Part 1
Keywords:Confidence Intervals for Means, Normal distribution, tstudent distribution
Exercise 1. (5 points) For the following data set
114 
112 
123 
111 
109 
103 
112 
113 
107 
129 
108 
126 
131 
129 
126 
115 
123 
108 
131 
115 
134 
133 
132 
114 
104 
127 
128 
137 
114 
139 
128 
117 
121 
110 
141 
121 
138 
140 
122 
130 
115 
102 
Table 1: Data set Exercise 1
Construct a
 90%confidence interval for the population mean;
 95%confidence interval for the population mean;
 99%confidence interval for the population mean;
 Whatassumption do you need to make about the population of interest to construct the confidence intervals ?
Exercise 2. (5 points) For the following data set
136 
141 
137 
145 
131 
135 
138 
136 
130 
132 
136 
134 
137 
133 
141 
139 
Table 2: Data set Exercise 2
Construct a
1. 90%confidence interval for the population mean;
2. 95%confidence interval for the population mean;
Exercise 3. (5 points) Two boxes, A and B contain equal numbers of balls (black and white). A sample of 100 balls selected with replacement from each of the boxes revealed 52 white balls form box A and 44 white balls from box B.
 Testthe hypothesis that the two boxes have equal ratio of white balls, using a significance level of 0.05
 Testthe hypothesis that box A has greater ratio of white balls than box B, using a significance level of 0.05
Exercise 4. (5 points) Consider the sample data for grades A, B, C & D of four different classes, which are tabulated in table 3. Test whether the four means are equal at the 0.05 level of significance.
Perform a test that will compare all 4 means at once. We are not interested in pairwise comparisons at this stage.
Please show all the steps in your calculations.
Table 3: Data set Exercise 4
80 
81 
83 
78 
83 
79 
72 
79 
80 
83 
80 
80 
65 
76 
81 
98 
92 
65 
87 
57 
81 
74 
68 
81 
65 
34 
68 
71 
34 
45 
78 
91 
57 
61 
51 
46 
71 
69 
66 
72 
71 
59 
63 
82 
81 
67 
61 
52 
72 
75 
64 

81 
80 
74 

36 
55 
78 

83 
67 


91 
94 


73 



51 



98 



Answer
Part 1.
Exercise 1:
Data 

114 
112 
123 
111 
109 
103 
112 
113 
107 
129 
108 
126 
131 
129 
126 
115 
123 
108 
131 
115 
134 
133 
132 
114 
104 
127 
128 
137 
114 
139 
128 
117 
121 
110 
141 
121 
138 
140 
122 
130 
115 
102 
Population Mean = 
121.2381 

Confidence intervals 
90% 





Confidence intervals 
95% 

Number of Sample = 
42 

Confidence intervals 
99% 

Standard deviation = 
11.08022 





Standard error = 
1.709715 

t(90%) 

1.644854 





t(95%) 

1.959964 




t(99%) 

2.575829 
1) 
Upper 90% confidence limit = 
124.0503 




Lower 90% confidence limit = 
118.4259 



2) 
Upper 95% confidence limit = 
124.5891 




Lower 95% confidence limit = 
117.8871 



3) 
Upper 99% confidence limit = 
125.642 




Lower 99% confidence limit = 
116.8342 


The population mean is 121.2381.
 90% confidence interval of the population mean =
Upper 90% confidence limit = 124.0503.
Lower 90% confidence limit = 118.4259.
 95% confidence interval of the population mean =
Upper 95% confidence limit = 124.5891 (Efron and Tibshirani 1986).
Lower 95% confidence limit = 117.8871.
 99% confidence interval of the population mean =
Upper 99% confidence limit = 125.642.
Lower 99% confidence limit = 116.8342.
 The assumptions needed to construct the confidence intervals are
 Each sample is independent to each other.
 All the samples are normally distributed (Payton, Greenstone and Schenker 2003).
 There exists no significant outlier in the data set.
Exercise 2:
Data 

136 
141 
137 
145 
131 
135 
138 
136 
130 
132 
136 
134 
137 
133 
141 
139 
Population Mean = 
136.3125 

Confidence intervals 
90% 





Confidence intervals 
95% 

Number of Sample = 
16 

Confidence intervals 
99% 

Standard deviation = 
3.961797 





Standard error = 
0.990449 

t(90%) 

1.644854 





t(95%) 

1.959964 




t(99%) 

2.575829 
1) 
Upper 90% confidence limit = 
137.9416 




Lower 90% confidence limit = 
134.6834 



2) 
Upper 95% confidence limit = 
138.2537 




Lower 95% confidence limit = 
134.3713 



3) 
Upper 99% confidence limit = 
138.8637 




Lower 99% confidence limit = 
133.7613 


(Source: Altman and Gardner 1988)
The population mean is 136.3125.
 90% confidence interval of the population mean =
Upper 90% confidence limit = 137.9416.
Lower 90% confidence limit = 134.6834.
 95% confidence interval of the population mean =
Upper 95% confidence limit = 138.2537.
Lower 95% confidence limit = 134.3713.
 99% confidence interval of the population mean =
Upper 99% confidence limit = 138.8637.
Lower 99% confidence limit = 133.7613.
 The assumptions needed to construct the confidence intervals are
 Each sample are independent to each other (Ci 1987).
 All the samples are normally distributed.
 There exists no significant outlier in the data set.
Exercise 3:
Null hypothesis: 
The proportions of white balls in two boxes are equal. 

Alternative hypothesis: 
The proportions of white balls in box A is greater than box B. 

Box A 


Box B 

Total balls (n1) = 
100 

Total balls (n2) = 
100 
White balls = 
52 

White balls = 
44 
Proportion (p1) = 
0.52 

Proportion (p2) = 
0.44 
Difference in proportions (p1  p2) = 
0.08 



Total samples (n1+n2) = 
200 



Total white balls = 

96 


Total proportion (pbar) = 
0.48 



Zstatistic = 

1.132277 


Level of significance = 

5% 


Zcritical = 

1.959964 


Sig. = 

Cannot reject null hypothesis 

Hypothesis:
The hypotheses are
Null hypothesis (H_{0}): Box A has equal proportion of white ball such as proportion of white ball in Box B.
Alternative hypothesis (H_{A}): Box A has higher proportion of white ball than the proportion of white ball in Box B.
Level of significance:
The level of significance is assumed to be 5%.
Test applied:
Two sample proportional Ztest.
Calculated Zstatistic:
Z = = 1.132277 (Sahai and Khurshid 1996)
(p1= proportion of white ball in box A; p2= proportion of white ball in box B; p=proportion of white ball together in box A and box B, n1= total number of balls in box A; n2= total number of balls in box B).
Decisionmaking:
1.959964>1.132277, so, Z_{sig.}>Z_{crit.} Therefore, the null hypothesis could not be rejected with 95% probability (Levine et al. 1999).
Interpretation:
Therefore, it could be interpreted that Box A has equal proportion of white ball of proportion of white ball in Box B.
Exercise 4:
Anova: Single Factor 






Summary 






Groups 
Count 
Sum 
Average 
Variance 


Grade A 
20 
1445 
72.25 
292.4079 


Grade B 
17 
1164 
68.47059 
217.2647 


Grade C 
15 
1074 
71.6 
95.68571 


Grade D 
12 
887 
73.91667 
238.6288 


ANOVA 






Source of Variation 
SS 
df 
MS 
F 
Pvalue 
F crit 
Between Groups 
236.9355 
3 
78.97851 
0.364614 
0.778776 
2.758078 
Within Groups 
12996.5 
60 
216.6084 










Total 
13233.44 
63 




Hypothesis:
The hypotheses are
Null hypothesis (H_{0}): The averages of four different grades A, B, C and D of four different classes are equal to each other (Heiberger and Neuwirth 2009).
Alternative hypothesis (H_{0}): There exists at least one inequality in the averages of four different grades A, B, C and D of four different classes.
Observations:
The average of grade A is 72.25, grade B is 68.47, grade C is 71.6 and grade D is 73.92.
Test applied:
Oneway ANOVA.
Level of Significance:
The level of significance is found to be 5%.
Test Statistic:
Calculated Fstatistic = 0.364614 with pvalue = 0.778776.
Decisionmaking:
As, calculated pvalue is greater than the 5% level of significance, therefore, null hypothesis could not be rejected with 95% probability (Christensen 1987). On the other hand, as F_{crit.}is greater than F_{cal.}(2.758078>0.364614), therefore, null hypothesis is accepted with 95% possibility (Ross and Willson 2017).
Interpretation:
The averages of four different grades A, B, C and D of four different classes are equal to each other.
References:
Altman, D.G. and Gardner, M.J., 1988. Statistics in Medicine: Calculating confidence intervals for regression and correlation. British medical journal (Clinical research ed.), 296(6631), p.1238.
Christensen, R., 1987. OneWay ANOVA. In Plane Answers to Complex Questions (pp. 5769). Springer, New York, NY.
Ci, B., 1987. Confidence intervals. Lancet, 1, pp.494497.
Efron, B. and Tibshirani, R., 1986. Bootstrap methods for standard errors, confidence intervals, and other measures of statistical accuracy. Statistical science, pp.5475.
Heiberger, R.M. and Neuwirth, E., 2009. Oneway anova. In R through excel (pp. 165191). Springer, New York, NY.
Levine, D.M., Berenson, M.L., Stephan, D. and Lysell, D., 1999. Statistics for managers using Microsoft Excel (Vol. 660). Upper Saddle River, NJ: Prentice Hall.
Payton, M.E., Greenstone, M.H. and Schenker, N., 2003. Overlapping confidence intervals or standard error intervals: what do they mean in terms of statistical significance?. Journal of Insect Science, 3(1).
Ross, A. and Willson, V.L., 2017. OneWay Anova. In Basic and Advanced Statistical Tests (pp. 2124). SensePublishers, Rotterdam.
Sahai, H. and Khurshid, A., 1996. Formulae and tables for the determination of sample sizes and power in clinical trials for testing differences in proportions for the two?sample design: a review. Statistics in medicine, 15(1), pp.121.
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