Question

Question 1) Company claims that packages of skittle contain 61.5 grams of skittles. Do you agree with this claim? use α = 0.05.

48 students participated in evaluating small package of skittles. The following table shows the result of this study. Assume the weight of skittles are normally distributed with standard deviation of 2 grams. Net Weight Green Red Yellow Brown Orange Count 2923 Total 2978.4 576 589 641 531 586 Sample Mean 62.05

Question 3) Consider question 1. Find the power of test for the case in which the true mean weight of skittles is 62.5 gram.

Question 4) Consider question 3. What sample size would be required to detect a true mean weight of 62.5 if you wanted the power of test to be at least 0.99?

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Answer 1

0.5656 (1) T Null Hypg Ho Alter. Hypo HA u = 61.5 ut 61.5 ; 7 Sumple duta = = 62.050 =2; n = 48 Since n730; we use z-test; & Test statistic = 2 you = X M = 1. Jos => p-value = 2x P. (2 x 1.905) = 0.05678 - Since p-value > x=0.05 we Accept Null Hypo il we agree with claim of 61.5 gm. - (3) Now, Power = 1 - Pr (Type II error) = 1 - Pr (Accept H. / Hois false) = 1 - Pr (Accept Holu = 62.5) Note: "Accept Ho region is when (-1.91) (Z < (1.96). lie. -1.90 (X- 61,5 41.96 => "Accept to region - - I 62.0656 ; si lower - Go. 9344 : So; accept Ho & 60,934< X < 62.0656 Pr (Accept Hol u = 62,5) = Pr 160. 9345 X < 62.0656 / 4 = 62.5) = p. ( -5.42 <Z 4 -1.5048) P, (Accept Holer-625)_ 01066 - Power = 1-0.066 = 0.934 Power = 0.934

(44) Now, we want Power 7, 0.99 - P/ Accept Ho / 4 = 62.5) 30.01 We want to find n such that, above is. true; Pr[ X < 62,0656 /4-625] <ool ? ? [ 25 62 dist-625 ] = 0.01 We know, Pr [Z < (22) cool =) z =(2.326) un (62.0656-62.5) = ~ 2.326 2) n = 10.70 g n = 114.68 = [n> 115/