Apply Correlation, Causation, Predication, Confidence, and Errors

Correlation, Causation, Prediction, Confidence, and Errors
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Correlation, Causation, Prediction, Confidence, and Errors
Chapter Seven
Problem 1
The graphic representation of a scatter diagram can help in indicating the relationship between two different variables (Bennett, Briggs, & Triola, 2014).

The scatter diagram illustrates a case of positive correlation since as the explanatory variable X increases, the response variable Y also increases. Additionally, as the X variable decreases, the Y variable decreases. Even though the two variables are positively correlated, there is one outlier entry at point (6.00, 22.5).
Problem 2

In this case, the relationship between the X and the Y variables reveals a situation of negative correlation due to the downward trend. The values of the variables move in opposite directions since as the value of X increases, that of Y decreases. Conversely, as the value of X decreases, the value of Y increases. The outlier entry, in this case, occurs at point (8.00, 10.00).
Problem 3
(Data attached in the Excel file)

The scatter diagram reveals that the two variables (mean daily calories and infant mortality rate) have a strong negative correlation. Mean daily calories are negatively related to infant mortality rate, that is, as mean daily calories increases, there is a reduction in the infant mortality rate. Also, a decrease in infant mortality rate can be noted as the mean daily calories intake increases.

The data entry for both variables has outliers at point (3429, 44) and (2671, 7).

From the data, the correlation coefficient is r = -0.901784376. Since the Pearson’s coefficient lies between -1 and -0.5, the two variables have a strong negative correlation. The Pearson correlation coefficient of -0.901784376 is significant and confirms the graphical outlook which indicates a strong negative relationship between mean daily calories and infant mortality.

From the data analysis, the P-value is 0.000361. Since this value is less than 0.5, the correlation is indeed statistically significant (Bennett, Briggs, & Triola, 2014). It means that the relationship between the two variables is caused by some external factors rather than random chance. Daily calories constitute 81% of the variance that is experienced in infant mortality. The figure is highly significant.
Problem 4
When carrying out an analysis concerning correlation between different variables, the Pearson’s coefficient (r) always lies in between -1 and 1 (Bennett, Briggs, & Triola, 2014). Since the value given by Bill is slightly higher than the acceptable limit, it is possible that he encountered some errors in his analysis and the results are not accurate.
Problem 5
In determining how variables are correlated, we consider the correlation coefficients that lie between -1 and 1. The extent to which variables are related can be determined by how close the correlation coefficient lies between -1 and 1. As the figure moves towards the midpoint (zero), the correlation becomes weaker (Bennett, Briggs, & Triola, 2014). In Judy’s case, the negative sign of the Pearson’s coefficient only indicates the type of relationship that exists between the variables she used in her analysis. The sign does not ascertain that the relationship is bad but rather, it shows that there is a negative correlation between the variables. Therefore, the sign is just an indication of the nature of the correlation between variables under study.
Problem 6
Correlation does not imply causality
Putting into consideration the correlation that exists between variables, we consider the extent to which a given variable increases or decreases in value as the other corresponding variable also increases or decreases (Bennett, Briggs, & Triola, 2014). On the contrary, correlation does not show if indeed the behavior of the first variable causes that of the second variable of if the behavior of the second variable causes that of the first. Also, correlation does not indicate if another external variable determines the behavior of both variables. Therefore, correlation does not imply causality.
Problem 7

Using the best line of fit for the data, when the quantity of product X is at 150, the value of Y is 250. Therefore, the predicted price at 150 units is 250.
On the other hand, when X is 100, the value of Y is 175. Therefore, the predicted price for the product X at this point is 175.
Problem 8
(Data attached in Excel file)

In this case, R=0.984274791. The value of the R indicates that there exists a strong relationship between the two variables under comparison, that is, GPA and ACT scores.
From the above results, the gradient is determined to be 13.763. On the other hand, the intercept on the Y axis is -19.336. Using the linear equation, we have
YACT=13.76XGPA-19.34In case Y=20, then we have
20=13.76X-19.3413.76X=20+19.34X=39.3413.76=2.859If Y=25, then
25=13.76X-19.3413.76X=25+19.34X=44.3413.76=3.222Y=34, then
34=13.76X-19.3413.76X=34+19.34X=53.3413.76=3.876Chapter Eight
Problem 1
The mean study time is given as ∑f(x)n . In this case, n=9 therefore we have
(18+7+10+13+12+16+5+20+21)9=1229=13.56
When the value is rounded off to the nearest tenth of an hour we have the mean study time being equal to 13.6
Problem 2
Sample mean = 63
Population mean = 64.6
Standard deviation = 1.9
The Z-Score for the data is given as
Z=Sample mean-Population meanStandard deviationSubstituting we obtain,
Z=63-64.61.9=-1.61.9=-0.8421Problem 3
The mean of the sample distribution is the same as that of the population. Since there is only a single sample, the best approach to estimate the population proportion is to use the sample proportion. The percentage without doctorates becomes
p=(110-66)110=410=0.4 or 40%Given the population N=349, to obtain the number of teachers who do not have doctorates we multiply N*p=349×4100=139.6~140 Teachers have no doctorates.
Problem 4
The formula for finding the marginal error is given as E=2S√n . Assuming a confidence interval of 95%, the Z-Score value is ±1.96 (Bennett, Briggs, & Triola, 2014). Since the standard deviation (S=16) and the sample size (n=400), the margin of error becomes
E=2*16√400=3220=1.6Problem 5
The estimate for the mean Mathematics ACT score can be obtained from the formula
M±(2×Standard devation)√nSince the population, mean has a confidence interval of 95%
M=28
Standard deviation = 4
n = 64
Substituting the values, we have
M±(2×Standard devation)√n=28±2*4√64=28±88=28±1It implies that the mean score on the ACT mathematics test lies between 27 and 29.
Problem 6
To estimate the minimum sample size, we use the formula
n≈(z×Standard deviationE)2Where z represents the Z-score (z = 1.96 at 95% confidence interval) and E is the marginal error (E = 10,000)
Substituting the values in the formula
n≈z×Standard deviationE2=1.96×70000100002=13.722=188.2384≈188 homesProblem 7
Since we are using a sample, the marginal error formula becomes
E=z×p1-pnThe confidence interval is 95% therefore, z = 1.96.
P = 0.33
n = 490
Substituting the values into the formula, we have
E=z×p1-pn=1.96×0.331-0.33490 =1.96×0.331-0.33490=1.96*0.2211490=1.96*0.000451224=1.96*0.021242=0.041634≈4.2%
Therefore, the marginal error is 4.2%.
References
Bennett, J. O., Briggs, W. L., & Triola, M. F. (2014). Statistical reasoning for everyday life.