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## Statistics- The 98% Ci for the difference between population

A1)a) The 98% Ci for the difference between population means is 🙁 x1 − x 2 ) Â± Z 0.01 Ã S Ã!1 1+n1 n2Where,x1 = 0.52, n1 = 10, s1 = 0.02x 2 = 0.5, n2 = 15, s 2 = 0.01Z 0.01 = 2.3263S=!2(n1 − 1) s12 + (n2 − 1) s 2n1 + n2 − 2Using all the values , the confidence interval is given by:(0.006, 0.034)b)Let ! Âµ1 denote the mean reaction time of non players of video games in the population and! Âµ 2 denote the mean reaction time of players of video games in the population.Null hypothesis, Ho: ! Âµ1 =! Âµ 2Alternative hypothesis, H1: ! Âµ1 >! Âµ 2( x1 − x 2 ) /{S ÃThe test statistic :Z= !11+ }n1 n2Where,x1 = 0.52, n1 = 10, s1 = 0.02x 2 = 0.5, n2 = 15, s 2 = 0.01S=!2(n1 − 1) s12 + (n2 − 1) s 2n1 + n2 − 2Thus Z=3.3226The tabulated value at 2.5 % LOS is 1.9596Thus HO is rejected as the calculated value is greater than tabulated value. Thus theresearcher can conclude his hypothesis.A2)a)The 97.5% CI for the difference in the population proportion is given by:Ë Ë( p1 − p 2 ) Â± Z 0.0125 Ã!ËËËËp1 Ã (1 − p1 ) p 2 Ã (1 − p 2 )+n1n2ËËWhere, ! p1 is the sample proportion of Americans and ! p 2 is the sample proportion ofCanadians.550= 0.551000225Ëp2 == 0.45500n1 = 1000Ëp1 =n2 = 500!Z 0.0125 = 2.2414Thus the CI is (.0389,0.1610)b)Let p1 and p2 be the population proportion for Americans and Canadiansrespectively/H0:p1=p2HI:p1! ≠ p2Ë Ë( p1 − p 2 ) /{Test stat: Z=!ËËËËp1 Ã (1 − p1 ) p 2 Ã (1 − p 2 )+}n1n2Using all the values from part a), Z =3.6698The calculated value is greater than tabulated value so we reject the null hypothesis.A3)Here we use the paired t test .Let xiâs denote the number of hours before program and yiâs denote the number ofhours after program. Then diâs denote the differences between xiâs and yiâs.dThe test statistic is t=! S d / n~ t n −1Where ,d =∑d!Sd =iin∑ (di− d )2in −1Thus using the given values , t=-1.04The tabulated value is 1.89Since abs t is less than tab t , we do no have sufficient evidence to reject thenull hypothesis. So we say there is no significant difference in the before and afterprogram mean hours.A4)Here we would set up a chi square testSEASONOBS FREQEXPECTED FREQ!Fall80701.4285Winter40502Spring70601.667Summer1020( o i − ei ) 2 / ei5The expected frequencies are calculated using the given percentage for each season andexpressing for 200 Like for Fall it is 35% and 35% of 200 is 70 and likewise for others. Now thetest statistic is 🙁 o i − ei ) 22χ =∑~ χ n −1eii!2So from the above table , the calculated value is 10.0955 (1.4285+2+1.667+5)Tabulated value is 7.814.Since calc value is greater than the tabulated one , we conclude thatobserved data contradicted the hypothesis.A5)We set up the contingency tables as below:OBSERVED VALUESOPINIONNEUTRAOPPOSED LUNDER 25INFAVOURTOTALOVER 555102010301050151053030AGE 35-5554525100!!EXPECTED VALUESOPINIONOPPOSEDUNDER 25AGENEUTRALIN FAVOURTOTAL622,512,55013,57,53030!!!209OVER 5551535-5594525100Expected values are calculated using{( row total *column total)/grand total}The testχ2 = ∑( o i − ei ) 2eiwith (r-1)*(c-1) df where is the # of rows and c is the #istatistic is given as: !of columns. Using the values from above and the formula, we get calculated value as17.352.The tabulated value at 1% LOS is: 13.2767.Since calculated value is greater than thetabulated value, we reject the hypothesis that there is no relation between age and opinionA6)a) The 95% CI for the population variance is given by:!(n − 1) S 2(n − 1) S 2>σ2 >χ (2 −α / 2 )χ (2α / 2 )1Here ,α = 0.05S 2 = 25n = 162χ 0.975,15 = 6.2622χ 0.025,15 = 27.4884!Using all the values, the CI is (13.64211,59.885)The CI for standard dev is(3.6935,7.7385)b)!HO : σ 2 = 1002! H 1 : σ < 100(n − 1) S 2~ χ (2n −1)σ2Test statistic:!Thus calculated value is: 3.75Tabulated value is 25.Since Calculated value is les than tabulated value we donâthave enough evidence to reject HOA7) Let !Âµ1 , Âµ 2 , Âµ 3 denote mean times to relieve pain for Brand A,B,C respectivelyH0: !Âµ1 = Âµ 2 = Âµ 3H1:Atleast 2 are differentWe run one way ANOVAAnova: Single FactorSUMMARYGroupsCountSumAverageVarianceBRAND A45112.75 4.916667BRAND B48120.2518.25BRAND C4761934ANOVASource ofVariationSSBetween Groups129.1667Within GroupsTotal171.5300.6667dfMSFP-valueF crit2 64.58333 3.389213 0.079947 5.7147059 19.0555611!From the table since the p value is greater than 0.025, we do not have sufficient evidence toreject the null hypothesis, thus the mean time is not statistically significant for 3 brands!

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