Statistics

Gross domestic product (GDP) refers to the market value of all final goods and services produced within a country over a given period of time. One of the ways it can be measured is by the expenditure approach: GDP = private consumption (C) + gross investment (I) + government spending (G) + exports (X) – imports (M). This immediately illustrates that investment, defined as an increase in capital stock (gross fixed capital formation), affects the level of GDP, all else held constant. Conversely, businesses tend to invest if they predict growth in GDP and hence potential profits. This positive relationship is confirmed by the positive coefficients of the variables and both growth rates tend to have matching signs.

Performing a regression enables us to find numerical values for theoretical parameters. The default method for estimating the parameters of an equation is ‘least squares’. The overall fit of the regression line is measured by R2, which calculates how close the points are to the estimated regression line (of best fit) in the scatter plot. Since a maximum value of 1 indicates a perfect fit and a value closer to 0 indicates no apparent relationship, a value of 0.76 suggests a reasonably good fit. However, macroeconomic theory explains that there are several other important factors which can influence real GDP, such as interest rates and savings, which have not been included in the simple regression.

This may engender autocorrelation, which is more likely to arise when utilising time series for regression analysis. Serial correlation, an equivalent term, occurs when the disturbance term picks up ‘outside influence’ from variables and can persist over time. The Durbin-Watson (DW) test computes this on a scale from 0 to 4, where positive correlation is closer to 0 and negative correlation closer to 4. A value closer to 2 means there is no correlation. In this instance, the DW stat is 1.52, exhibiting slightly positive correlation, though it is also above the upper bound value and hence it is insignificant.

The small positive correlation can be seen on the residual graphs which tend to display small patterns. In general, autocorrelation does not indicate that the OLS coefficient estimates are biased, although BLUE is no longer valid since one of the Gauss-Markov assumptions is no longer satisfied. However, if the serial correlation is due to variables being incorrectly omitted from the model then bias is likely. If there is positive serial correlation then it is very likely that the OLS estimates of the coefficient standard errors will be biased downwards. This will mean that the t-ratios are likely to be too high.

In order to carry out certain tests we need to work out the number of degrees of freedom (df), that is, the number of observations in the sample minus the number of parameters estimated (the number of rows in the middle panel). In our example, df= 43-2 =41. By observing the t-statistic against the critical values, we can see that it is large. Therefore we could reduce the risk of a Type I error (of falsely rejecting the null hypothesis) to 0.1 percent by using the 0.1 percent significant level.

The lower the significance level, the higher the power of the test will be. Given that the t-statistic is larger than the given critical value, the estimated coefficient is said to be ‘statistically significant’ (i.e. coefficient is nonzero). The p-value (Prob), on the other hand, is not literally 0.0000 as it has been rounded to 4 decimal places which demonstrates that its magnitude is negligible.

One of the ways of working out which is a good estimator is to assess heteroscedasticity. This means ‘differing dispersion’ and occurs when the variance of the disturbance terms is not the same for all observations. Its presence makes the OLS estimators inefficient because, in principle, we could find alternative estimators with smaller variances. Similarly to autocorrelation, the estimators of the standard errors of the regression coefficients will be wrong.

They are computed on the assumption that the distribution term is homoscedastic. Thus there will be bias and accordingly, the t and F tests are invalid.1 It is likely that the standard errors will be underestimated, so the t statistics will be overestimated and the precision of the regression coefficients will be deceptive. In order to see if our regression is affected, we perform a White test. It looks more generally for evidence of an association between variance of the disturbance term and the regressors. We can use R2 as a proxy. Since its value of 0.12 is far less than the 5% significance level even at 30 degrees of freedom, we can assume there is no heteroscedasticity. However, this form of testing lacks power because of its generality.

As a preface, Waite progresses through various statistics fostering today’s pattern of decreased marriages. She states clearly that, “The decline in marriage is directly connected to the rise in cohabitation-living with someone in a sexual relationship without being married. ” Statistics showed a vast decrease in marriages between both black and white marriages. This seems to be an epidemic in today’s society providing examples which might reflect people or situations In our lives.

Another one, of the many, shocking statistics show that about “one third” of births occur outside of wedlock. Waiter’s worry is that marriages are statistically more beneficial to the children who are conceived and born with a stable set of parents. The first argument poses a stance that health between a family is of greater quality when the family Is complete. That being said, Waite never quite defined what a complete family Is, but the term Is connotatively a derived of a mother, father, and could possibly consist of children.

From the paper, a family Is anything Inside matrimony. Waiter’s first argument for health is that marriage appears to reduce risky and unhealthy behaviors. Marriage will also increases material well-being such as income, assets, or wealth. The last component, which I think is the most beneficial, is moral support. All of these ideals are intricate measures needed for a healthy life-all of which stem and are heightened through marriage. The second argument for a pro marriage lifestyle is that life Is easier financially through a stable Income.

Incomes can either be shared or enhanced with marriage. Waite argues that single parent households are in Jeopardy due to lack of sufficient funds and energy where dual parent homes are either amplified by double income or saved by production at home. Wives tend to lead the stay at home Job of doing the duties with house work while men go to work. This leaves more time to pay attention to family as well as diligence In their career while the wife rears the kids and does house work; leaving much more time to relax and recoup.

The third argument supported by Waiter’s article supports a greater intimacy between a couple in marriage. When people think of the word intimacy, it is attached (generally) to sex. Waite argues that not only are sexual needs exceeded, but emotionally the bond is wound tighter. “The long term contract implicit in marriage- which is not implicit in cohabitation- facilitates emotional investment in the relationship, which should affect both frequency of and needs are met. The final argument is the impact of marriage on the children.

Statistics show that two times as many children that are raised in one-parent families than children from two-parent families drop out of high school. A startling fact but is upheld to be true. Almost all cases of poverty were recorded by cases of children growing up in single parent homes. It summates that children are superbly affected by the role models which are designated in their lives. The last page of the article persuades the reader to “reverse the trend” and all of the casualties invested by overdeveloped monogamous relationships.

Multiple sources are accredited with the foundation of a rubber standard. The ideals in society fluctuate because of public policy and acceptance. Policies and standards must be enforced throughout society to rectify change. In summate, Waite argues that a positive lifestyle is that of which is inside of marriage. She resolves that “marriage produces individuals who drink less, smoke less, abuse substances less, live longer, earn more, are wealthier, and have children who do better- need to give more thought and effort to supporting this valuable social institution. “

Instructions

Use this page as your cover page, and attach your group work behind your work. Your assignment answers should be incomplete and grammatically correct sentences.

1. According to the National Center for Health Statistics (2004), 22. 4% of adults are smokers. A random sample of 300 adults is obtained.

• (a) Describe the sampling distribution of phat, the sample proportion of adults who smoke.
• (b) In a random sample of 300 adults, what is the probability that at least 50 are smokers?
• c) Would it be unusual if a random sample of 300 adults results in 18% or less being smokers? Explain your answer.

2. A machine at K&A Tube & Manufacturing Company produces a certain copper tubing component in a refrigeration unit. The tubing components produced by the manufacturer have a mean diameter of 0. 75 inches with a standard deviation of 0. 004 inches. The quality-control inspector takes a random sample of 30 components once each week and calculates the mean diameter of these components. If the mean is either less than 0. 748 inches or greater than 0. 752 inches, the inspector concludes that the machine needs an adjustment.

• a) Describe the sampling distribution of the sample means diameter for a random sample of 30 such components.
• (b) What is the probability that, based on a random sample of 30 such components, the inspector will conclude that the machine needs an adjustment when, in fact, the machine is correctly calibrated?

3. In a random sample of 678 adult males 20 to 34 years of age, it was determined that 58 of them have hypertension (high blood pressure). Source: The Centers for Disease Control.

• (a) Obtain a point estimate for the proportion of adult males 20 to 34 years of age who have hypertension.
• (b) Construct a 95% confidence interval for the proportion of adult males 20 to 34 who have hypertension. Interpret the confidence interval. You wish to conduct your own study to determine the proportion of adult males 20 to 34 years old who have hypertension.
• (c) What sample size would be needed for the estimate to be within 3 percentage points (interval length is 0. 06) with 95% confidence if you use the point estimate obtained in part (a)?
• (d) What sample size would be needed for the estimate to be within 3 percentage points with 95% confidence if you don’t have a prior estimate (use phat=. )?

4. A random sample of 60 married couples who have been married 7 years was asked the number of children they have.

• (a) Obtain a point estimate for the mean and standard deviation number of children of all couples who have been married 7 years.
• (b) What is the shape of the distribution of the sample mean? Why?
• (c) Compute a 95% confidence interval for the mean number of children of all couples who have been married 7 years. Interpret this interval.
• (d) Compute a 99% confidence interval for the mean number of children of all couples who have been married 7 years. Interpret this interval.
• (e) You wish to conduct your own study to determine the mean number of children of all couples who have been married for 7 years. What sample size would be needed for the estimate to be within 0. 25 (interval length is 0. 5) with 99% confidence if you use the point estimates obtained in part (a)?

QSO 510: Module 2 Homework

Notes:

1. Before doing this assignment, do the practice problem posted under Apply and Discover.
2. Word-process your solutions within this template. Do not create a new file.
3. Show all steps used in arriving at the final answers. Incomplete solutions will receive partial credit.
4. Word-process formulas using Equation Editor and diagrams using Drawing Tool.

Problem 1
If many samples of size 15 (that is, each sample consists of 15 items) were taken from a large normal population with a mean of 18 and a variance of 5, what would be the mean, variance, standard deviation, and shape of the distribution of sample means? Give reasons for your answers. Note: Variance is the square of the standard deviation. Adapted from Statistics for Management and Economics by Watson, Billingsley, Croft, and Huntsberger. Fifth Edition. Chapter 7 Page 308. Allyn and Bacon. 1993x = 18  2x=5 shape = normal
Therefore x= 2x = 5 = 2.2360
If a large number of samples, each of size 15, are selected from the population and a sample mean is selected computed for each sample, the mean, variance, and shape of the distribution of sample means would be as follows: x= 18. The standard deviation of the sample means x can be computed as given below: x = X = 2.2360 = 0.5773 n 15.  Variance is the square of the standard deviation. Therefore, the variance of the sample means ²x can be computed as follows:

²x = ²x = 5 = 0.3333 n 15.

The shape of the distribution of sample means = normal since sample size n 30.

Problem 2
If many samples of size 100 (that is, each sample consists of 100 items) were taken from a large non-normal population with a mean of 10 and a variance of 16, what would be the mean, variance, standard deviation, and shape of the distribution of sample means? Give reasons for your answers. Note: Variance is the square of the standard deviation. Adapted from Statistics for Management and Economics by Watson, Billingsley, Croft, and Huntsberger. Fifth Edition. Chapter 7 Page 308. Allyn and Bacon.

1993 x = 10 2x=16 shape =non- normal
Therefore x= 2x = 16 = 4.
If a large number of samples, each of size 100, are selected from the population and a sample mean is selected computed for each sample, the mean, variance, and shape of the distribution of sample means would be as follows: x= 10
The standard deviation of sample means x can be computed as given below:

x = X = 4 = 0.4 n 100.

Variance is the square of the standard deviation. Therefore, the variance of the sample means ?²x? can be computed as follows:

²x = ²x = 16 = 0.16 n 100.

The shape of the distribution of sample means = normal since sample size n 30.

Problem 3
Time lost due to employee absenteeism is an important problem for many companies. The human resources department of Western Electronics has studied the distribution of time lost due to absenteeism by individual employees. During a one-year period, the department found a mean of 21 days and a standard deviation of 10 days based on data for all the employees.

• a) If you pick an employee at random, what is the probability that the number of absences for this one employee would exceed 25 days?
• b) If many samples of 36 employees each are taken and sample means computed, distribution of sample means would result. What would be the mean, standard deviation, and shape of the distribution of sample means for samples of size 36? Give reasons for your answers.
• c) A group of 36 employees is selected at random to participate in a program that allows a flexible work schedule, which the human resources department hopes will decrease employee absenteeism in the future. What is the probability that the mean for the sample of 36 employees randomly selected for the study would exceed 25 days?

Source: Statistics for Management and Economics by Watson, Billingsley, Croft and Huntsberger. Chapter 7 Page 305. Fifth Edition. Allyn and Bacon 1993.

• a) For x = 25, z = 25-21/10 = 0.4 From the z table p (0 < z < .4) = .1554

Therefore, p (x > .4) = .5 – p(0 < z < .4) = .5 – .1554 = .3446 = 34.46%

• b) x = 21 x=10
  If a large number of samples, each of size 36, are selected from the population and a sample mean is selected computed for each sample, the mean, variance and shape of the distribution of sample means would be as follows: x= 21.

The standard deviation of sample means x can be computed as given below:

x = X = 10 = 1.6666 n 36.

Variance is the square of the standard deviation. Therefore, the variance of the sample means ²x can be computed as follows:

²x = ²x = 10² = 2.7777 n 36.

The shape of the distribution of sample means = normal since sample size n 30.

• c) This question concerns the distributions of sample means. For the distribution of sample means: x = 21 and x = X = 10 = 1.6666 n 36, and the shape would be normal.

For x = 25 z = 25 – 21 = 2.4, 1.6666.  From the z-table P(021) = 0.5 – P(0(3-3.1)/0.4)  =P(z>-0.25) =1-0.4013 (from z table) =0.5987

• d) Sample size,n= 64 mean remains same 3.1
  standard error of the mean (standard deviation of sample means) expected to be = s/vn = 0.4/v64 = 0.4/8 =0.05 expected shape of the distribution of sample means will be bell shaped as this is normally distributed.

• g) If a random sample of 64 customers is selected, the probability that the sample mean would exceed 3 minutes=P(x>3) =P((x-µ)/e >(3-3.1)/0.05) =P(z>-2) =1-0.0228 (from z table) =0.9772.

The distributions were reviewed comparing the data to see whether men and omen differ in their mean when it comes to their socioeconomic index and their age when their first child was born. It should be noted that each group has a fairly large sample size. Looking at distributions of socioeconomic index (set) by gender, it indicates distributions look similar between men and women. The histograms show that the distributions are not normal, but similar to one another. The same with the box plot. There doesn’t seem to be an issue to run a t-test on these two groups.

The striations are not identical when looking at the respondent’s age when 1st child was born (stranger) by gender. The mean should be an appropriate measure of central tendency, and is lower in females, as seen in the box plot. However, because the distributions are similar enough, the t-test can still be run. Respondent Socioeconomic Index Group Statistics SST. Error Mean Respondent Socioeconomic Index Independent Samples Test Eleven’s Test for Equality of Variances t-test for Equality of Means SST.

In order to draw relevant conclusions from any given data set, the researcher needs to address the following points:

1. There must be clear delineation between continuous and discrete variables. Continuous variables have values that can be measured in intervals. Discrete variables, on the other hand, have values that cannot be measured in intervals (in whole numbers or integers). Examples of continuous variables are temperature, weight, income, height, and depth. Examples of discrete variables are age, number of months, and number of children;
2. The second consideration is related to the type of statistical tests used in specific analysis. The Student’s t-test is used for comparing the variance of ‘two’ sample groups. The ANOVA or the Analysis of Variance is used for more than two sample groups (it can also be used in analyzing the variance of two sample groups. Regression analysis provides a synthetic equation which describes the relationship between one or more variables;
3. In group analysis, correlation is of little value. The important thing is the perpetuity of the variance or the existence of disparity;
4. It would be irresponsible for the researcher to conduct multiple statistical tests when only one or two is required to establish the validity of any given hypothesis;
5. The researcher should always consider the normality of any given data set. If the data set is normally distributed, he/she may use the ANOVA or the t-test. If, however, the data set is not normally distributed, he/she should use nonparametric tests like the Kruskal-Wallis Test.
6. Lastly, the researcher must maintain internal validity and reliability in his/her methodologies. This is important because research is grounded on validity and reliability. Manipulation or undue ‘dropping’ of variables is always the ‘mortal sin’ of quantitative research.

In the data set provided, note that continuous and discrete variables are not grouped. The researcher should therefore provide a necessary format for an easier analysis of data. Items with blank ‘content’ are automatically assigned a ‘0’ value. If the associated variable is an ordinal one, retaining the blank item is acceptable. If, however, the associated variable is a nominal one, then the researcher should remove that sample from the data set. Retaining ‘empty’ values in rank analysis often leads to confusion and misinterpretation of data. The size of any sample always follows from the size of the universe or the population.

If the population is about 120 000, then the sample size should be around 4 to 5 percent. If the population is more or less 1000, then the sample size should be around 40 to 100. Note that if the sample size is equal or less than 25, then the researcher should use the Student’s t-test. Suppose the sample size is greater than 25 (congruent to the population), the researcher may use the Z-test (assuming that the data is normally distributed). Now, the so-called p-value should not be confused with the P(X) – the former indicates the critical area in a distribution while the latter the probability of any given event. Finally, data analysis is insufficient if not supplemented by discourses on related literature, theories, and hypotheses.

• Hypothesis Testing Hypothesis I: The age of residents of region I is greater than the age of residents of region
• VII. Null Hypothesis: There is no difference in the age of residents of region I and the age of residents of region
• VII. Alternative Hypothesis: The age of residents of region I is greater than the age of residents of region
• VII. Preliminary Analysis The variables ‘age’ and ‘region’ are discrete variables. It is better to plot discrete variables in bar graphs than in line graphs. ANOVA may be used in the analysis. Note that analysis is one-tailed.

Statistical Test

The Student’s t-test is used to determine the mean variance of the two groups.

Results

The resulting p-value is equal to 0. 191056. Because the p-value is greater than ? = 0. 05, we fail to reject the null hypothesis. In short, there is no difference in the age of residents of region I and the age of residents of region

• VII. Hypothesis II. The wealth score of mail donors is greater than the wealth score of mail non-donors.
• Null Hypothesis: There is no difference in the wealth score of mail donors and mail non-donors.
• Alternative Hypothesis: The wealth score of mail donors is greater than the wealth score of mail non donors.
• Preliminary Analysis The variables ‘age’ and ‘region’ are discrete variables. It is better to plot discrete variables in bar graphs than in line graphs. ANOVA may be used in the analysis. Note that analysis is one-tailed.

Statistical Test

The Kruskal-Wallis test is used to determine the mean variance of the two groups. This test is used because the values of the data set are not normally distributed.

Results

The resulting p-value is equal to 0. 20158. Because the p-value is greater than ? = 0. 05, we fail to reject the null hypothesis. In short, there is no difference in the wealth score of mail donors and mail non-donors (consequently, the wealth score of mail donors is not greater than the wealth score of mail non-donors).

Continuous and Discrete Variables

Discrete Variables

Some of the discrete variables in the data set are age, region, and number of children. Age Measures of Central Tendency The mean is equal to 46. 202. In short, the average age of most respondents is 46. The median age is 44. The mode is 46. The largest number of respondents is aged 46. The distribution is relatively skewed to the left.

Region Measures of Central Tendency

The mean is equal to 4. 69. In short, most of the respondents came from regions IV and V. The mode is equal to 7. The largest number of respondents came from region VII. The median is equal to 4. The distribution is relatively skewed to the right. Number of Children Measures of Central Tendency The mean is equal to . 586. In short, most of the respondents have at most one child The mode is equal to 0. The largest number of respondents has no children. The median is equal to 2. The distribution is relatively skewed to the right. Continuous Variables Some of the continuous variables in the data set are income and wealth score. Income

The mean is equal to USD 25000- 43, 000 In short, on the average, most of the respondents have incomes ranging from 25000 to 43000. The mode is equal to USD 40000. The largest number of respondents has income of about USD 40000. The median is equal to USD 50000. The distribution is relatively skewed to the left. Wealth Score Measures of Central Tendency The mean is equal to 301. 84. In short, on the average, most of the respondents have wealth score equal to 301. The mode is equal to 235. The largest number of respondents have wealth score equal to 235. The median is equal to 252. The distribution is relatively skewed to the right.

Simple linear regression is the statistic method used to make summary of and provide the association between variables that are continues and quantitative ,basically it deals with two measures that describes how strong the linear relationship we can compute in data .Simple linear regression consist of one variable known as the predictor variable and the other variable denote y known as response variable .

It is expected that when we talk of simple linear regression to touch on deterministic relationship and statistical relationship, the concept of least mean square .the interpretation of the b0 and b1 that they are used to interpret the estimate regression . There is also what is known as the population regression line and the estimate regression line .

This linearity is measured using the correlation coefficient (r), that can be -1,0,1.The strength of the association is determined from the value of r .( https://onlinecourses.science.psu.edu/stat501/node/250). History of simple linear regression Karl Pearson established a demanding treatment of Applied statistical measure known as Pearson Product Moment Correlation .

This come from the thought of Sir Francis Galton ,who had the idea of the modern notions of correlation and regression ,Sir Galton contributed in science of Biology ,psychology and Applied statistics . It was seen that Sir Galton is fascinated with genetics and heredity provided the initial inspiration that led to regression and Pearson Product Moment Correlation .

The thought that encouraged the advance of the Pearson Product Moment Correlation began with vexing problem of heredity to understand how closely features of generation of living things exhibited in the next generation. Sir Galton took the approach of using the sweet pea to check the characteristic similarities. ( Bravais, A. (1846).

The use of sweet pea was motivated by the fact that it is self- fertilize ,daughter plants shows differences in genetics from mother with-out the use of the second parent that will lead to statistical problem of assessing the genetic combination for both parents .The first insight came about regression came from two dimensional diagram plotting the size independent being the mother peas and the dependent being the daughter peas.

He used this representation of data to show what statisticians call it regression today ,from his plot he realised that the median weight of daughter seeds from a particular size of mother seed approximately described a straight line with positive slope less than 1. “Thus he naturally reached a straight regression line ,and the constant variability for all arrays of character for a given character of second .It was ,perhaps best for the progress of the correlational calculus that this simple special case should promulgated first .It so simply grabbed by the beginner (Pearson 1930,p.5).

Then it was later generalised to more complex way that is called the multiple regression. Galton, F. (1894),Importance of linear regressionStatistics usually uses the term linear regression in interpretation of data association of a particular survey, research and experiment .The linear relationship is used in modelling .The modelling of one explanatory variable x and response variable y will require the use of simple linear regression approach .

The simple linear regression is said to be broadly useful in methodology and the practical application. This method on simple linear regression model is not used in statistics only but it is applied in many biological, social science and environmental research. The simple linear regression is worth importance because it gives indication of what is to be expected, mostly in monitoring and amendable purposes involved on some disciplines(April 20, 2011 , plaza ,).

Description of linear regression The simple linear regression model is described by Y=(?0 + ?1 +E), this is the mathematical way of showing the simple linear regression with labelled x and y .This equation gives us a clear idea on how x is associated to y, there is also an error term shown by E. The term E is used to justification for inconsistency in y, that we can be able to detect it by the use of linear regression to give us the amount of association of the two variables x and y .

Then we have the parameters that are use to represent the population (?0 + ?1x) .We then have the model given by E(y)= (?0 + ?1x), the ?0 being the intercept and ?1 being the slope of y ,the mean of y at the x values is E(y) . The hypothesis is assumed is we assume that there is a linear association between the two variables ,that being our H0 and H1 we assume that there is no linear relationship between H0 and H1. Background of simple linear regression Galton used descriptive statistics in order for him to be able to generalise his work of different heredity problems .

The needed opportunity to conclude the process of analysing these data, he realised that if the degree of association between variables was held constant,then the slope of the regression line could be described if variability of the two measure were known . Galton assumed he estimated a single heredity constant that was generalised to multiple inherited characteristics .

He was wondering why, if such a constant existed ,the observed slopes in the plot of parent child varied too much over these characteristics .He realise variation in variability amongst the generations, he attained at the idea that the variation in regression slope he obtained were solely due to variation in variability between the various set of measurements .

In resent terms ,the principal this principal can be illustrated by assuming a constant correlation coefficient but varying the standard deviations of the two variables involved . On his plot he found out that the correlation in each data set. He then observe three data sets ,on data set one he realised that the standard deviation of Y is the same as that of X , on data set two standard deviation of Y is less than that of X ,third data set standard deviation of Y is great than that of X .

The correlation remain constant for three sets of data even though the slope of the line changes as an outcome of the differences in variability between the two variables.The rudimentary regression equation y=r(Sy / Sx)x to describe the relationship between his paired variables .He the used an estimated value of r , because he had no knowledge of calculating it The (Sy /Sx) expression was a correction factor that helped to adjust the slope according to the variability of measures .

He also realised that the ratio of variability of the two measures was the key factor in determining the slope of the regression line .The uses of simple linear regression Simple linear regression is a typical Statistical Data Analysis strategy. It is utilized to decide the degree to which there is a direct connection between a needy variable and at least one free factors. (e.g. 0-100 test score) and the free variable(s) can be estimated on either an all out (e.g. male versus female) or consistent estimation scale.

There are a few different suppositions that the information must full fill keeping in mind the end goal to meet all requirements for simple linear regression. Basic linear regression is like connection in that the reason for existing is to scale to what degree there is a direct connection between two factors.

The real contrast between the two is that relationship sees no difference amongst the two variables . Specifically, the reason for simple linear regression “anticipate” the estimation of the reliant variable in light of the estimations of at least one free factors. https://www.statisticallysignificantconsulting.com/RegressionAnalysis.htm

Reference

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• “Tables to Help Students Grasp Size Differences in Simple Correlations,” Teaching of Psychology, 5, 219-221.FitzPatrick, P. J. (1960),
• “Leading British Statisticians of the Nineteenth Century,” Journal of the American Statistical Association, 55, 38-70.Galton, F. (1894),
• Natural Inheritance (5th ed.), New York: Macmillan and Company.
• https://onlinecourses.science.psu.edu/stat501/node/250.https://www.statisticallysignificantconsulting.com/RegressionAnalysis.htmGhiselli, E. E. (1981),
• Measurement Theory for the Behavioral Sciences, San Francisco: W. H. Freeman.Goldstein, M. D., and Strube, M. J. (1995), “Understanding Correlations: Two Computer Exercises,” Teaching of Psychology, 22, 205-206.Karylowski, J. (1985),
• “Regression Toward the Mean Effect: No Statistical Background Required,” Teaching of Psychology, 12, 229-230.Paul, D. B. (1995),
• Controlling Human Heredity, 1865 to the Present, Atlantic Highlands, N.J.: Humanities Press.Pearson, E. S. (1938),
• Mathematical Statistics and Data Analysis (2nd ed.), Belmont, CA: Duxbury.Pearson, K. (1896),
• “Mathematical Contributions to the Theory of Evolution. III. Regression, Heredity and Panmixia,” Philosophical Transactions of the Royal Society of London, 187, 253-318.Pearson, K. (1922),
• Francis Galton: A Centenary Appreciation, Cambridge University Press.Pearson, K. (1930),
• The Life, Letters and Labors of Francis Galton, Cambridge University Press.Williams, R. H. (1975), “A New Method for Teaching Multiple Regression to Behavioral Science Students,” Teaching of Psychology, 2, 76-78.