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Chapter 5: Problem 79

Consider the four \((x, y)\) pairs \((0,0),(1,1)\), \((1,-1)\), and \((2,0)\) a. What is the value of the sample correlation coefficient \(r\) ? b. If a fifth observation is made at the value \(x=6\), find a value of \(y\) for which \(r>0.5\). c. If a fifth observation is made at the value \(x=6\), find a value of \(y\) for which \(r<0.5\).

Short Answer

Expert verified
a. The value of the sample correlation coefficient \(r\) for the given pairs is 0. b. For a fifth observation at \(x=6\), the value of \(y\) should be around 5 to make \(r > 0.5\). c. For a fifth observation at \(x=6\), the value of \(y\) should be around -5 to make \(r < 0.5\).

Step by step solution

01

Calculate the Mean Values

Start by calculating the mean values of \(x\) and \(y\). The set of \(x\)-values are \{0, 1, 1, 2\} and \(y\)-values are \{0, 1, -1, 0\). Mean of \(x\), \(\overline{x} = (0+1+1+2) / 4 = 1\) and mean of \(y\), \(\overline{y} = (0+1-1+0) / 4 = 0\).
02

Compute Sample Correlation Coefficient \(r\)

The sample correlation coefficient \(r\) is calculated as \(r = \frac{1}{n-1} \sum_{i=1}^{n} \left( \frac{x_i-\overline{x}}{s_x} \right) \left( \frac{y_i-\overline{y}}{s_y} \right)\) where \(s_x\) and \(s_y\) are the sample standard deviations of \(x\) and \(y\). Here \(s_x = \sqrt{ \frac{1}{n-1} \sum_{i=1}^{n} (x_i-\overline{x})^2 }\) and similarly for \(s_y\). After calculating these values, you would find that \(r = 0\).
03

Determine Fifth Point For \(r > 0.5\)

To make \(r > 0.5\), we need the new point to increase the positive correlation. So, pick a point for \(y\) such that it is similar to \(x=6\), for example, \(y=5\). After calculating, you will see that \(r\) is much larger than 0.5 thus validating our choice.
04

Determine Fifth Point For \(r < 0.5\)

To make \(r < 0.5\), consider adding a point for \(y\) that decreases the positive correlation. So, select a \(y\)-value that is lower and opposite from \(x=6\), for example, \(y=-5\). After calculating, you will observe that \(r\) is much lower than 0.5 thus validating our selection.

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Key Concepts

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

Mean Values
The concept of mean values or averages is central to many statistical calculations. It gives us a measure of the central tendency of a data set. To find the mean value of a set of numbers, you simply add them all up and divide by the number of elements in the set. This applies to both the \(x\) and \(y\) values in our exercise. By calculating the mean, we are essentially finding a balance point for the data.
  • For our \(x\)-values \(\{0, 1, 1, 2\}\), the mean is calculated as: \[ \overline{x} = \frac{0 + 1 + 1 + 2}{4} = 1 \]
  • For our \(y\)-values \(\{0, 1, -1, 0\}\), the mean is: \[ \overline{y} = \frac{0 + 1 - 1 + 0}{4} = 0 \]
Mean values help in further calculations such as the sample standard deviation and correlation coefficient by providing a baseline value for comparison.
Sample Standard Deviation
The sample standard deviation is a critical measure that indicates the extent of the variation or dispersion in a set of values. It helps in understanding how spread out the numbers are from their mean value. The formula for the sample standard deviation \(s\) is:\[ s = \sqrt{ \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \overline{x})^2 } \]This considers the squared differences from the mean and averages them, giving us a useful metric for variability.
  • A large standard deviation indicates that the values are spread widely around the mean.
  • A small standard deviation suggests that the values are close to the mean.
Calculating the sample standard deviation is a step forward in finding the correlation, as it measures how each element of the data differs from the mean.
Positive Correlation
A positive correlation between two variables implies that as one variable increases, the other tends to increase as well. This relationship suggests a direct association between the variables. The values of the correlation coefficient \(r\) range from \(-1\) to \(1\):
  • An \(r\) value closer to \(1\) indicates a strong positive correlation.
  • A perfectly positive correlation, where \(r = 1\), means all data points lie exactly on a line with positive slope.
Understanding positive correlation is useful for predicting the potential behavior of a variable based on the movement of another. In our exercise, selecting a \(y\)-value greater than 5 at \(x=6\) increases the degree of this positive correlation, thus elevating \(r\) above \(0.5\).
Negative Correlation
Negative correlation points to an inverse relationship between two variables: as one variable increases, the other decreases. Just like positive correlation, the value of the correlation coefficient \(r\) is key here:
  • An \(r\) value closer to \(-1\) denotes a strong negative correlation.
  • A perfectly negative correlation, where \(r = -1\), indicates that one variable perfectly reduces as the other rises.
In the context of the given problem, to achieve a correlation coefficient \(r\) of less than \(0.5\), we can introduce a data point where the \(y\)-value is negative, such as \(y = -5\) at \(x=6\). This action diminishes the positive correlation strength due to the inverse relationship introduced by the new data point.

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Most popular questions from this chapter

The accompanying data are \(x=\) cost (cents per serving) and \(y=\) fiber content (grams per serving) for 18 high-fiber cereals rated by Consumer Reports (www .consumerreports.org/health). $$ \begin{array}{cc} \begin{array}{c} \text { Cost per } \\ \text { serving } \end{array} & \begin{array}{c} \text { Fiber per } \\ \text { serving } \end{array} \\ \hline 33 & 7 \\ 46 & 10 \\ 49 & 10 \\ 62 & 7 \\ 41 & 8 \\ 19 & 7 \\ 77 & 12 \\ 71 & 12 \\ 30 & 8 \\ 53 & 13 \\ 53 & 10 \\ 67 & 8 \\ 43 & 12 \\ 48 & 7 \\ 28 & 14 \\ 54 & 7 \\ 27 & 8 \\ 58 & 8 \\ \hline \end{array} $$ a. Compute and interpret the correlation coefficient for this data set. b. The serving size differed for the different cereals, with serving sizes varying from \(1 / 2\) cup to \(1^{1 / 4}\) cups. Converting price and fiber content to "per cup" rather than "per serving" results in the accompanying data. Is the correlation coefficient for the per cup data greater than or less than the correlation coefficient for the per serving data? $$ \begin{array}{cc} \text { Cost per Cup } & \text { Fiber per Cup } \\ \hline 9.3 & 44 \\ 10 & 46 \\ 10 & 49 \\ 7 & 62 \\ 6.4 & 32.8 \\ 7 & 19 \\ 12 & 77 \\ 9.6 & 56.8 \\ 8 & 30 \\ 13 & 53 \\ 10 & 53 \\ 8 & 67 \\ 12 & 43 \\ 7 & 48 \\ 28 & 56 \\ 7 & 54 \\ 16 & 54 \\ 10.7 & 77.3 \\ \hline \end{array} $$

The relationship between hospital patient-tonurse ratio and various characteristics of job satisfaction and patient care has been the focus of a number of research studies. Suppose \(x=\) patient-to-nurse ratio is the predictor variable. For each of the following potential dependent variables, indicate whether you expect the slope of the least-squares line to be positive or negative and give a brief explanation for your choice. a. \(y=\) a measure of nurse's job satisfaction (higher values indicate higher satisfaction) b. \(y=\) a measure of patient satisfaction with hospital care (higher values indicate higher satisfaction) c. \(y=\) a measure of patient quality of care.

Explain why it can be dangerous to use the leastsquares line to obtain predictions for \(x\) values that are substantially larger or smaller than those contained in the sample.

For each of the following pairs of variables, indicate whether you would expect a positive correlation, a negative correlation, or a correlation close to \(0 .\) Explain your choice. a. Maximum daily temperature and cooling costs b. Interest rate and number of loan applications c. Incomes of husbands and wives when both have fulltime jobs d. Height and IQ e. Height and shoe size f. Score on the math section of the SAT exam and score on the verbal section of the same test g. Time spent on homework and time spent watching television during the same day by elementary school children h. Amount of fertilizer used per acre and crop yield (Hint: As the amount of fertilizer is increased, yield tends to increase for a while but then tends to start decreasing.)

Anabolic steroid abuse has been increasing despite increased press reports of adverse medical and psychiatric consequences. In a recent study, medical researchers studied the potential for addiction to testosterone in hamsters (Neuroscience \([2004]: 971-981)\). Hamsters were allowed to self-administer testosterone over a period of days, resulting in the death of some of the animals. The data below show the proportion of hamsters surviving versus the peak self-administration of testosterone \((\mu \mathrm{g}) .\) Fit a logistic regression equation and use the equation to predict the probability of survival for a hamster with a peak intake of \(40 \mu \mathrm{g}\). \begin{tabular}{cccc} \multicolumn{4}{c} { Survival } \\ Peak Intake (micrograms) & Proportion \((p)\) & \(\frac{p}{1-p}\) & \(y^{\prime}=\ln \left(\frac{p}{1-p}\right)\) \\ \hline 10 & \(0.980\) & \(49.0000\) & \(3.8918\) \\ 30 & \(0.900\) & \(9.0000\) & \(2.1972\) \\ 50 & \(0.880\) & \(7.3333\) & \(1.9924\) \\ 70 & \(0.500\) & \(1.0000\) & \(0.0000\) \\ 90 & \(0.170\) & \(0.2048\) & \(-1.5856\) \\ \hline \end{tabular}
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