Introduction to Psychological Research Methods Free Essay Example

A community living close to an industrial area has raised concerns about contaminants from industry causing health problems in local families. Based on current research, the public health guidelines state that children should not have blood lead levels above 10 micrograms per decilitre, because elevated lead levels may cause long-term cognitive impairment.

As part of the investigation, parents of all children in the 2- to 6-year-old age range who live within a 10km radius of the industrial area are contacted and strongly encouraged to have their children assessed.

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Lead levels in the children’s blood of are measured. Blood samples are taken by a qualified phlebotomist at children’s homes. A psychologist is employed to assess the children’s level of cognitive development. Each child is assessed individually at the psychologist’s office, using a standardised developmental scale. For each child, their attained score is compared to the score expected for children their age (in years and months), based on norms for the test gathered on a nationally representative sample of 200 children per age group in 2009. A discrepancy score is calculated, which indicates how many months ahead of, or behind, the developmental norm for their age each child is. Positive scores indicate being ahead of same-age peers. Negative scores indicate developmental delay. The table below shows the lead levels (in  g/dL) of the 20 children tested and their developmental discrepancy scores (in months). For statistical calculations, please round your final answers to two (2) decimal places.

What is the independent variable in this study? What is the dependent variable in this study? Explain your answer.

Blood lead levels, age and location distance are the independent variables but in this case the main is blood lead levels this is influenced by the industrial area based on people radius to the location of the industry and age of the children.

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Cognitive development is the dependent it is influenced by the levels of lead levels in blood.

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Complete the following table: Table 1.

Participant	Lead level	Lead 2	Development score	Development 2	Lead x Development
1	10	100	-5	25	-50
2	11	121	-10	100	-110
3	10	100	-10	100	-100
4	11	121	-4	16	-44
5	10	100	-11	121	-110
6	9	81	2	4	18
7	12	144	-15	225	-180
8	10	100	-18	324	-180
9	10	100	-9	81	-90
10	11	121	-26	676	-286
11	6	36	-13	169	-78
12	3	9	-5	25	-15
13	12	144	-10	100	-120
14	13	169	-11	121	-143
15	11	121	-8	64	-88
16	7	49	4	16	28
17	9	81	-2	4	-18
18	17	289	-21	441	-357
19	8	64	3	9	24
20	9	81	5	25	45
	199	2131	/192/	2646	/1930/

Calculate the Mean, Standard deviation and Standard error of the mean for lead levels and development scores.

	Lead level	Development score
Mean	199/20 =9.95	192/20 9.60
Standard deviation ( SD )	√ (2131-1980.05)/19 =2.82	√(2646-1843.2)/19 =6.5
Standard error of the mean ( SE )	2.82/ √20 0.63	6.5/ √20 =1.45

Calculate single sample t to compare the children’s lead levels to the recommended maximum of 10 micrograms per decilitre. Calculate single sample t to compare the children’s developmental discrepancy scores to a mean of 0 (indicating normal rate of cognitive development). Complete the following table. Please show your working on a separate sheet.

	Lead	Development
t obtained	(9.95-10)/0.63 =-0.08	(9.6-10)/1.45 =--0.28
Degrees of freedom (df)	20-1 =19	20-1 =19
t critical (for a 2-tailed test with alpha = .05)	t,19,0.025 =2.09	t,19,0.025 =2.09
Significance test result ( circle one )	Significant / Not significant	Significant / Not significant

Write the results of the t-tests in APA format. For lead the absolute value of t (obtained) =0.08 which is greater than t (critical) =2.09 therefore we reject the null hypothesis and conclude it is not significant.

_______________________________________________________________________ for development lead the absolute value of t (obtained) =0.28 which is greater than t(critical)=2.09 therefore we reject the null hypothesis and conclude it is not significant. ________________________________________________________________________

What conclusions would you feed back to the investigation about the lead levels of these children compared to the recommended maximum, and their level of cognitive development compared to the norm?

________________________________________________________________________ The blood lead levels which are greater than 10  g/dL has no effect on the level of ________________________________________________________________________ cognitive development among children compared to the normal. ________________________________________________________________________

Draw a scatterplot showing the relationship between blood lead levels and developmental discrepancy scores. You may do this by hand on graph paper or using Excel if you prefer.

Calculate Pearson’s r for the association between blood lead levels and developmental discrepancy scores. Complete the following table. Please show your working on a separate sheet.

Pearson’s r obtained	(38600-38208)/ √((42620-39601)*(52920-36864) 392/6962.26 =0.06
Degrees of freedom (df)	20-2 18
Pearson’s r critical (for a 2-tailed test with alpha = .05)	r 18,0.05/2 =0.44
Significance test result (circle one)	Significant / Not significant

Write the result of this Pearson’s r in APA format. A Pearson correlation was run to assess the relationship between blood levels and discrepancy development scores in children aged between 6-10 years. There was low positive association between blood lead level and development discrepancy scores, r(18,0.05/2)=0.44, p-value<0.025 with the blood lead levels explaining 6% of the development discrepancy scores. _______________________________________________________________________

From your calculation of Pearson’s r and your scatterplot, what conclusions would you report back to the investigation about the relationship between children’s blood lead levels and developmental delay? Indicate which specific observations lead you to these conclusions. there is an agreement on both the scatter plot and the Pearson correlation coefficient that

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there is an association between blood lead levels and discrepancy development scores.

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Since the children with blood levels of over 10  g/dL have a negative delay based on other

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Children of there age that is (11,-10), (11,-4), (12,-15), (11,-26), (12,-10), (13,-11), (11,-18)

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And (17,-21) per the scatter plot and the Pearson shows positive association..

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the scatter plot shows that there are more variables to be studied in order to determine the

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True nature of the association since same amount of blood lead levels do not reflect same

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development discrepancy scores that is (11,-10), (11,-4) and (11,-26) for children with blood

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Lead levels of 11  g/dL and considering that some of the children with less than 10  g/dL

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blood lead levels have also the situation of cognitive development that is (3,-5).

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What are the main methodological STRENGTHS of this study? Explain briefly why they are strengths.

______________________________________________________________________________

They are inexpensive and easy to carry out. This will help for the cases where there is inadequate funds, for certainty of other people’s research etc this may be the pinpoint of achieving the correct result as there are no worries of funds or personnel.

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They are used to study rare diseases and diseases that takes time to manifest. A follow up of large group of people is needed for the development of the outcome and since it looks at past exposures of people same case as the scenario for a disease that takes long time to develop.

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The idea for using t-test and Pearson correlation to draw conclusion was ideal as it helps in generating the general overview of the studied variables that is if there is an association which is good in many scenarios.

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L) What are the main methodological WEAKNESSES of this study? Explain briefly why they are weaknesses and how you could improve them.

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This was a case study

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The result might be questionable. it can be hard for people to recall there past this will affect since the point of interest is people with exposure to industrial contaminants and there would not be the certainty of that. It can be improved by finding the people who remembers the past in order to have the certainty.

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Since it relies on the past memories can be biased this may have effect since for the case of children they may not spend lots of the time around the stipulated location radius. This may be avoided by seeking detailed information from their parents of their specific details about there life that may affect the study..

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PSY173 Introduction to Psychological Research Methods

Statistical Analysis Exercise 3

Statistical Formulae

C entral Tendency

Mean:

Variability :

Variance:

S tandard deviation:

Standard error of the mean:

Single-sample t-test

Step 1 : Calculate t obtained

Where   comparison value or mean of comparison population

Step 2 : Use table to find t critical

df = n –1

Select probability (alpha) level (e.g. .05, .01 etc)

Step 3 : Decide whether or not t is significant

If t obtained > t critical reject null hypothesis (H 0 ), difference is significant

If t obtained </= t critical retain null hypothesis (H 0 ), difference is not significant

Pearson’s r (Correlation coefficient)

Step 1 : Calculate r

Raw score formula:

Step 2 : Use table to find r critical

df = n -2

Select probability (alpha) level

Step 3 : Decide whether or not r obtained is significant.

If r obtained > r critical reject null hypothesis (H 0 ), correlation is significant

If r obtained </= r critical retain null hypothesis (H 0 ), correlation is not significant

Introduction to Psychological Research Methods

Lead x Development

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