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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90

69

73

81

90

71

74

82

93

71

74

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83

96

72

75

84

97

72

75

84

99

72

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76

85

102

72

76

86

114

Female shoulder heights:

FEMALE

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50

75

67

75

68

75

71

75

72

76

73

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79

73

81

73

82

73

82

73

83

74

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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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