# Sociology Statistical Assessment Essay A+
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ASSESSMENT 1

Question 1 ( 5 Markss )

Statisticss show that alumnuss in New Zealand make \$ 250,000 more in their life-time than non-graduates. 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, statistician’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 non-overlapping 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:

1. 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.

1. For each of the variables listed above, place the measuring graduated table.

Education degree – ordinal graduated table

Smoking position – nominal graduated table

Physical activity – 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 – intervalBlood sugar – Ratio

Weight – interval Sys BP – Ratio

Height – intervalSerum cholesterin – Ratio

Body mass index – 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.

1. 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
1. Pull the undermentioned graphs:a. saloon graph for educational degrees

b. pie-chart for smoking position. 1. 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
1. 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:

1. 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 in-between 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= s2

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:

 Mean 152.14 Median 143.5 Manner 161 Scope 384 Discrepancy 2998.202 South dakota 54.75584 Coefficient of fluctuation 0.3599
1. 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:

1. 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

1. is a tobacco user

P = 37/100

1. is physically inactive

P = 49/100

1. is a physically inactive tobacco user

P ( A and B ) = 18/100

1. 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 = ?

1. 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 ) – P ( A and B ) = 1/100+13/100 – 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
1. P ( A ) = 6/100 = 0.06
2. P ( B ) = 63/100 = 0.63
3. P ( C ) = 37/100 = 0.37
4. P ( A | B ) = P ( A and B ) / P ( B ) = 2/100/ 0.63 = 0.03
5. P ( A | C ) = P ( A and C ) / P ( C ) = 4/100/ 0.37 = 0.10
6. 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

1. 7 different drugs be arranged in a row?

7! = 7x6x5x4x3x2x1 = 5040

1. a combination of 5 different drugs be chosen out of 10 different drugs?

P ( n, R ) = n! / ( n-r ) ! = 10! / ( 10- 5 ) ! = 30,240

1. 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! ( n-r ) ! = 10! /3! ( 10-3 ) ! = 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- µ/= 190-220/30 = -1.0, Internet Explorer. Area is 0.3413

Z2 = x- µ/= 205-220/30 = 0.50, Internet Explorer, country is 0.1915

Probability between 190 and 205 = 0.3413-0.1915 = 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.

1. 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 = 132.286 Standard divergence = 22.9442

95 % of CI of µ = 132.2 +or- 1.96 ( 22.9/) = ( 126.4, 137.9 )

99 % of CI of µ = 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 = 132.231 Standard divergence = 25.078

95 % of CI of µ = 132.2 +or- 1.96 ( 25.07/ ) = ( 124.4, 139.9 )

99 % of CI of µ = 132.2 +or- 2.58 ( 25.07/40 ) = ( 120.8, 143.5 )

1. 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