Question

A company tested 11 random brands of vanilla yogurt and found the number of calories per serving given below. Complete parts a through c below. 124 168 160 121 118 111 177 156 102 102 139 91 D a) Plot the data. Use the plots to check the assumptions and conditions for inference Choose the correct histogram below. ОА, OB Oc. OD 5- 3- kouerber ਦੇ ਨ · 2 80 120 160 200 Caliper Serving 80 120 180 200 Calories par serving BA 120 160 200 Calore per Marving BOT201602 Calores par Barving Choose the correct Normal probability plot below. OA ОВ. Oc. OD. a Q a 200 E 200- 1801 180 160 Celores per serving 200 180 160 140 120 100 80 Celore per Serang 140 120 1 100 80 Calores per Sering 200 190] 160 140 120 9 100 80 380 res per Serving 160 140] 120] 100) 80 50+ 2 0 0 Normal Scores Normal Scores Normal Scores Normal Scores Are the assumptions and conditions for inference satisfied?

Answer 1

a) Histogram: Answer D.

OD. Souabos 1 0- 120 160 200 Calories per serving

Normal probability plot:

OB. 200 180 160 Calore per Sering 120 100+ 80 60 Normal Scores

A. The Randomization Condition is met. The Normal probability plot is relatively straight, but the histogram is bimodal and asymmetric, so the Nearly Normal Condition is not met.

b)

∑x = 1467

∑x² = 203757

n = 11

Mean , x̅ = Ʃx/n = 1467/11 = 133.3636

Standard deviation, s = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(203757-(1467)²/11)/(11-1)] = 28.4825

95% Confidence interval :

At α = 0.05 and df = n-1 = 10, two tailed critical value, t-crit = T.INV.2T(0.05, 10) = 2.228

Lower Bound = x̅ - t-crit*s/√n = 133.3636 - 2.228 * 28.4825/√11 = 114.23

Upper Bound = x̅ + t-crit*s/√n = 133.3636 + 2.228 * 28.4825/√11 = 152.50

114.23 < µ < 152.50

c)

C. The confidence interval contains the diet guide's claim of 140 calories, so this claim is supported by the confidence interval.