SSS1 Class Test 2 Confidence Intervals (2009/2010) (Version: 8 th Oct 2009) (This test does not contribute towards your final mark)

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1 SSS1 Class Test 2 Cofidece Itervals (2009/2010) (Versio: 8 th Oct 2009) (This test does ot cotribute towards your fial mark) 1. Fid the followig probability: Pr(-2 < z < 2) This is the same as 1 mius the probability of z lyig i the two tails, which is the same as double the area associated with the lower tail (z < -2): Pr(-2 < z < 2) = 1 (2 x Pr(z < -2)) = 1 (2 x ) = = Fid z i such that Pr(z < z i ) = 0.06 From the body of the z-table, we ca see that the lower tail area 0.06 is associated with a z value of aroud Pr(z < -1.56) = Fid the value of z i such that Pr(-z i < z < z i ) = 0.99 If we take away the cetral 99% of the z-distributio, we are left with two equal tails that accout for 1%. Sice both these tails are equal i size, the area associated with each is 0.5% (= 0.005). We therefore wat to fid the value of z associated with the lower tail probability of 0.005: Pr(-z i < z < z i ) = 0.99 Pr(z < z i ) = We ca see from Table A that the lower tail area is associated with a z value of aroud So, Pr(z < -2.58) = Pr(-2.58 < z < 2.58) = Fid the value of z i such that Pr(-z i < z < z i ) = 0.95 If we take away the cetral 95% of the z-distributio, we are left with two equal tails that accout for 5%. Sice both these tails are equal i size, the area associated with each is 0.25% (= 0.025). We therefore wat to fid the value of z associated with the lower tail probability of 0.025: Pr(-z i < z < z i ) = 0.95 Pr(z < z i ) = We ca see from Table A that the lower tail area is associated with a z value of aroud So, Pr(z < -1.96) = Pr(-1.96 < z < 1.96) =

2 5. Fid the value of t i such that Pr(t < t i ) = if df = 23. Lookig at Table B, goig dow the colum headed by the lower-tail probability of 0.025, ad goig alog the row headed 23 degrees of freedom, we get a critical value of t of Pr(t < ) = Fid the value of t i such that Pr(t < t i ) = if df = 6. Lookig at Table B, goig dow the colum headed by the lower-tail probability of 0.005, ad goig alog the row headed 6 degrees of freedom, we get a critical value of t of Pr(t < ) = Is this the correct formula for computig the cofidece iterval for the mea whe you have a large sample? * μ = x i ± z s Yes. 8. What are the lower ad upper values of debt that boud the 95% cofidece iterval for the populatio mea household debt durig the credit cruch? (Sample mea debt = 10,000, sample stadard deviatio = 1,000, sample size = 100). * Step 1: Level of cofidece =.95. Large sample so ca use: μ = x i ± z s Step 2: Fid the critical values of z that boud the cetral 95%: Pr(-1.96 < z < 1.96) = 0.95 Step 3: Substitute z* = 1.96, = 100, s = 1,000, ad xbar = 10,000 ito the formula: 95% CI for μ = 10,000 ± 1.96 x (1000/10) = 10,000 ± 196 = ( 9,804, 10,196) Macro sytax ad results are give below: CI_L1M =(100) x_bar=(10000) s=(1000) c=(0.95). Large sample cofidece iterval for the populatio mea x_bar ZiL SE err Lower Upper

3 9. Suppose the distributio of scores i a statistics test have the followig distributio. Desity Desity Fuctio Estimate for SSS1 Class Test Results Results of SSS1 Test o 28 Sept TEST1 kerel = epaechikov, badwidth = 2 You have a sample of 9 test results with a average score of 10, ad a stadard deviatio of 3. How would you compute the 90% cofidece iterval for the populatio mea score? Step 1: Level of cofidece =.90. You have a small sample, ad variable is fairly ormal, so ca use * s the t-distributio to compute the CI: μ = x i ± t where df = -1 = 8 Step 2: Fid the critical values of t that boud the cetral 90%: Pr(-t* < t < t*) = 0.90 Pr(t < t*) = 0.05 Look dow the colum headed 0.05, ad look alog the row headed df = 8, which gives a value of t = Step 3: Substitute t* = -1.86, = 9, s = 3, ad xbar = 10 ito the formula: 95% CI for μ = 10 ± 1.86 x (3/3) = 10 ± 1.86 = (8.14, 11.86) Macro sytax ad results are give below: CI_S1M =(9) x_bar=(10) s=(3) c=(0.90). Small sample cofidece iterval for the populatio mea x_bar TiL SE err Lower Upper

4 10. O Moday 5 th October 2009, the frot page of the Metro ewspaper reported that the majority of adults i the UK support the reitroductio of capital puishmet. The fidig was based o the results of a survey of 1,100 adults, 54 per cet of whom said they backed the death pealty. a. Usig the traditioal method of approximatig the stadard error of the proportio (formula give below), calculate the 95% ad 99% cofidece itervals for the proportio of adults i the populatio that support the death pealty. π = p ± z * σ, where σ p = stadard error of the proportio is approximated by p ( 1 p) /. p Step 1: Levels of cofidece =.95 ad.99. You have a large sample, ad the variable is biary, so you ca use the z-distributio to compute the CI for the populatio proportio: π = p ± z * σ, where σ p = stadard error of the proportio is approximated by p ( 1 p) /. p Step 2: Fid the critical values of z that boud the cetral 95%, Pr(-z* < z < z*) = 0.95 Pr(z < z*) = 0.025; From the table we ca see that z = bouds the lower tail probability of 2.5%. ad the critical values that boud 99%: Pr(-z* < z < z*) = 0.99 Pr(z < z*) = 0.005; From the table we ca see that z = bouds the lower tail probability of 0.5%. Step 3: For 95% CI, substitute z* = 1.96, = 1,100, σ p = (0.54(0.46)/1100) = ito the formula for π: 95% CI for π = 0.54 ± 1.96 x = 0.54 ± = (0.511, 0.569) For 99% CI, substitute z* = 2.58, = 1,100, σ p = (0.54(0.46)/1100) = ito the formula for π: 99% CI for π = 0.54 ± 2.58 x = 0.54 ± = (0.501, 0.579) b. Give your results, how valid is the Metro s claim that more tha half of voters back the reitroductio of the death pealty (p.1)? The claim holds true eve at the 99% level of cofidece, but oly just: the lower boud of the 99% CI suggests that the proportio of voters i favour of reitroducig the death pealty could be as low as 50.2%. c. Is your result likely to be much differet to the result you would obtai from usig the Wilso method? Ulikely to be much differece because the proportio is ot close to zero or oe. See below: CI_L1P =(1100) x=(594) c=(.95). Traditioal Large sample CI for oe proportio ptrad zstar se_trad etrad low_trad up_trad Wilso Large sample CI for oe proportio pwls zstar se_wls ewls low_wls up_wls

5 CI_L1P =(1100) x=(594) c=(.99). Traditioal Large sample CI for oe proportio ptrad zstar se_trad etrad low_trad up_trad Wilso Large sample CI for oe proportio pwls zstar se_wls ewls low_wls up_wls d. Does it matter that the sample used excludes those aged 65 ad over? Strictly speakig, yes. People aged 65 ad over ca vote, so they should be icluded if we are tryig to make estimates for the populatio of voters. However, were there actually to be a referedum o reitroducig the death pealty, those that tur up to vote may ot be a radom sample, so the estimates reported i the Metro article might ot be a good guide to the outcome of a vote (ot that the ewspaper explicitly makes this claim). 5