The Essay On Importance Of Questionnaire Method, Simple Random Sampling, And Data Analysis In Surveying And Analyzing Data.

Question (1a)

Question (1a)

Questionnaire method of survey could be used for 100 sample of students because the method is economical to obtain quantitative data for analysis and predictions. The questionnaire method of survey is preferred for this study in order to reach the sample of 100 students easily within a short period of time, obtain first and effective answers based on individuals.

Question (1b)

Simple random sampling. The research targets students as the respondents, whose population are available (Aladag, 2015). Simple random sampling gives every respondent an equal opportunity to be chosen, therefore, tries to eliminate the bias. Due to this reasons, simple random sampling is preferred method to be used.

Question (1c)

Based on the given data, we have two variables namely; preparation time and marks.  Our independent variable for this kind of data is preparation time while students’ marks is our dependent variable.

Student’s marks is a variable we are interested in and which is believed to be affected by the preparation time.  According to this study, preparation time is a variable we believe to affect the marks. Based on the above reason, we conclude that students’ marks and preparation time is our dependent and independent variable respectively.

Both the preparation time and the students’ marks are quantitative data since the data is presented numerically.

Question (1d)

The method requires the population to be grouped (Saeed, 2017), therefore, doing so, requires time.

Our population sample is 100, which is not large enough to draw a better conclusion; this presents a great challenge.

Class interval	Frequency	Relative frequency	Cumulative relative frequency
20-29	1	0.01	0.01
30-39	8	0.08	0.09
40-49	16	0.16	0.25
50-59	20	0.20	0.45
60-69	20	0.20	0.65
70-79	17	0.17	0.82
80-89	12	0.12	0.94
90-100	6	0.06	1

Frequency Histogram for preparation of time

The frequency histogram for the preparation of time is displaying a normal distribution (Khan, 2015) with most of the data scatter around the mean. This reveals that the data follow a normal distribution

Relative Frequency Histogram for the preparation of time

The distribution of the data as per the relative frequency histogram above shows that the data is negatively skewed.  The data is skewed towards to the left.

Cumulative Relative Frequency Histogram for the preparation of time

The cumulative relative frequency histogram increases as the class interval increases.

DISTRIBUTION TABLE FOR MARKS

Class interval	Frequency	relative frequency	Cumulative frequency
20-29	1	0.01	0.01
30-39	5	0.05	0.06
40-49	10	0.10	0.16
50-59	17	0.17	0.33
60-69	20	0.20	0.54
70-79	22	0.22	0.76
80-89	14	0.14	0.90
90-100	10	0.10	1.00

Frequency Histogram the Marks

The frequency histogram for the marks is non-symmetrical (Stoykov, 2013) with most of the data skewed towards the left. This reveals that the data do not a follow the normal distribution.

Relative Frequency Histogram for the Marks

The distribution of the students’ marks as per the relative frequency histogram above shows that the students’ marks is negatively skewed.  The data is skewed towards to the left.

Question (1b)

Cumulative Relative Frequency Histogram for the Marks

The cumulative relative frequency histogram increases as the class interval increases

Question (1f)

I identified the preparation time to be the cause of the students’ marks increase or decrease and therefore becomes our x axis. The students’ marks experience the effect, this becomes our Y axis.

Question (1g)

Regression Statistics
Multiple R	0.546556431
R Square	0.298723932
Adjusted R Square	0.291568054
Standard Error	14.65409389
Observations	100

ANOVA
	Df	SS	MS	F	Significance F
Regression	1	8964.478169	8964.478169	41.7452508	4.04091E-09
Residual	98	21044.76183	214.7424677
Total	99	30009.24

	Coefficients	Standard Error	t Stat	P-value	Lower 95%	Upper 95%
Intercept	28.98427749	5.874519858	4.933897271	3.2994E-06	17.32648402	40.64207096
X Variable 1	0.583053974	0.090241275	6.461056477	4.04091E-09	0.403973101	0.762134847

Model Summary
Model	R	R Square	Adjusted R Square	Std. Error of the Estimate
1	.547a	.299	.292	14.65409
a. Predictors: (Constant), preparation time

Y= 28.984 + 0.5831 * x

Marks = 28.984 + 0.5831 * Preparation Time in hours

An increase of Preparation time by one, results to change in marks as follows;

Marks = 28.984 + 0.5831 * 1

=28.984 + 0.5831

= 29.567

Question (1h)

Statistics
		Preparation time	Marks
N	Valid	100	100
	Missing	0	0
Mean		63.0400	65.7400
Median		64.0000	68.0000
Mode		64.00	70.00
Std. Deviation		16.32060	17.41045
Variance		266.362	303.124
Range		65.00	75.00
Minimum		25.00	25.00
Maximum		90.00	100.00
Percentiles	25	49.0000	54.0000
	30	54.0000	58.0000
	50	64.0000	68.0000
	75	76.7500	78.0000

Question (1i)

Correlation

Descriptive Statistics
	Mean	Std. Deviation	N
Preparation time	63.0400	16.32060	100
marks	65.7400	17.41045	100

Correlations
		Preparation time	marks
Preparation time	Pearson Correlation	1	.547**
	Sig. (2-tailed)		.000
	Sum of Squares and Cross-products	26369.840	15375.040
	Covariance	266.362	155.303
	N	100	100
marks	Pearson Correlation	.547**	1
	Sig. (2-tailed)	.000
	Sum of Squares and Cross-products	15375.040	30009.240
	Covariance	155.303	303.124
	N	100	100
**. Correlation is significant at the 0.01 level (2-tailed).

The correlation of preparation time and marks is r =0.547

Therefore, preparation time and marks show a statistically significant relationship since the Pearson Correlation (r =0.547) is statistically significant (p< 0.0, for a 2- tailed test).

The direction of marks and preparation time is positive, meaning that both variables (dependent and independent move in the same direction. An increase of the preparation time by one unit will result in a corresponding increase of the marks (Stergiou, 2015).  A decrease of the preparation time by one unit will result in a corresponding decrease of the marks by one unit.

We can, therefore, approximate the strength or the magnitude of the association by;

(.3 < | r | < .5)

Question (2a)

Standard error tells the extent to which sample mean deviates from the population mean. The small the standard error, the close the sample mean to the population mean

Question (2b)

 Our coefficient of determination is 0.2672

It tells us that about 26.72% of the dependent variable (y) source of variation is being explained by the independent variable (x1 and x2).

Question (2c)

The adjusted R Square = 0.2635.

The adjusted R Square (0.2635) is the proportion of the overall variance that is explained by the model. Therefore, it is an improved version of R Square (0.2672) after the variables which are statistically insignificant have been removed (excluded in the model) (Suslov Mark Yur’evich, 2015).

Both the adjusted R Square and adjusted R Square measures the fitness of the model. The lower the values of the adjusted R Square, the better the fitness of the model.

Question (2d)

F-test measures the utility of the model. Therefore, we postulate a hypothesis that

Ho: The intercept fitness of the model and the model are equal

H1: The intercept fitness of the model is reduced than the model

Question (1c)

The test will proceed as follows;

The significance F is 0.000 and the p-value = 0.05. Since the p-value is greater than the significance F, we do not reject the null hypothesis and conclude that the intercept fitness of the model and the model are equal.

Question (2e)

The coefficients presented in the output are Intercept = 93.8993, X1 = 0.4849, X2 = -0.0229

The intercept (p-value = 0.0000) is less than 0.05; revealing that the intercept is statistically significant. The p-value for the x1 variable is 0.0000 which is less than 0.05; revealing that the x1 variable is statistically significant.

The x2 variable is 0.5615which is greater than 0.05; revealing that the x2 variable is not statistically significant (Laber, 2017).

The coefficient implies that y variable is predicted by the change of the coefficient as follows

Y = 93.8993 + 0.4849 * x1 – 0.0229 * x2

The coefficients are interpreted as follows;

An increase in one unit of the x1 variable will result to the corresponding increase in y variable by 0.4849 units and an increase in one unit of the x2 variable will result to the corresponding decrease in y variable by 0.0229 units. The variable x2 is negatively correlated to y, as implied by the negative sign

Question (2f)

Yes, the heights of the sons and the fathers are linearly related. This is justified by the positive correlation as shown by the x1 intercept

Question (2g)

No, the heights of the sons and the mothers are not linearly related. This is justified by the negative correlation as shown by the x2 intercept.

References

Aladag, S. C. (2015). Improvement in Estimating the Population Median in Simple Random Sampling and Stratified Random Sampling Using Auxiliary Information. Communications in Statistics - Theory and Methods, 44(5), 78-86.

Khan, M. F. (2015). Image contrast enhancement using normalized histogram equalization. Optik - International Journal for Light and Electron Optics, 126(24), 72-81.

Laber, E. B. (2017). Statistical Significance and the Dichotomization of Evidence: The Relevance of the ASA Statement on Statistical Significance and p-Values for Statisticians. Journal of the American Statistical Association, 112(519), 124-134.

Saeed, A. D. (2017, 08 21-25). Proceedings of the Conference of the ACM Special Interest Group on Data Communication - SIGCOMM '17 - Carousel. ACM Press the Conference of the ACM Special Interest Group - Los Angeles, CA, USA (2017.08.21-2017.08.25, pp. 1-13. doi:10.1145/3098822.3098852

Stergiou, C. (2015). Explaining Correlations by Partitions. Foundations of Physics D, 45(12), 45-56.

Stoykov, S. R. (2013). Non-linear vibrations of beams with non-symmetrical cross sections. International Journal of Non-Linear Mechanics, 10(55), 112-119.

Suslov Mark Yur’evich, T. I. (2015). Ordinary least squares and currency exchange rate. International scientific review, 2(3), 22-33.