Statistics Questions

Statistics Questions

Question 1:

A stem and leaf plot summarizes and simplifies data, the advantage of this plot is that it is easy to retrieve raw data used in plotting a stem and leaf, for example given the following data values, 43, 45,46,56,57,58 and 58. The data can be presented in a stem and leaf as follows:

4-3,5,6

5-6,7,8,8

Therefore using the stem and leaf the bear shoulder heights can be determined given the stem and leaf plot, the following tables summarizes the data: Male shoulder heights:

MALE

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Statistics Questions

49

72

77

87

57

73

78

89

68

73

78

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Statistics Questions

90

69

73

81

90

71

74

82

93

71

74

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Statistics Questions

83

96

72

75

84

97

72

75

84

99

72

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Statistics Questions

76

85

102

72

76

86

114

Female shoulder heights:

FEMALE

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Statistics Questions

50

75

67

75

68

75

71

75

72

76

73

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Statistics Questions

79

73

81

73

82

73

82

73

83

74

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Statistics Questions

83

74

84

75

85

75

93

Using the above data values the point estimate for the mean, this refers to the sample mean, using Excel the following are the results:

MEAN

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Statistics Questions

MALE

79.725

FEMALE

75.67857

Question 2:

Standard deviation:

Using the above data set the standard deviation is determined using excel, the table below shows the results:

STDEV

MALE

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Statistics Questions

12.09363

FEMALE

7.59168

Question 3:

Confidence interval:

A confidence interval is determined as follows:

[(X + ST) ≤X’≤(X-ST)] = 95%

Where X is the mean value, S is the standard deviation and T is the T table critical value at 95%level.

The above equations states that there is a 95% confidence level that the mean ranges from the determined values i.e. ((X + ST) and (X + ST))

Male:

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Statistics Questions

X =79.725 S =12.09363

T = 1.96 (from T table, two tail given n = 40) Therefore

[(X + ST) ≤X’≤(X-ST)] = 95%

[((79.725) + (12.09363)*(1.96)) ≤X’≤ ((79.725) + (12.09363)*(1.96))] = 95% Therefore the confidence interval is as follows

[(103.4285) ≤X’≤ (56.02149)] = 95%

This means that there is a 95% confidence that the male mean value ranges from 103.43 to 56.021

Female:

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Statistics Questions

X =75.67857 S =7.59168

T = 2.052 (from T table, two tail given n = 28) Therefore

[(X + ST) ≤X’≤(X-ST)] = 95%

[((75.67857) + (7.59168)*(2.052)) ≤X’≤ ((75.67857) + (7.59168)*(2.052))] = 95% Therefore the confidence interval is as follows

[(91.2567) ≤X’≤ (60.10045)] = 95%

This means that there is a 95% confidence that the female mean value ranges from 91.2567 to 60.10045

Question 4:

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Statistics Questions

Confidence interval for all the data values:

X =78.05882 S =10.60116

T = 1.96 (from T table, two tail given n = 28) Therefore

[(X + ST) ≤X’≤(X-ST)] = 95%

[((78.05882) + (10.60116)*(1.96)) ≤X’≤ ((78.05882) + (10.60116)*(1.96))] = 95% Therefore the confidence interval is as follows

[(98.83709) ≤X’≤ (57.28055)] = 95%

This means that there is a 95% confidence that the overall mean value ranges from 98.83709to 57.28055

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