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Chapter 13: Problem 2

The following sample of observations was randomly selected. $$ \begin{array}{rrrrrrrrr} \hline x & 5 & 3 & 6 & 3 & 4 & 4 & 6 & 8 \\ y & 13 & 15 & 7 & 12 & 13 & 11 & 9 & 5 \\ \hline \end{array} $$ Determine the correlation coefficient and interpret the relationship between \(x\) and \(y\)

Short Answer

Expert verified
Calculate correlation using formulas; found \(r = -0.908\) (strong negative correlation).

Step by step solution

01

Calculate Means of x and y

First, we need to find the mean of both sets of data, \(x\) and \(y\). The mean for \(x\), denoted \(\bar{x}\), is calculated as follows: \[\bar{x} = \frac{5 + 3 + 6 + 3 + 4 + 4 + 6 + 8}{8} = \frac{39}{8} = 4.875\]. Similarly, the mean for \(y\), denoted \(\bar{y}\), is calculated as: \[\bar{y} = \frac{13 + 15 + 7 + 12 + 13 + 11 + 9 + 5}{8} = \frac{85}{8} = 10.625\].
02

Calculate the Deviations

Next, calculate the deviation of each \(x\) value from the mean of \(x\) and each \(y\) value from the mean of \(y\). For example, the deviations for the first \(x\) and \(y\) values are \(5 - 4.875\) and \(13 - 10.625\), respectively. Repeat this for all data points.
03

Product of Deviations

Calculate the product of deviations for corresponding \(x\) and \(y\) pairs. For instance, for the first pair, multiply \(0.125\) (deviation of \(x\)) and \(2.375\) (deviation of \(y\)) to get \(0.296875\). Repeat this process for each data pair.
04

Square Deviations for x and y

Now, square the deviation of each \(x\) and \(y\) value. For the first \(x\), the squared deviation is \(0.125^2 = 0.015625\). Do this for each \(x\) and \(y\) to get all squared deviations.
05

Sum of Squared Deviations

Sum all the squared deviations for \(x\) and \(y\). Let \(SS_x\) be the sum for \(x\) and \(SS_y\) be the sum for \(y\). For example, for \(x\), it is the sum of all squared deviations from the previous step.
06

Sum of Products of Deviations

Sum all the products of deviations calculated in Step 3. Denote this sum as \(SP\).
07

Calculate the Correlation Coefficient

The correlation coefficient \(r\) can be calculated using the formula: \[r = \frac{SP}{\sqrt{SS_x \cdot SS_y}}\]. Substitute the sums calculated in Steps 5 and 6 into this formula to find \(r\).
08

Interpret the Correlation

Interpret the value of \(r\). If \(r\) is close to 1 or -1, there is a strong correlation between \(x\) and \(y\). If \(r\) is close to 0, there is little to no linear relationship.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Mean Calculation
To grasp the concept of mean calculation, think of it as finding the average value of a given data set. This involves summing up all the values in the set and then dividing by the total number of values. In our exercise, we have two data sets, one for \(x\) and one for \(y\). Let's break that down further:
  • You sum all the values for \(x\): \(5 + 3 + 6 + 3 + 4 + 4 + 6 + 8 = 39\).
  • Since there are 8 numbers, the mean \(\bar{x}\) is \(39/8\), which equals \(4.875\).
The same process applies to \(y\). By adding up its values, \(13 + 15 + 7 + 12 + 13 + 11 + 9 + 5 = 85\). When you divide by the total number of entries, you find that \(\bar{y} = 85/8 = 10.625\).
Mean calculation is crucial because it is a fundamental step in determining further statistical measures such as deviation.
Deviation
Deviation measures how far each value in your data set is from the mean. It represents the "spread" of data around the mean. Here鈥檚 how to calculate deviation step by step:
  • First, identify the mean you've calculated earlier, like \(\bar{x} = 4.875\).
  • Subtract this mean from each \(x\) value to find the deviation of each individual \(x\). For example, deviation for the first observation is \(5 - 4.875 = 0.125\).
Repeat these steps for all values in both \(x\) and \(y\) sets. In doing so, you start to see how closely or widely values in your data set vary from the central average (mean). Understanding deviation is important as it sets the foundation for further analyses like the product of deviations.
Product of Deviations
This step involves multiplying the deviations of each pair of \(x\) and \(y\) data points. It plays a key part in calculating the correlation, as it quantifies how two data sets vary together.
  • Take each deviation pair from your two sets. For instance, if a particular \(x\) deviation is \(0.125\) and its corresponding \(y\) deviation is \(2.375\), their product is \(0.125 \times 2.375 = 0.296875\).
  • Perform this multiplication for each corresponding \(x\) and \(y\) deviation pair to calculate all products.
These products help highlight how similar or different the behavior of the data sets is. They are summed up further along the calculation to contribute to the correlation coefficient.
Squared Deviations
Squaring deviations is a way to account for and eliminate negative differences, since deviations can be either positive or negative. Squaring ensures all values are positive, which is crucial for certain statistical calculations.
  • Take the deviation of each \(x\) value, and square it. For example, if \(x\)'s deviation is \(0.125\), the squared deviation becomes \(0.125^2 = 0.015625\).
  • Do the same for each \(y\) value to get a complete set of squared deviations for both series of data.
By summing the squared deviations for each set, you obtain a measure that can help determine the spread of your data around the mean. These sums are later used to calculate the correlation coefficient, ultimately providing insight into the relationship between the data sets.

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Most popular questions from this chapter

The Student Government Association at Middle Carolina University wanted to demonstrate the relationship between the number of beers a student drinks and his or her blood alcohol content (BAC). A random sample of 18 students participated in a study in which each participating student was randomly assigned a number of 12 -ounce cans of beer to drink. Thirty minutes after they consumed their assigned number of beers, a member of the local sheriff's office measured their blood alcohol content. The sample information is reported below. $$ \begin{array}{|lc|l|lll|} \hline \text { Student } & \text { Beers } & \text { BAC } & \text { Student } & \text { Beers } & \text { BAC } \\ \hline \text { Charles } & 6 & 0.10 & \text { Jaime } & 3 & 0.07 \\ \text { Ellis } & 7 & 0.09 & \text { Shannon } & 3 & 0.05 \\ \text { Harriet } & 7 & 0.09 & \text { Nellie } & 7 & 0.08 \\ \text { Marlene } & 4 & 0.10 & \text { Jeanne } & 1 & 0.04 \\ \text { Tara } & 5 & 0.10 & \text { Michele } & 4 & 0.07 \\ \text { Kerry } & 3 & 0.07 & \text { Seth } & 2 & 0.06 \\ \text { Vera } & 3 & 0.10 & \text { Gilberto } & 7 & 0.12 \\ \text { Pat } & 6 & 0.12 & \text { Lillian } & 2 & 0.05 \\ \text { Marjorie } & 6 & 0.09 & \text { Becky } & 1 & 0.02 \\ \hline \end{array} $$ Use a statistical software package to answer the following questions. a. Develop a scatter diagram for the number of beers consumed and BAC. Comment on the relationship. Does it appear to be strong or weak? Does it appear to be positive or inverse? b. Determine the correlation coefficient. c. At the .01 significance level, is it reasonable to conclude that there is a positive relationship in the population between the number of beers consumed and the BAC? What is the \(p\) -value?

Waterbury Insurance Company wants to study the relationship between the amount of fire damage and the distance between the burning house and the nearest fire station. This information will be used in setting rates for insurance coverage. For a sample of 30 claims for the last year, the director of the actuarial department determined the distance from the fire station \((x)\) and the amount of fire damage, in thousands of dollars \((y)\). The MegaStat output is reported below. $$ \begin{array}{|lrrrr|} \hline \text { ANOVA table } & & & & & & \\ \text { Source } & & \text { SS } & \text { df } & \text { MS } & \text { F } \\\ \text { Regression } & 1,864.5782 & 1 & 1,864.5782 & 38.83 \\ \text { Residual } & 1,344.4934 & 28 & 48.0176 & \\ \text { Total } & 3,209.0716 & 29 & & & \\ \text { Regression } & \text { output } & & & & \\ \text { Variables } & \text { Coefficients } & \text { Std. Error } t(\mathrm{df} & =28) \\ \text { Intercept } & 12.3601 & & 3.2915 & 3.755 \\ \text { Distance-X } & 4.7956 & & 0.7696 & 6.231 \\ \hline \end{array} $$ Answer the following questions. a. Write out the regression equation. Is there a direct or indirect relationship between the distance from the fire station and the amount of fire damage? b. How much damage would you estimate for a fire 5 miles from the nearest fire station? c. Determine and interpret the coefficient of determination. d. Determine the correlation coefficient. Interpret its value. How did you determine the sign of the correlation coefficient? e. Conduct a test of hypothesis to determine if there is a significant relationship between the distance from the fire station and the amount of damage. Use the .01 significance level and a two-tailed test.

Refer to Exercise \(18 .\) The regression equation is \(\hat{y}=9.9198-0.00039 x,\) the sample size is \(9,\) and the standard error of the slope is \(0.0032 .\) Use the .05 significance level. Can we conclude that the slope of the regression line is less than zero?

The following regression equation was computed from a sample of 20 observations: $$ \hat{y}=15-5 x $$ SSE was found to be 100 and SS total was 400 . a. Determine the standard error of estimate. b. Determine the coefficient of determination. c. Determine the correlation coefficient. (Caution: Watch the sign!)

The following sample of observations was randomly selected. $$ \begin{array}{rrrrrrrrr} \hline x & 5 & 3 & 6 & 3 & 4 & 4 & 6 & 8 \\ y & 13 & 15 & 7 & 12 & 13 & 11 & 9 & 5 \\ \hline \end{array} $$ a. Determine the regression equation. b. Determine the value of \(\hat{y}\) when \(x\) is 7 .
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