Steps of quantitative research process
Problem identification. This involves identifying a research problem and developing research questions.
Literature review. Once the problem is identified, the research learns about the topic under investigation. This is done in order to provide foundational knowledge in the topic.
Clarity of the problem. Often the identified problem is broad. Once existing literature is reviewed, the researcher narrows study scope by identifying gaps in the existing literature.
Development of hypothesis and instrumental plan. Based on the gaps in the literature, the researcher develops hypothesis to be tested including quantitative variables to be used. A means of quantifying the variables is then developed.
Data collection. The data is then collected based on the defined instrumentation plan.
Processing data. The collected data is then cleansed for analysis by removing outliers and any other components that do not meet inclusion criteria
Data analysis. The processed data is then subjected to statistical analysis
Findings and conclusion. The analysed data is then interpreted to derive conclusions based on the research questions.
Statistics |
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INFANT DEATH RATE* |
||
N |
Valid |
19 |
Missing |
2 |
|
Mean |
30.00 |
|
Median |
7.00 |
|
Mode |
4 |
|
Skewness |
2.479 |
|
Std. Error of Skewness |
.524 |
|
Kurtosis |
6.995 |
|
Std. Error of Kurtosis |
1.014 |
|
Minimum |
3 |
|
Maximum |
182 |
|
Percentiles |
25 |
4.00 |
50 |
7.00 |
|
75 |
45.00 |
Mean=30.0
Mode=4.0
Median=7.00
The distribution is right skewed since skewness (2.479>+1)
Five number summary
Minimum=3.0
Q1=4.0
Median=7
Q3 =45
Max=182
Tobler’s first law of geography - states that "everything is related to everything else, but near things are more related than distant things.”
Measures of central tendency – this is the measure of central location in a set of data which the measures are mean, median and mode.
Five number summary – this provides a concise summary of the distributions of the observations (minimum, 1st quartile, median, 3rd quartile and maximum).
Box plot –A graphical depiction of a group of numerical data through their quartiles with lines showing variability outside the upper and lower quartiles
Histogram –this is a graph that provide visual interpretation of data by indicating data points it lies within a range of values.
Normal distribution – arrangement in statistical data where most values cluster in the middle of the range with few others at the ends.
Non-normality- data follows specific type of non-normal distribution like bacterial growth which is exponential.
Empirical rule – states that in a normal data set, virtually every piece of data will fall within three standard deviations.
Stratified sample –a sample from dividing population into strata and drawing from them.
Standard normal distribution –normal distribution with a mean of \(0\) and standard deviation of \(1\).
From normal distribution table, , z = -2.3263
Now, -2.3263 = (X-1200)/120
Hence, X = 920.844 hours
From normal distribution table, z = 0
Hence, 0 = (X-1200)/120
X = 1200 hours
From normal distribution table, z = 1.6449
Now, 1.6449 = (X-1200)/120
Hence, X = 1397.388 hours
Correlations |
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Freedom to make life choices |
Social support |
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Freedom to make life choices |
Pearson Correlation |
1 |
.418** |
Sig. (2-tailed) |
.000 |
||
Sum of Squares and Cross-products |
32.392 |
11.125 |
|
Covariance |
.021 |
.007 |
|
N |
1533 |
1526 |
|
Social support |
Pearson Correlation |
.418** |
1 |
Sig. (2-tailed) |
.000 |
||
Sum of Squares and Cross-products |
11.125 |
22.058 |
|
Covariance |
.007 |
.014 |
|
N |
1526 |
1549 |
|
Correlation is significant at the 0.01 level (2-tailed). |
The correlation coefficient =+0.418 indicating a weak positive linear relationship between freedom to make life choices and social support.
One-Sample Statistics |
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N |
Mean |
Std. Deviation |
Std. Error Mean |
|
Number of older siblings |
38 |
1.26 |
1.005 |
.163 |
One-Sample Statistics |
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N |
Mean |
Std. Deviation |
Std. Error Mean |
|
Number of older siblings |
38 |
1.26 |
1.005 |
.163 |
The t-value =1.614
t-critical =1.687
Therefore p>0.05
Hypothesis
Since p>0.05, do not reject the null hypothesis that says that
Group Statistics |
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Gender |
N |
Mean |
Std. Deviation |
Std. Error Mean |
|
Number of older siblings |
female |
28 |
1.18 |
.905 |
.171 |
male |
10 |
1.50 |
1.269 |
.401 |
Independent Samples Test |
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Levene's Test for Equality of Variances |
t-test for Equality of Means |
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F |
Sig. |
t |
df |
Sig. (2-tailed) |
Mean Difference |
Std. Error Difference |
95% Confidence Interval of the Difference |
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Lower |
Upper |
|||||||||
Number of older siblings |
Equal variances assumed |
1.657 |
.206 |
-.865 |
36 |
.393 |
-.321 |
.371 |
-1.075 |
.432 |
Equal variances not assumed |
-.737 |
12.427 |
.475 |
-.321 |
.436 |
-1.268 |
.626 |