Question

A large buyer of household batteries wants to decide which of two equally priced brands to purchase. To do this, he takes a random sample of 100 batteries of each brand. The lifetimes, measured in hours, of the batteries are recorded in the file P09_16.xlsx. Before testing for the difference between the mean lifetimes of these two batteries, he must first determine whether the underlying population variances are equal.

a. Perform a test for equal population variances. Report a p-value and interpret its meaning. b. Based on your conclusion in part a, which test

statistic should be used in performing a test for the difference between population means? Perform this test and interpret the results.

calculate a 99% confidence interval for the difference between the mean lifetimes of brand 1 and brand 2 batteries

Brand 2

1 99.11 110.65

2 99.45 92.24

3 98.39 96.63

4 97.07 99.45

5 99.97 102.55

6 100.06 109.60

7 98.20 96.53

8 98.13 104.64

9 107.73 88.03

10 95.58 96.87

11 96.98 96.02

12 100.47 97.33

13 101.23 105.14

14 100.39 99.86

15 106.07 93.81

16 98.02 92.26

17 100.26 103.67

18 102.48 93.61

19 97.88 97.28

20 102.58 102.51

21 101.00 106.59

22 97.01 100.90

23 103.91 98.67

24 96.23 102.64

25 99.84 99.76

26 102.49 108.11

27 97.80 101.96

28 101.10 98.36

29 99.51 101.93

30 99.49 112.25

31 101.63 105.35

32 104.85 100.95

33 99.69 85.00

34 97.68 95.50

35 100.31 111.24

36 95.29 109.34

37 101.00 99.67

38 92.23 101.57

39 99.87 105.21

40 98.70 92.84

41 97.70 99.89

42 102.67 94.24

43 100.30 96.43

44 101.76 93.64

45 100.11 97.44

46 103.05 109.93

47 100.82 100.20

48 99.94 101.34

49 98.89 97.41

50 95.22 109.89

51 99.55 107.33

52 102.67 95.24

53 98.43 91.33

54 101.69 102.49

55 100.86 106.89

56 98.86 103.68

57 99.19 103.97

58 95.90 93.47

59 101.85 94.93

60 96.05 101.94

61 103.36 98.81

62 102.31 103.66

63 103.15 102.03

64 102.03 99.62

65 99.00 100.96

66 101.36 99.05

67 98.75 95.72

68 99.03 90.28

69 98.75 88.87

70 94.80 110.15

71 101.26 113.99

72 99.08 97.37

73 103.60 102.17

74 102.56 95.82

75 95.44 104.44

76 98.26 108.21

77 95.05 100.28

78 100.65 92.43

79 99.86 91.92

80 105.21 81.47

81 101.26 92.58

82 103.88 106.49

83 106.71 102.46

84 100.07 92.72

85 99.30 101.97

86 103.74 96.85

87 99.60 96.92

88 104.24 103.56

89 100.59 87.85

90 101.17 100.65

91 96.65 96.33

92 100.27 89.44

93 98.99 103.91

94 103.24 103.96

95 100.29 87.99

96 93.25 98.86

97 102.86 96.47

98 102.65 101.32

99 102.74 103.46

100 98.78

96.70

Answer 1

a)

p-value = 1

b)

we will use equal variance assumed

t = 0.78023

since TS < critical value

we fail to reject the null hypothesis

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