Statistics V: Confidence Intervals and Hypothesis Testing with More Accurate Margins of Error; p-values

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1 14 Statistics V: Cofidece Itervals ad Hypothesis Testig with More Accurate Margis of Error; p-values To kow ad uderstad the differece betwee the approximate 1 formula for margi of error,, ad the more accurate stadard 1 error formula from the booklet of formulae ad tables. To kow how to calculate ad apply the more accurate formulae for stadard error whe costructig a 95% cofidece iterval. To apply the more accurate cofidece itervals whe carryig out hypothesis tests. To calculate ad apply the p-value for a test statistic as a alterative approach to hypothesis testig. A more accurate stadard error formula The backgroud The cetral limit theorem i the previous chapter has itroduced us to the stadard error of the mea, sx (sometimes writte s x ). Whe we take oe sample of size elemets ad calculate its proportio, pn, we are oly fidig the proportio from oe of may samples of size. If we were to cosider all such proportios, we would have the samplig distributio of the proportio. The stadard deviatio of the samplig distributio of the proportio is called the stadard error (SE) of the proportio, writte as spn. The formula for spn is give i the booklet of formulae ad tables as spn p(1 p) where is the sample size ad p is the populatio proportio. C Hece, if the true populatio proportio, p, is kow, we ca use either the empirical rule or the stadard ormal tables to estimate the probability that a particular sample proportio will lie withi a certai distace of the populatio proportio. From the previous chapter we kow that the empirical rule gives a rough guide, while the stadard ormal tables give a more accurate estimate.

2 2 LESS STRESS MORE SUCCESS However, i most cases we will ot kow the true populatio proportio. This is what we are required to estimate. We reverse the process: istead of writig that there is a 95% chace that a sample proportio, Np, lies i the iterval [p E, p E], we write that there is a 95% chace that p, the populatio proportio, lies i the iterval [ Np E, Np E]. Hece, we use the stadard error (SE) of the proportio: s Np C The 95% cofidece iterval usig the stadard error At the 95% level of cofidece the value of z is: 2 usig the empirical rule or 1 96 from the stadard ormal tables z 0 z 0 95 p ˆ E pˆ p ˆ + E z is the umber of stadard deviatio (SE) that the margi of error is from the mea. Hece, the margi of error, E, is give by: E (z)( ) At the 95% cofidece iterval, E We the state that the 95% cofidece iterval for the true proportio, p, is: s Np s Np Np E p Np E Np 1 # 96 C p Np 1 # 96 C

3 STATISTICS V: CONFIDENCE INTERVALS AND HYPOTHESIS TESTING 3 Example 1 Durig the makig of a movie, a survey foud that out of 724 extras, oly 181 were suitable for parts i a major movie. (i) Fid the 95% cofidece iterval for the proportio of all film extras that may be suitable for parts i a major movie. (ii) With 95% cofidece, what is the highest proportio of extras who would be suitable for parts i a major movie? (iii) If oly 400 extras were surveyed. (a) What effect would this have o the margi of error? (b) Are there ay implicatios of takig this actio? (i) The Np # 25 (= 25%) 1 Np # s Np C C 0 # 25 0 # Hece, the stadard error SE s Np 1 # 6%. Remember, the margi of error, E, at the 95% level of cofidece (1 96) s Np (1 96) (1 6) 3 136% 3 1%. The 95% cofidece iterval for the true proportio, p, is the: Np E p Np E p % 3 1% p 25% 3 1% 21 9% p 28 1% That is, the true proportio, with 95% cofidece, lies betwee 21 9% ad 28 1%. 95% cofidece iterval 0 # # % 25% 28 1% (ii) From part (i) we are 95% cofidet that the highest proportio of extras who would be suitable for parts i a major movie is 28 1%.

4 4 LESS STRESS MORE SUCCESS (iii) (a) With fewer extras surveyed, the stadard error would be greater, i.e. s Np C C (0 # 25)(0 # 75) # # # 2% ad 2 2% 1 6%. Hece, the margi of error would be greater. (b) The ew survey would be less accurate. It would be quicker ad less expesive to carry out. Example 2 A poll shows that the govermet s approval ratig is at 70%. The poll is based o a radom sample of 896 voters with a margi of error of 3%. Show that the poll used a 95% level of cofidece. Cofidece limits (z)(s Np ) 0 # 03 (z) C 0 # 03 (z) C 0 # 7(1 0 # 7) # 03 (z)(0 # 153) 0 # 03 0 # z # 96 z Hece, the poll is usig the 95% level of cofidece. 95% cofidece iterval for the populatio mea The cetral limit theorem may also be applied to form a cofidece iterval for the mea of a populatio, give the mea of a large eough sample, ad a stadard deviatio. A populatio has a mea of m ad a stadard deviatio of s. Suppose a sample of size Ú 30 has a mea of x. The the 95% cofidece iterval for the populatio mea, m, is: x 1 # 96 s 1 m x 1# s 96 1

5 STATISTICS V: CONFIDENCE INTERVALS AND HYPOTHESIS TESTING 5 I practice, if the stadard deviatio of the populatio, s, is ot kow, the we use the stadard deviatio, s, of the sample i its place. A survey was carried out to fid the weekly retal costs of holiday apartmets i a certai coutry. A radom sample of 400 apartmets was take. The mea of the sample was 320 ad the stadard deviatio was 50. Form a 95% cofidece iterval for the mea weekly retal costs of holiday apartmets i that coutry. Let m the mea weekly retal of all holiday apartmets. For the sample, x 320, the stadard deviatio s 50 ad 400. We use s 50 because the stadard deviatio, s, of the populatio is ot available. s 50 SE of the mea # 5 The the 95% cofidece iterval for the mea weekly retal of all apartmets is give by: x 1 # 96a s 1 b m x 1# 96a s 1 b # 96(2 # 5) m # 96(2 # 5) 315 # 1 m 324 # 9 95% cofidece iterval Hece, the 95% cofidece iterval for the mea weekly retal cost is from to Hypothesis testig of the populatio proportio usig the more accurate stadard error Hypothesis testig is a techique used i statistics to test whether a claim that is made is cosistet with the data obtaied. You could refer back to the material o hypothesis testig i the previous chapter before tacklig the ext two examples.

6 6 LESS STRESS MORE SUCCESS Example 1 Natioal data i 1970 showed that 58% of the adult populatio had ever smoked cigarettes. I 2010, a atioal health survey iterviewed a radom sample of 880 adults ad foud that 52% had ever smoked cigarettes. (i) Costruct a 95% cofidece iterval for the proportio of adults i 2010 who had ever smoked cigarettes. (ii) Does this provide evidece of a chage i behaviour amog the Irish? Write appropriate hypotheses. Usig your cofidece iterval, test a appropriate hypothesis ad state your coclusio. (i) Use Np E p Np E where Np 52% Np z 1 96 The Based o these data, we are 95% cofidet that the proportio of adults i 2010 who had ever smoked cigarettes is betwee 48 7% ad 55 3%. (ii) H 0 : p 58% H A : p Z 58% 0 # 52 1 # 96 B (0 # 52)(0 # 48) # 52 0 # 033 p 0 # 52 0 # % cofidece iterval 48 # 7% p 55 # 3% 58% 44% 46% 48 7% 52% 55 3% 60% p 0 # 52 1 # 96 B (0 # 52)(0 # 48) 880 Sice 58% is ot i the cofidece iterval (see the diagram above), we reject H 0. We coclude that the proportio of adults i 2010 who had ever smoked was less tha i 1970.

7 STATISTICS V: CONFIDENCE INTERVALS AND HYPOTHESIS TESTING 7 Example 2 A study addressed the issue of whether pregat wome ca correctly guess the sex of their baby. Amog a radom sample of 312 pregat wome, 171 correctly guessed the sex of the baby. (i) Costruct a 95% cofidece iterval from the give data. (ii) Use these sample data to test the claim, at the 5% level of sigificace, that the success rate of such guesses is o differet from the 50% success rate expected with radom chace guesses. (i) Np # 548 (= 54 # 8%) 1 Np 1 0 # # z 1 # 96 Use Np E p Np E 0 # # 96 B (0 # 548)(0 # 452) # # 055 p 0 # # # 493 p 0 # # 3% p 60 # 3% 49 3% 95% cofidece iterval 60 3% p 0 # # 96 B (0 # 548)(0 # 452) % 50% 52% 55% 58% 63% Based o the give data, we are 95% cofidet that the proportio of pregat wome who correctly predicted the sex of their baby was betwee 49 3% ad 60 3%. (ii) H 0, the ull hypothesis: the success rate is o differet from 50%. H 0 : p % H A : p Z 0 5 As 50% is icluded i the cofidece iterval above we fail to reject the ull hypothesis. Hece, we coclude that there is ot sufficiet evidece to warrat rejectio of the claim that wome ca correctly guess the sex of their baby.

8 8 LESS STRESS MORE SUCCESS I the fial aalysis, testig the ull hypothesis, H 0, simply ivolves a cofidece iterval ad a red dot. Either Or Cofidece iterval Cofidece iterval If the red dot is iside the cofidece iterval, we fail to reject H 0. If the red dot is outside the cofidece iterval, we reject H 0. p-values: A alterative approach to hypothesis testig A p-value is: a probability used to make a decisio o the ull hypothesis, H 0 a measure of the stregth of evidece to reject or fail to reject the ull hypothesis. p-value p-value = Sum of two equal shaded regios = 2 (shaded area to the right) Critical values for R (at the 5% level of sigificace) Reject Fail to reject Reject % % z z The decisio to reject, or fail to reject, H 0 is based o the compariso of the p-value with the level of sigificace. O our course we oly use a two-tailed test at the 5% level of sigificace. It is vital to kow that the critical p-value 0 05 at the 5% sigificace level. If p 0 # 05, there is strog evidece to reject H 0. If p 7 0 # 05, there is strog evidece to fail to reject H 0.

9 STATISTICS V: CONFIDENCE INTERVALS AND HYPOTHESIS TESTING 9 How to perform a hypothesis test usig p-value 1. State H 0 ad H A. 2. Calculate the z score (this is ofte called the test statistic, T ). 3. Determie the p-value (a diagram is useful). 4. If p 0 # 05, reject H 0. If p 7 0 # 05, fail to reject H State the coclusio i words. Example 1 A machie produces metal rods which have a mea legth of 500 cm with a stadard deviatio of 4 cm. After a service to the machie, it is claimed that the machie ow produces rods with legths that are ot equal to 500 cm. To test the claim, a radom sample of 100 rods from the serviced machie are measured ad foud to have a mea legth of cm. (i) Write dow H 0 ad H A. (ii) Calculate the test statistic for this sample mea. (iii) Calculate a p-value for this sample mea. (iv) At the 5% level of sigificace, is there evidece to show that the mea legth of the metal rod from the serviced machie is ot 500 cm? Justify your aswer. (i) The ull hypothesis, H 0 : m 500 cm. The alterative hypothesis, H A : m Z 500 cm. (ii) x 500 # 5, m 500, s 4 ad 100 The test statistic is give by: T x m 500# s 4 #

10 10 LESS STRESS MORE SUCCESS (iii) p(t 7 1 # 25) = 1 P(z 1 # 25) = 1 0 # 8944 (from tables) = 0 # 1056 The p-value sum of the two shaded regios. The p-value 2 (0 1056) (iv) Sice (or 21 12% 5%), we coclude there is strog evidece ot to reject the ull hypothesis We state that we fail to reject the claim that the mea legth of the metal rods from the serviced machies is ot 500 cm. Example 2 A compay claims that the average weight of a packet of cereal it produces is 400 g with a stadard deviatio of 12 g. To test this claim, a radom sample of 64 of these packets were weighed ad foud to have a mea value of 403 g. (i) Write dow H 0 ad H A. (ii) Calculate the test statistic for this sample mea. (iii) Calculate a p-value for this sample mea. (iv) At the 5% level of sigificace, is there evidece to show that the mea weight of the packets of cereal is ot 400 g? Justify your aswer. (i) The ull hypothesis, H 0 : m 400 g The alterative hypothesis, H A : m Z 400 g. (ii) x 403, m 400, s 12 ad 64 The test statistic is give by: T x m s

11 STATISTICS V: CONFIDENCE INTERVALS AND HYPOTHESIS TESTING 11 (iii) P(T 7 2) = 1 P(z 2) = 1 0 # 9772 (from tables) = 0 # 0228 The p-value 2 (0 0228) (iv) Sice (or 4 56% 5%), we coclude there is strog evidece to reject the ull hypothesis, H 0. We state that there is strog evidece to reject the claim by the compay that the average weight of a packet of cereal is 400 g. If p is low, H 0 must go. The lower the p-value, the stroger the evidece agaist H 0. The larger the sample size, the more precise the estimate.