A1) a) The 98% Ci for the difference between population means is : ( x1 x 2 ) Z 0.01 S ! 1 1 + n1 n2 Where, x1 = 0.52, n1 = 10, s1 = 0.02 x 2 = 0.5, n2 = 15, s 2 = 0.01 Z 0.01 = 2.3263 S= ! 2 (n1 1) s12 + (n2 1) s 2 n1 + n2 2 Using all the values , the confidence interval is given by: (0.006, 0.034) b) Let ! 1 denote the mean reaction time of non players of video games in the population and ! 2 denote the mean reaction time of players of video games in the population. Null hypothesis, Ho: ! 1 =! 2 Alternative hypothesis, H1: ! 1 >! 2 ( x1 x 2 ) /{S The test statistic :Z= ! 1 1 + } n1 n2 Where, x1 = 0.52, n1 = 10, s1 = 0.02 x 2 = 0.5, n2 = 15, s 2 = 0.01 S= ! 2 (n1 1) s12 + (n2 1) s 2 n1 + n2 2 Thus Z=3.3226 The tabulated value at 2.5 % LOS is 1.9596 Thus HO is rejected as the calculated value is greater than tabulated value. Thus the researcher can conclude his hypothesis. A2) a)The 97.5% CI for the difference in the population proportion is given by: ( p1 p 2 ) Z 0.0125 ! p1 (1 p1 ) p 2 (1 p 2 ) + n1 n2 Where, ! p1 is the sample proportion of Americans and ! p 2 is the sample proportion of Canadians. 550 = 0.55 1000 225 p2 = = 0.45 500 n1 = 1000 p1 = n2 = 500 ! Z 0.0125 = 2.2414 Thus the CI is (.0389,0.1610) b) Let p1 and p2 be the population proportion for Americans and Canadians respectively/ H0:p1=p2 HI:p1! p2 ( p1 p 2 ) /{ Test stat: Z=! p1 (1 p1 ) p 2 (1 p 2 ) + } n1 n2 Using all the values from part a), Z =3.6698 The calculated value is greater than tabulated value so we reject the null hypothesis. A3) Here we use the paired t test . Let xi's denote the number of hours before program and yi's denote the number of hours after program. Then di's denote the differences between xi's and yi's. d The test statistic is t=! S d / n ~ t n 1 Where , d = d ! Sd = i i n (d i d )2 i n 1 Thus using the given values , t=-1.04 The tabulated value is 1.89 Since abs t is less than tab t , we do no have sufficient evidence to reject the null hypothesis. So we say there is no significant difference in the before and after program mean hours. A4) Here we would set up a chi square test SEASON OBS FREQ EXPECTED FREQ ! Fall 80 70 1.4285 Winter 40 50 2 Spring 70 60 1.667 Summer 10 20 ( o i ei ) 2 / ei 5 The expected frequencies are calculated using the given percentage for each season and expressing for 200 Like for Fall it is 35% and 35% of 200 is 70 and likewise for others. Now the test statistic is : ( o i ei ) 2 2 = ~ n 1 ei i ! 2 So 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 that observed data contradicted the hypothesis. A5) We set up the contingency tables as below: OBSERVED VALUES OPINION NEUTRA OPPOSED L UNDER 25 IN FAVOUR TOTAL OVER 55 5 10 20 10 30 10 50 15 10 5 30 30 AGE 35-55 5 45 25 100 ! ! EXPECTED VALUES OPINION OPPOSED UNDER 25 AGE NEUTRAL IN FAVOUR TOTAL 6 22,5 12,5 50 13,5 7,5 30 30 ! ! ! 20 9 OVER 55 5 15 35-55 9 45 25 100 Expected values are calculated using{( row total *column total)/grand total}The test 2 = ( o i ei ) 2 ei with (r-1)*(c-1) df where is the # of rows and c is the # i statistic is given as: ! of columns. Using the values from above and the formula, we get calculated value as 17.352.The tabulated value at 1% LOS is: 13.2767.Since calculated value is greater than the tabulated value, we reject the hypothesis that there is no relation between age and opinion A6) a) The 95% CI for the population variance is given by: ! (n 1) S 2 (n 1) S 2 >2 > (2 / 2 ) (2 / 2 ) 1 Here , = 0.05 S 2 = 25 n = 16 2 0.975,15 = 6.262 2 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 : 2 = 100 2 ! H 1 : < 100 (n 1) S 2 ~ (2n 1) 2 Test statistic:! Thus calculated value is: 3.75 Tabulated value is 25.Since Calculated value is les than tabulated value we don't have enough evidence to reject HO A7) Let ! 1 , 2 , 3 denote mean times to relieve pain for Brand A,B,C respectively H0: ! 1 = 2 = 3 H1:Atleast 2 are different We run one way ANOVA Anova: Single Factor SUMMARY Groups Count Sum Average Variance BRAND A 4 51 12.75 4.916667 BRAND B 4 81 20.25 18.25 BRAND C 4 76 19 34 ANOVA Source of Variation SS Between Groups 129.1667 Within Groups Total 171.5 300.6667 df MS F P-value F crit 2 64.58333 3.389213 0.079947 5.714705 9 19.05556 11 ! From the table since the p value is greater than 0.025, we do not have sufficient evidence to reject the null hypothesis, thus the mean time is not statistically significant for 3 brands