Sociology Statistical Assessment Essay
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Get AccessASSESSMENT 1
Question 1 ( 5 Markss )
Statisticss show that alumnuss in New Zealand make $ 250,000 more in their lifetime than nongraduates. If you were to oppugn the cogency of this observation, what would be your footing for making so?
It is hard to oppugn the cogency of this observation because statistics proof includes aggregation, organisation, analysis, reading and coverage. It contains all informations which proves the statement is right. That is the chief intent of statistics in any research. In every research, statisticianâ€™s helps researcher from the preliminary phase, that is, design of research to till stop that is, describing. Statistics leads a research worker to do valid decisions.
If I question cogency of this observation, it will be on the footing of sample aggregation method and prejudice. These are the things which may impact the consequence of research.
Question 2 ( 6 Markss )
In what ways are a random sample, convenience sample, and systematic sample different? In what ways are they similar?
When every sample in a population has an equal opportunity of being selected, random sampling technique is normally used because it has limited opportunities of prejudice.
In some researches, particularly in clinical tests, we need specific figure of samples and have to place more expeditiously and fast. Here, convenience sampling is more suited method of trying. We are choosing suited samples from the population harmonizing to our demand. As like the name we are choosing convenient samples.
In systematic sampling, foremost we are choosing one random sample and returns with choosing everyNth sample. It normally used when a sample frame is available. That means, population consists of complete and nonoverlapping samples.
Similarity of these all sampling technique is that all of these are easy to analyses and limited prejudice.
Question 3 ( 4 Markss )
Describe the trying attack you would take if you were asked to find the figure of pregnant adult females non obtaining antenatal attention in your community.
Stratified sampling is more suited technique here because by this technique we can split the population in to different groups based on different factors which affects antenatal attention of adult females. So this helps to cut down the prejudice. From these groups we can take the samples by random, convenience or systematic sampling.
Question 4 ( 5+5+6+6+12+6 Markss )
With mention to the informations on the Honolulu Heart Study, answer the undermentioned inquiries:
 List all the variables represented in the information, and province whether they are qualitative or quantitative variables.
Height, weight, age, instruction degree, physical activity at place, smoking position, blood glucose, serum cholesterin, Sys BP & A ; organic structure mass index are the different variables represented in informations. In this Education degree, Physical activity and smoke position are qualitative variables because these variable can non stand for in Numberss. All others, tallness, weight, age, blood glucose, serum cholesterin, sysBP and organic structure mass index are quatitative variables because all these we can stand for in Numberss.
 For each of the variables listed above, place the measuring graduated table.
Education degree â€“ ordinal graduated table
Smoking position â€“ nominal graduated table
Physical activity â€“ Ordinal graduated table
In ordinal graduated table variables represents in a consecutive increasing or decreasing form and they relate to each other. But in nominal graduated table the variables represent in an independent observation.
Age â€“ intervalBlood sugar â€“ Ratio
Weight â€“ interval Sys BP â€“ Ratio
Height â€“ intervalSerum cholesterin â€“ Ratio
Body mass index â€“ Ratio
In an interval graduated table, there is no absolute nothing. Here age, tallness, weight do non hold the value nothing. But in ratio, the nothing is absolute one.
 Construct a simple frequence tabular array for the undermentioned variables:
a. educational degree
EducationLevel 

Frequency 
Percentage 
Valid Percentage 
Accumulative Percentage 

Valid 
none 
25 
25.0 
25.0 
25.0 
primary 
32 
32.0 
32.0 
57.0 

intermediate 
24 
24.0 
24.0 
81.0 

senior 
9 
9.0 
9.0 
90.0 

proficient 
10 
10.0 
10.0 
100.0 

Entire 
100 
100.0 
100.0 

b. smoke position
SmokingStatus 

Frequency 
Percentage 
Valid Percentage 
Accumulative Percentage 

Valid 
No 
63 
63.0 
63.0 
63.0 
Yes 
37 
37.0 
37.0 
100.0 

Entire 
100 
100.0 
100.0 

 Pull the undermentioned graphs:a. saloon graph for educational degrees
b. piechart for smoking position.
 Fix a frequence tabular array that includes category interval and frequence for the undermentioned variables:a. weight ( use equal category intervals of 10, get downing with 45 )
weightclassinterval 

Frequency 
Percentage 
Valid Percentage 
Accumulative Percentage 

Valid 
1.00 
10 
10.0 
10.0 
10.0 
2.00 
43 
43.0 
43.0 
53.0 

3.00 
34 
34.0 
34.0 
87.0 

4.00 
11 
11.0 
11.0 
98.0 

5.00 
2 
2.0 
2.0 
100.0 

Entire 
100 
100.0 
100.0 

B. ( with category interval get downing with 45 and with a breadth of 10 )
serum cholesterin ( use equal category intervals of 50, get downing with 130 ) .
Serumcholesterolclassinterval 

Frequency 
Percentage 
Valid Percentage 
Accumulative Percentage 

Valid 
1.00 
15 
15.0 
15.0 
15.0 
2.00 
50 
50.0 
50.0 
65.0 

3.00 
28 
28.0 
28.0 
93.0 

4.00 
6 
6.0 
6.0 
99.0 

6.00 
1 
1.0 
1.0 
100.0 

Entire 
100 
100.0 
100.0 

 Pull a histogram for the weight and serum cholesterin based on the frequence tabular arraies created in ( vitamin E ) above.
Question 5 ( 14+3 Markss )
With mention to the informations on the Honolulu Heart Study, answer the undermentioned inquiries:
 Find the mean, average, manner, scope, standard divergence, discrepancy and coefficient of fluctuation for the variables age and blood glucose.
Mean:
Median: It is the inbetween value of informations, that means the centre of informations which it divide in to equal parts.
Manner: The value which shows more often in a information.
Scope: It is the difference between the highest and lowest value.
Standard divergence:
Discrepancy= s^{2}
Coefficient of fluctuation=
Age: 

Mean 
53.67 

Median 
52.5 

Manner 
52 

Scope 
21 

Discrepancy 
26.02131 

South dakota 
5.101109 

Coefficient of fluctuation 
0.09504 

Blood Sugar:

 Is the variable SysBP positively skewed, negatively skewed or symmetric? Justify your reply.
Descriptive Statisticss 

Nitrogen 
Minimum 
Maximum 
Mean 
Std. Deviation 
Lopsidedness 

Statistic 
Statistic 
Statistic 
Statistic 
Statistic 
Statistic 
Std. Mistake 

SysBP 
100 
92.0 
208.0 
130.100 
21.2068 
.884 
.241 
Valid N ( listwise ) 
100 

Here, SysBP is positively skewed because the statistics have a positive figure.
Question 6 ( 1+1+1+1+1+1 Markss )
Suppose a individual is selected indiscriminately from theinformationsof the Honolulu Heart Study. Find the chance that he or she:
 has completed high school or proficient school
Probability of high school P ( A ) = 9/100
Probability of high school P ( A ) = 9/100
Probability of proficient school P ( B ) = 10/100
P ( A or B ) = P ( A ) + P ( B ) – Phosphorus ( A and B ) = 9/100+ 10/100 – 0 = 19/100
 is a tobacco user
P = 37/100
 is physically inactive
P = 49/100
 is a physically inactive tobacco user
P ( A and B ) = 18/100
 has serum cholesterin greater than 250 and systolic blood force per unit area above 130
A = Cholesterol more than 250 = 17/100
B= Systolic blood force per unit area above 130 = 42/100
P ( A and B ) = 9/100
P ( A or B ) = 42/100 + 17/100 – 9/100 = 50/100 = ?
 has a blood glucose degree of 100 or less.
Probability of glucose degree 100 P ( A ) = 1/100
Probability of glucose degree less than 100 P ( B ) = 13/100
P ( A or B ) = P ( A ) +P ( B ) â€“ P ( A and B ) = 1/100+13/100 â€“ 0 = 14/100
P = 14/100
Question 7 ( 1+1+1+1+1+1 Markss )
Using the theinformationsof the Honolulu Heart Study, fix a new frequence tabular array of SysBP for tobacco users and nonsmokers. Use equal category intervals of 20, get downing with 90. Using this new tabular array, allow the events
A = a systolic blood force per unit area of 170 or greater
B = a nonsmoker
C = a tobacco user
Non tobacco users with SysBp & gt ; or = 170
Smoking Status 
Sys BP 

0 
172 

0 
170 

Smokers with SysBP & gt ; or = 170
Smoking Status 
Sys BP 
1 
178 
1 
190 
1 
176 
1 
208 
SmokingStatus 

Frequency 
Percentage 
Valid Percentage 
Accumulative Percentage 

Valid 
No 
63 
63.0 
63.0 
63.0 
Yes 
37 
37.0 
37.0 
100.0 

Entire 
100 
100.0 
100.0 

Sys BP 
178 
190 
176 
208 
172 
170 
Systolic BP & gt ; or = 170
BPclassinterval 

Frequency 
Percentage 
Valid Percentage 
Accumulative Percentage 

Valid 
1.00 
15 
15.0 
15.0 
15.0 
2.00 
39 
39.0 
39.0 
54.0 

3.00 
28 
28.0 
28.0 
82.0 

4.00 
12 
12.0 
12.0 
94.0 

5.00 
4 
4.0 
4.0 
98.0 

6.00 
2 
2.0 
2.0 
100.0 

Entire 
100 
100.0 
100.0 

 P ( A ) = 6/100 = 0.06
 P ( B ) = 63/100 = 0.63
 P ( C ) = 37/100 = 0.37
 P ( A  B ) = P ( A and B ) / P ( B ) = 2/100/ 0.63 = 0.03
 P ( A  C ) = P ( A and C ) / P ( C ) = 4/100/ 0.37 = 0.10
 Compare ( vitamin D ) and ( vitamin E ) , and remark. Are smoking position and blood force per unit area degree independent events?
The above consequences show that the conditional chance is increasing in tobacco users as compared to nonsmokers. That means both these variables are dependent each other.
Question 8 ( 1+2+2 Markss )
Answer the following. In how many ways can
 7 different drugs be arranged in a row?
7! = 7x6x5x4x3x2x1 = 5040
 a combination of 5 different drugs be chosen out of 10 different drugs?
P ( n, R ) = n! / ( nr ) ! = 10! / ( 10 5 ) ! = 30,240
 3 patients be selected from a list of 10 malignant neoplastic disease patients so that the first 1 selected will be treated for chemotherapy ; the 2nd, radiation ; and the 3rd, surgery?
C ( n, R ) = n! / R! ( nr ) ! = 10! /3! ( 103 ) ! = 120
Question 9 ( 3 Markss )
If the cholesterin degree of work forces in a community is usually distributed with a mean of 220 and a standard divergence of 30, what is the chance that a indiscriminately selected sample of 49 work forces will hold a mean between 190 and 205?
Z1 = x Âµ/= 190220/30 = 1.0, Internet Explorer. Area is 0.3413
Z2 = x Âµ/= 205220/30 = 0.50, Internet Explorer, country is 0.1915
Probability between 190 and 205 = 0.34130.1915 = 0.1498 or 15 %
Question 10 ( 6+2 Markss )
Using the the informations of the Honolulu Heart Study, select a random sample of 40 topics and a random sample of 60 topics.
 Calculate the 95 % and the 99 % assurance intervals for the variable SysBP.
60 topics
SysBP 

Frequency 
Percentage 
Valid Percentage 
Accumulative Percentage 

Valid 
94.0 
1 
1.7 
1.7 
1.7 
96.0 
1 
1.7 
1.7 
3.3 

104.0 
1 
1.7 
1.7 
5.0 

106.0 
1 
1.7 
1.7 
6.7 

108.0 
2 
3.3 
3.3 
10.0 

112.0 
4 
6.7 
6.7 
16.7 

114.0 
3 
5.0 
5.0 
21.7 

116.0 
4 
6.7 
6.7 
28.3 

118.0 
3 
5.0 
5.0 
33.3 

120.0 
1 
1.7 
1.7 
35.0 

122.0 
3 
5.0 
5.0 
40.0 

124.0 
1 
1.7 
1.7 
41.7 

128.0 
7 
11.7 
11.7 
53.3 

130.0 
1 
1.7 
1.7 
55.0 

132.0 
2 
3.3 
3.3 
58.3 

134.0 
5 
8.3 
8.3 
66.7 

136.0 
1 
1.7 
1.7 
68.3 

138.0 
2 
3.3 
3.3 
71.7 

140.0 
4 
6.7 
6.7 
78.3 

142.0 
1 
1.7 
1.7 
80.0 

146.0 
1 
1.7 
1.7 
81.7 

152.0 
1 
1.7 
1.7 
83.3 

154.0 
4 
6.7 
6.7 
90.0 

156.0 
1 
1.7 
1.7 
91.7 

162.0 
2 
3.3 
3.3 
95.0 

170.0 
1 
1.7 
1.7 
96.7 

172.0 
1 
1.7 
1.7 
98.3 

178.0 
1 
1.7 
1.7 
100.0 

Entire 
60 
100.0 
100.0 

Mean = 
132.2857 
Standard divergence = 
22.94415 
95 % of CI of Âµ = 132.2 +or 1.96 ( 22.9/) = ( 126.4, 137.9 )
99 % of CI of Âµ = 132.2 +or 2.58 ( 22.9/60 ) = ( 124.5, 139.9 )
40 topics
SysBP 

Frequency 
Percentage 
Valid Percentage 
Accumulative Percentage 

Valid 
94.0 
1 
2.5 
2.5 
2.5 
98.0 
1 
2.5 
2.5 
5.0 

100.0 
1 
2.5 
2.5 
7.5 

102.0 
1 
2.5 
2.5 
10.0 

108.0 
3 
7.5 
7.5 
17.5 

112.0 
1 
2.5 
2.5 
20.0 

114.0 
3 
7.5 
7.5 
27.5 

116.0 
3 
7.5 
7.5 
35.0 

118.0 
1 
2.5 
2.5 
37.5 

120.0 
1 
2.5 
2.5 
40.0 

122.0 
2 
5.0 
5.0 
45.0 

126.0 
1 
2.5 
2.5 
47.5 

128.0 
3 
7.5 
7.5 
55.0 

132.0 
2 
5.0 
5.0 
60.0 

134.0 
2 
5.0 
5.0 
65.0 

138.0 
1 
2.5 
2.5 
67.5 

140.0 
3 
7.5 
7.5 
75.0 

142.0 
1 
2.5 
2.5 
77.5 

144.0 
1 
2.5 
2.5 
80.0 

146.0 
2 
5.0 
5.0 
85.0 

150.0 
1 
2.5 
2.5 
87.5 

154.0 
1 
2.5 
2.5 
90.0 

162.0 
1 
2.5 
2.5 
92.5 

172.0 
1 
2.5 
2.5 
95.0 

176.0 
1 
2.5 
2.5 
97.5 

190.0 
1 
2.5 
2.5 
100.0 

Entire 
40 
100.0 
100.0 

Mean = 
132.2308 
Standard divergence = 
25.07797 
95 % of CI of Âµ = 132.2 +or 1.96 ( 25.07/) = ( 124.4, 139.9 )
99 % of CI of Âµ = 132.2 +or 2.58 ( 25.07/40 ) = ( 120.8, 143.5 )
 What happens to the assurance interval as the sample size additions?
The consequences show that when the sample size increases it generates smaller intervals. So sample size and intervals are reciprocally relative to each other.
Mention: Basic Statisticss for the Health Sciences by Kuzma