Drawing of Random Samples

The dataset contains 8525 samples of various measures based on health maintaining processes and scales of lifestyle, exercise and eating habit. The report involves the analysed data of food habit, exercise level and health quality of randomly selected 2000 variables from 8525 variables.

The researcher had chosen 2000 random variables from 8525 samples by the following method-

• Select “Data” option in the menu-bar and then go to “Select cases”.
• We choose independent variable “ID” and right click on the bullet of random samples of cases.
• Then for drawing random samples, we click on the second bullet. We write 2000 in first box and 8525 in second white box.
• Lastly, we clicked “Continue” in first drop down-box and then “OK” in second drop down-box.
• Our random sample is finally generated.

The variable “SA_Health” refers the self-assessed health status reported by participants.

Self-Assessed Health Status
	Frequency Save Time On Research and Writing Hire a Pro to Write You a 100% Plagiarism-Free Paper. Get My Paper	Percent	Valid Percent	Cumulative Percent
Valid	Excellent	402	20.1	20.1	20.1
Very Good	710	35.5	35.5	55.6
Good	609	30.5	30.5	86.1
Fair	198	9.9	9.9	96.0
Poor	81	4.1	4.1	100.0
Total	2000	100.0	100.0

Statistics
Self-Assessed Health Status
N	Valid	2000
Missing	0
Minimum	1
Maximum	5
Percentiles	25	2.00
50	2.00
75	3.00

Descriptive Statistics
	N	Range	Minimum	Maximum	Mean	Std. Deviation	Variance	Skewness	Kurtosis
Statistic	Statistic	Statistic	Statistic	Statistic	Statistic	Statistic	Statistic	Std. Error	Statistic	Std. Error
Self-Assessed Health Status	2000	4	1	5	2.42	1.043	1.088	.471	.055	-.232	.109
Valid N (listwise)	2000

 The Frequency table of self-assessed Health Status indicate that among 2000 variables, the health status is “Very Good” for 710 (35.5%) participants followed by “Good” for 609 (30.5%) participants. The least frequency of health status is “Poor” for 81 (4.1%) participants. The descriptive statistics indicates that mean of Self Assessed health status is 2.42 and standard deviation is 1.043 (Argyrous, 1997). The mode is “Very Good” (level = 4). The distribution of self-assessed health-status is positively skewed. The Kurtosis value indicates that the distribution is “Leptokurtic”. The bar-plot refers that self-assesses health status has maximum frequency for “Very Good” and minimum frequency for “Poor” according to the height of bars (Data, 1988). The 1^st quartile, median and 3^rd quartile values of the SA_health is 2, 2 and 3. The minimum and maximum values of the SA_health are 1 and 5. The IQR (inter quartile range) is = (3^rd quartile – 1^st quartile) = (3-2) = 1.

The variable “FRUIT” measures the number of serves of fruit eaten per day by respondents.

Number of serves of Fruit per day
	Frequency	Percent	Valid Percent	Cumulative Percent
Valid	0	433	21.7	21.7	21.7
1	623	31.2	31.2	52.8
2	577	28.9	28.9	81.7
3	257	12.9	12.9	94.5
4	79	4.0	4.0	98.5
5	31	1.6	1.6	100.0
Total	2000	100.0	100.0

Statistics
Number of serves of Fruit per day
N	Valid	2000
Missing	0
Minimum	0
Maximum	5
Percentiles	25	1.00
50	1.00
75	2.00

Descriptive Statistics
	N	Range	Minimum	Maximum	Mean	Std. Deviation	Variance	Skewness	Kurtosis
Statistic	Statistic	Statistic	Statistic	Statistic	Statistic	Statistic	Statistic	Std. Error	Statistic	Std. Error
Number of serves of Fruit per day	2000	5	0	5	1.51	1.168	1.364	.593	.055	.027	.109
Valid N (listwise)	2000

Among the six values 0 to 5, most of the people eat 1 fruit per day followed by 2 fruits per day. Only 110 people eat 4 or 5 days per day. A significant number of 433 people do not eat fruit. The mean of the number of serves of fruit per day for 2000 people is 1.51 and standard deviation is 1.168. The distribution of fruits is positively distributed. The mode of the fruits served per day is 1. The number of serves of Fruits per day is plotted in histogram-plot. The highest height indicates that most of the people serve 1 fruit per day and lowest height of refers that minimum number of people eat 5 fruits per day. The minimum and maximum fruit consumption per day is 0 to 5. The 1^st, 2^nd and 3^rd quartile values are respectively 1, 1 and 2. The IQR (inter quartile range) is = (3^rd quartile – 1^st quartile) = (2-1) = 1.

The data set informs the exercise levels from 2011 National Health Survey in UK. It shows that 32% participants have an exercise level considered “sedentary”. Australians might be more active due to better weather conditions and suggest that the percentage of sedentary people in Australia is less than 32%. The variable “Ex_Level” indicates the level of exercise of participants.

Calculation and Analysis

We conduct a binomial test utilizing the “Ex_Level” variable for testing the claim of researchers.

Null Hypothesis (H₀): The proportion of sedentary cases is less than 0.32.

Alternative Hypothesis (H_A): The average score of BMI is greater than or equal to 0.32.

Binomial Test
	Category	N	Observed Prop.	Test Prop.	Exact Sig. (1-tailed)
ABC	Group 1	non-sedentary	1290	.65	.32	.000
Group 2	sedentary	710	.36
Total		2000	1.00

We took 0 as “sedentary” and other variables (1, 2 and 3) as “Non-sedentary” variable.  Then, we transformed “sedentary” as 1 and “Non-sedentary” as 0 for making success probability “1” and failure probability “0” (Chan, 2003). Among 2000 variable, we found 710 sedentary cases and 1290 non-sedentary cases. The observed proportion of sedentary cases is 0.36. Our, hypothetically assumed proportion is 0.32. The calculated significant one-tailed p-value is 0.0. The value being less than 0.05, we reject the null hypothesis of proportion of sedentary cases to be 0.32 at 95% confidence limit.

The variable “BMI” refers the Body Mass Index of the respondent at the time of survey. 2011 National Health Survey shows that the BMI score of UK was 27. Australian Health reports have consistently stressed that Australian obesity rates are getting higher, so the researcher expects that mean BMI score for Australians is higher than 27.

For testing of hypothesis, researcher is interested to conduct one-sample t-test utilizing the “BMI” variable to verify this claim.

Hypothesis:

Null Hypothesis (H₀): The average score of BMI is 27.

Alternative Hypothesis (H_A): The average score of BMI is not equal to 27.

Descriptive Statistics:

Among 2000 variable, 1671 variables contain BMI score. The average BMI is calculated as 27.5553 with the standard deviation 5.83796.

One-Sample Statistics
	N	Mean	Std. Deviation	Std. Error Mean
Body Mass Index	1671	27.5553	5.83796	.14281

One-Sample Test
	Test Value = 27
t	df	Sig. (2-tailed)	Mean Difference	95% Confidence Interval of the Difference
Lower	Upper
Body Mass Index	3.888	1670	.000	.55531	.2752	.8354

Inferential Statistics:

The one-sample t-statistic is 3.888 with degrees of freedom 1670. The p-value of this one sample t-test is 0.0.

Confidential Statistics:

The 95% confidence interval found in the differences of means of BMI indicates that the lower confidence interval is 0.2752 and upper confidence interval is 0.8354.

Conclusion:

Therefore, we reject the null hypothesis. Hence, the mean of BMI score is not equal to 27 with the probability of 95% (Norušis, 2006).

As per report, women are being on a diet. The researcher thought that females eat more fruit and vegetables than males do. For, testing the hypothesis, we conduct an independent samples t-test utilizing two variables “GENDER” and “FRUIT_VEG_COMBINED” to test this claim. Among 2000 people, 1026 females eat fruit and vegetables regularly whereas 974 males eat fruit and vegetables regularly.  

Null Hypothesis (H₀): Females intake equal level of fruit and vegetables as Males

Alternative Hypothesis (H_A): Females intake more fruit and vegetables as Females

Descriptive Statistics:

Among 2000 people, 974 males and 1026 females intake fruit & vegetable combined per day. The average fruit & vegetables for males and females are 3.71 and 4.08.

Group Statistics
	Gender	N	Mean	Std. Deviation	Std. Error Mean
Fruit & Vegetable Intake combined [per day]	Male	974	3.71	2.457	.079
Female	1026	4.08	2.397	.075

Independent Samples Test
	Levene’s Test for Equality of Variances	t-test for Equality of Means
F	Sig.	t	df	Sig. (2-tailed)	Mean Difference	Std. Error Difference	95% Confidence Interval of the Difference
Lower	Upper
Fruit & Vegetable Intake combined [per day]	Equal variances assumed	.256	.613	-3.349	1998	.001	-.363	.109	-.576	-.151
Equal variances not assumed			-3.346	1986.333	.001	-.363	.109	-.577	-.150

The Levene’s test for equality of variances indicates F-statistic = 0.256 with significant p-values 0.613 (Mehta and Patel, 2010). Therefore, we can conclude that variances are not equal for male and female intake level of fruits and vegetables. The t-statistic of equality of means indicate that t-values for equal variance (-3.349) and unequal variance (-3.346). The two-tails significant p-value is 0.001. Hence, we reject the null hypothesis of equality of means of fruit and vegetables intake for both the genders.

The 95% confidence intervals of “fruit and vegetables combine taken per day” with equal variances assumed have interval of differences (-0.576 and -0.151). Assuming unequal variances, the 95% confidence intervals of “fruit and vegetables combine taken per day” have interval (-0.577 and -0.150).

Conclusion:

We can accept the alternative hypothesis that refers female intake more fruits and vegetable than male with 95% probability. The assumption of report of diet of women could be granted as true.

References:

Argyrous, G. (1997). Descriptive statistics on SPSS. In Statistics for Social Research (pp. 60-78). Palgrave, London.

Chan, Y. H. (2003). Biostatistics 102: quantitative data–parametric & non-parametric tests. blood pressure, 140(24.08), 79-00.

Data, C. E. (1988). Basic statistical concepts.

Mehta, C. R., & Patel, N. R. (2010). IBM SPSS exact tests. SPSS Inc., Cambridge, MA.

Norušis, M. J. (2006). SPSS 14.0 guide to data analysis. Upper Saddle River, NJ: Prentice Hall.

Muijs, D. (2010). Doing quantitative research in education with SPSS. Sage