Question

7. Coke Cola Inc. approach your professor that they have invented a smart energy drink that they believe it can make the students calm and clear about doing statistics. The professor took this drink to the two sections of BUQU1230. One section took the sugar drinks (fake drinks) and the other section took the real drinks.

Sugar drinks	Smart drinks
97	64
95	85
89	72
79	64
78	74
87	93
83	70
94	79
76	79
79	75
83	66
84	83
76	74
82	70
85	82
85	82
91	75
72	78
86	99
70	57
91	91
82	78
73	87
96	93
64	89
74	81
88	84
88	63
60	78
73	66
	84
	85
	85

The result of the tests were listed in excel file. At 5% significance level, can we conclude that the average mark is higher in the section with smart drinks than the section with the sugar drink?

Answer 1

For Sugar drinks :

∑x = 2460

∑x² = 204186

n1 = 30

Mean , x̅1 = Ʃx/n = 2460/30 = 82

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(204186-(2460)²/30)/(30-1)] = 9.2214

For Smart drinks :

∑x = 2585

∑x² = 205597

n2 = 33

Mean , x̅2 = Ʃx/n = 2585/33 = 78.3333

Standard deviation, s2 = √[(Ʃx² - (Ʃx)²/n)/(n-1)] = √[(205597-(2585)²/33)/(33-1)] = 9.8510

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Null and Alternative hypothesis:

Ho : µ1 ≥ µ2

H1 : µ1 < µ2

Pooled variance :

S²p = ((n1-1)*s1² + (n2-1)*s2² )/(n1+n2-2) = ((30-1)*9.2214² + (33-1)*9.851²) / (30+33-2) = 91.3333

Test statistic:

t = (x̅1 - x̅2) / √(s²p(1/n1 + 1/n2 ) = (82 - 78.3333) / √(91.3333*(1/30 + 1/33)) = 1.5209

df = n1+n2-2 = 61

p-value = T.DIST(1.5209, 61, 1) = 0.9333

Decision:

p-value > α, Do not reject the null hypothesis

Conclusion:

There is not enough evidence to conclude that the average mark is higher in the section with smart drinks than the section with the sugar drink at 0.05 significance level.