequetly, the test has sigificace level α. I the special case i which α = 0.05, we get that t α/2, Thus, the ull hypothesis ca be rejected at the 0.05 sigificace level if X μ o > I this problem, X = 10.4, = 0.85, ad = 16. We the have that X μ o / 1.88, which is ot bigger tha 2.13, thus we caot reject the ull hypothesis at the 0.05 sigificace level. 7. Fid the p value of observed data cosistig of 7 successes i 10 Beroulliθ) trials i a test of H o : θ = 1 versus H 1 : θ > Solutio: Let Y deote the umber of successes i the = 10 trials. The Y biomial10, θ). This is the test statistic. The p value is the probability that, if the ull hypothesis is true, we will see the observed value of the statistic or more extreme oes. I this case, if the ull hypothesis is true, Y biomial10, 0.5) ad the p value is p value = PY 7) = 10 k=7 ) 10 1 k

11 Math 152. Rumbos Fall Three idepedet observatios from a Poissoλ) distributio yield the values x 1 = 3, x 2 = 5 ad x 3 = 1. Explai how you would use these data to test the hypothesis H o : λ = 1 versus the alterative H 1 : λ > 1. Come up with a appropriate statistic ad rejectio criterio ad determie the p value give by the data. What do you coclude? Solutio: Deote the observatios by X 1, X 2, X 3. The, X 1, X 2 ad X 3 are idepedet Poissoλ) radom variables. Defie the test statistic Y = X 1 + X 2 + X 3. The, Y Poisso3λ). The p value is the probability that the test statistic will take o the observed value, or more extreme oes, uder the assumptio that H o is true; that is, Y Poisso3). Thus, p value = PY 9) = 1 PY 8) = 1 8 k= k k! e 3 A rejectio regio is determied by the sigificace level that we set. For istace, if the sigificace level is α, the we ca have the rejectio criterio p value < α Reject H o. Thus, i this case, we ca reject H o at the α = 0.01 sigificace level, ad coclude that the data support the hypothesis that λ > 1.