Question

A psychologist wants to test whether there is any difference in puzzle-solving abilities between boys and girls. Independent samples of 11 boys and 13 girls were chosen at random. The boys took a mean of 42 minutes to solve a certain puzzle with a standard deviation of 5 minutes. The girls took a mean of 43 minutes to solve the same puzzle with a standard deviation of 3 minutes. Assume that the two populations of completion times are normally distributed and that the population variances are equal. Construct a 90% confidence interval for the difference , -u, between the mean puzzle-solving times for boys (,) and for girls (u2). Then complete the table below. Carry your intermediate computations to at least three decimal places. Round your responses to at least two decimal places. What is the lower limit of the 90% confidence interval? What is the upper limit of the 90% confidence interval? x 6 ?

Please answer neatly and correctly!

Answer 1

boys girls 42.000 5.000 11.000 X2 Sz ng 43.0000 3.000 13.0000 pool. var s= ((n-1)s?,+(nz-1)*s?:)/(n,+nz-2)= 16.2727 Point estimate: X₂-Xz -1.0000 std error of difference=Se= 5 *V(1/n,+1/nz) 1.653 for 90 % CI & 22 df value of t= margin of error E=t*std error lower confidence bound=mean difference-margin of error = Upper confidence bound=mean differnce +margin of error= 1.7170 2.838 -3.84 1.84

lower limit of the 90% confidence interval =-3.84

upper limit of the 90% confidence interval =1.84