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STAT2009 Research Methods in The Social Sciences

Added on - 13 Dec 2021

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STATISTICS
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Question 1
a) The relevant frequency table is as indicated below.
It is apparent that 84.2% of the respondents live in England.
b) The requisite graph is indicated below.
Based on the above graph, it is apparent that the shape of the above graph is asymmetric
which indicates to the presence of skew. Considering that there are some individuals with
exceptionally low life satisfaction, hence negative skew is present here. Owing to the
presence of skew, it would be fair to conclude that the above distribution is non-normal. Also,
it may be concluded that the mean life satisfaction score of the sample is 7.6 which indicates
high degree of satisfaction. Also, the variance seems to be on the lower end with majority of
individuals having satisfaction level higher than 7.
c) The requisite hypotheses to be tested are stated below.
Null Hypothesis: μ2015= 7.2
Alternative Hypothesis: μ2015> 7.2
Considering that the population standard deviation is not known, hence the relevant test
statistic would be T. Further the test would be right tail one sample test. It has been assumed
that the level of significance is 5%. Based on the requisite input, the relevant output obtained
from SPSS is stated below.
Based on the above output, it may be fair to conclude that the p value (0.00) is lower than the
assumed significance level. As a result, the available evidence is sufficient to cause rejection
of null hypothesis and acceptance of alternative hypothesis. Hence, it may be concluded that
the mean satisfaction levels in 2015 is greater than the level of 7.2 witnessed in 2010.
d) The relevant output for the average satisfaction levels of the two genders is shown below.
From the above table, it is apparent that the mean satisfaction level of males is 7.53 in
comparison to 7.66 for females.The confidence interval can be computed from the above
table.
Higher level of 95% confidence level for male satisfaction level = 7.53 + 1.96*0.047 =7.62
Lower level of 95% confidence level for male satisfaction level = 7.53 – 1.96*0.047 = 7.44
Higher level of 95% confidence level for female satisfaction level = 7.66 + 1.96*0.04 = 7.74
Lower level of 95% confidence level for female satisfaction level = 7.66 – 1.96*0.04 = 7.58
The above confidence intervals highlight that with 95% it can be predicted that the population
average life satisfaction score would lie in the interval.
e) The requisite output from SPSS for the life satisfaction of various marital status is
indicated below.
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