Question

Companies that sell groceries over the Internet are called e-grocers. Customers enter their orders, pay by credit card, and receive delivery by truck. A potential e-grocer analyzed the market and determine that the average order would have to exceed $85 if the e-grocer were to be profitable. To determine whether an e-grocery would be profitable in one large city, she offered the service and recorded the size of the order for a random sample of customers. Can we infer from these data that an e-grocery will be profitable in this city? (assume 5% significance when applicable)

Orders

83.3

91.13

78.65

84.25

91.87

86.14

76.97

111.59

90.06

101.31

117.58

104.91

148.69

68.21

97.77

88.53

86.7

88.65

120.23

114.18

112.94

102.41

75.76

119.43

88.04

88.76

87.34

113.6

92.36

91.32

75.06

84.5

61.77

95.6

77.56

71.73

86.69

96.4

100.28

118.78

78.46

101.05

76.88

86.51

84.98

50.74

95.9

108.1

73.28

72.18

102.09

73.4

64.66

59.43

73.65

79.45

101.94

88.61

95.93

94.52

102.08

105.82

87.81

95.38

93.82

72.04

93.91

93.33

50.55

68.66

95.69

75.59

120.18

69.77

53.73

107.1

72.51

91.79

96.3

82.15

100.46

77.63

66.69

84.07

97.74

Answer 1

Mean X̅ = Σ Xi / n
X̅ = 7587.61 / 85 = 89.266

Sample Standard deviation SX = √ ( (Xi - X̅ )2 / n - 1 )
SX = √ ( 25129.8231 / 85 -1 ) = 17.2964

To Test :-

H0 :- µ <= 85
H1 :- µ > 85

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 89.266 - 85 ) / ( 17.2964 / √(85) )
t = 2.2739

Test Criteria :-
Reject null hypothesis if t > t(α, n-1)
Critical value t(α, n-1) = t(0.05 , 85-1) = 1.663
t > t(α, n-1) = 2.2739 > 1.663
Result :- Reject null hypothesis

Decision based on P value
P - value = P ( t > 2.2739 ) = 0.0128
Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.0128 < 0.05 ,hence we reject null hypothesis
Conclusion :- Reject null hypothesis

There is sufficient evidence to infer from these data that an e-grocery will be profitable in this city.