2 An experiment was conducted to determinewhether giving candy to dining partiesresulted in greater tips. The mean tippercentages and standard deviations aregiven in the accompanying table along withthe sample sizes. Assume that the twosamples are independent simple randomsamples selected from normallv distributedNo candyTwo candies H23318.131.393320.78 2.45The test statistic, t, is - 5.40 . (Round to two decimal places as needed.)The P-value is 0.000. (Round to three decimal places as needed.) An experiment was conducted to determinewhether giving candy to dining partiesresulted in greater tips. The mean tippercentages and standard deviations aregiven in the accompanying table along withthe sample sizes. Assume that the twosamples are independent simple randomsamoles selected from normally distributedNo candyH13318.131.39Two candies H233 20.78 2.45giving candy does result in greater tipsb. Construct the confidence interval suitable for testing the claim in part (a).

Given Data : For Sample 1 x̄1 = 18.13 n1= 33 s1= 1.39 For Sample 2 x̄2= 20.78 n2 = 33 s2= 2.45B) 95% confidence interval : Degree of freedom = 50.6665 = 50 Point estimate: x̄1 - x̄2 = 18.13 - 20.78 = -2.65 Critical value critical value at α= 0.05 and df= 50 is tα/2,df= 2.009 .....from student's t distribution table Margin of error: ME=tα/2,df × V(s1^2/n1+s2^2/n2) = 2.009 ×V( 1.39 ^2/ 33.0 + 2.45 ^2/ 33.0 ) = 0.9851 Margin of error is 0.9851 Now 95% confidence interval is CI=(x̄1 - x̄2) ± ME =-2.65 ± 0.9851 = (-3.64, -1.66) Therefore -3.64 < u1-u2 < -1.66