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In Exercise 12.15 (data set EX1215), we measured the armspan and height of eight people with the following results: $$ \begin{array}{l|clll} \text { Person } & 1 & 2 & 3 & 4 \\ \hline \begin{array}{l} \text { Armspan (inches) } \\ \text { Height (inches) } \end{array} & 68 & 62.25 & 65 & 69.5 \\ & 69 & 62 & 65 & 70 \\ \text { Person } & 5 & 6 & 7 & 8 \\ \hline \text { Armspan (inches) } & 68 & 69 & 62 & 60.25 \\ \text { Height (inches) } & 67 & 67 & 63 & 62 \end{array} $$ a. Does the data provide sufficient evidence to indicate that there is a linear relationship between armspan and height? Test at the \(5 \%\) level of significance. b. Construct a \(95 \%\) confidence interval for the slope of the line of means, \(\beta\). c. If Leonardo da Vinci is correct, and a person's armspan is roughly the same as the person's height, the slope of the regression line is approximately equal to \(1 .\) Is this supposition confirmed by the confidence interval constructed in part b? Explain.

Short Answer

Expert verified
Answer: The main goals of this analysis are to perform a hypothesis test to determine if there is a linear relationship between armspan and height, construct a 95% confidence interval for the slope of the line of means, and check if Leonardo da Vinci's supposition that the slope of the regression line is approximately equal to 1 is confirmed.

Step by step solution

01

Calculate the Sample Correlation Coefficient

We will use the Pearson's correlation coefficient to determine the strength and direction of the linear relationship between armspan and height. Let's calculate the correlation coefficient, \(r\), using the following formula: $$ r = \frac{\Sigma(xy) - \frac{\Sigma x \cdot \Sigma y}{n}}{\sqrt{(\Sigma x^2 - \frac{(\Sigma x)^2}{n})(\Sigma y^2 - \frac{(\Sigma y)^2}{n})}} $$ Where: - \(xy\) means the product of each armspan and height data point - \(x\) means armspan - \(y\) means height - \(n\) is the sample size
02

Perform the Hypothesis Test

Now that we have calculated the correlation coefficient, we will test if there is a linear relationship between armspan and height. The null and alternative hypotheses for part (a) are: - Null hypothesis (\(H_0\)): There is no linear relationship between armspan and height (\(蟻=0\)). - Alternative hypothesis (\(H_1\)): There is a linear relationship between armspan and height (\(蟻鈮0\)). To perform the hypothesis test, we will use the test statistic \(t\) calculated as follows: $$ t = \frac{r}{\sqrt{\frac{1-r^2}{n-2}}} $$ We will compare the calculated test statistic, \(t\), with the critical value from the t-table at a 5% level of significance and with \(n - 2\) degrees of freedom.
03

Construct a 95% Confidence Interval for the Slope

In part (b), we will construct a 95% confidence interval for the slope of the line of means, 尾, which is given by: $$ \text{CI} = \left[b_1 \pm t^\ast\cdot SE(b_1)\right] $$ Where: - \(b_1\) is the estimated slope. - \(SE(b_1)\) is the standard error of the slope. - \(t^\ast\) is the critical value from the t-table with \(n - 2\) degrees of freedom and a 95% confidence level. We need to first calculate the slope, \(b_1\): $$ b_1 = \frac{r\cdot(\Sigma(x - \bar{x})\cdot(y - \bar{y}))}{(\Sigma(x - \bar{x})^2)} $$ Next, find the standard error of the slope: $$ SE(b_1) = \sqrt{\frac{\Sigma(y_i - \hat{y_i})^2}{(n-2)\cdot\Sigma(x_i - \bar{x})^2}} $$ Then, calculate the confidence interval for the slope.
04

Check if Leonardo da Vinci's Supposition is Confirmed

In part (c), we will use the confidence interval calculated in part (b) to check if Leonardo da Vinci's supposition that the slope of the regression line is approximately equal to 1 is confirmed. If the confidence interval contains 1, then his supposition is confirmed. Otherwise, it is not confirmed. In conclusion, by following these steps, we can analyze the relationship between armspan and height, test the hypotheses, construct the confidence interval for the slope, and check if Leonardo da Vinci's supposition about the relationship between armspan and height is confirmed.

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Key Concepts

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

Pearson's Correlation Coefficient
When studying the strength and direction of a linear relationship between two continuous variables, the Pearson's correlation coefficient, often symbolized as \( r \), is an invaluable tool.

Consider you have a dataset containing two variables of interest, like armspan and height from the exercise. To understand whether the size of one can predict the other and how strongly they are connected, \( r \) provides a numerical measure ranging from -1 to +1. A coefficient of +1 indicates a perfect positive linear relationship, -1 a perfect negative linear relationship, and 0 signifies no linear relationship at all.

To calculate \( r \), you would use the formula given in the exercise solution. It involves summing up the product of paired scores, the sum of individual scores, and then denominating it by their standard deviations. This calculation reflects how much two variables change together, compared to how much they vary individually.

In the given exercise, you would cross multiply armspan and height for each person, sum these products up, and follow the sequence of the formula. If the computed \( r \) is close to +1 or -1, it can be inferred that a strong linear relationship exists.
Hypothesis Testing
Hypothesis testing is a statistical method used to make inferences about the population based on sample data. In the context of the Pearson's correlation coefficient, it allows us to test the significance of the relationship between two variables.

The null hypothesis \( (H_0) \) typically proposes that no relationship exists鈥攎eaning that the true correlation \( \rho \) in the population is zero. Conversely, the alternative hypothesis \( (H_1) \) suggests that there is a relationship, and \( \rho \) is not equal to zero.

To decide whether to reject \( H_0 \), one calculates the test statistic \( t \), which tells us how extreme the observed correlation is assuming \( H_0 \) is true. If the calculated \( t \) is greater than the critical value from t-distribution, we reject \( H_0 \) and conclude that there is a significant linear relationship. By performing these steps as outlined in the solution with a 5% level of significance, the exercise seeks to demonstrate if armspan and height are related in a statistically significant way.
Confidence Intervals
Confidence intervals provide a range of values within which we can expect the true parameter to lie, with a certain degree of confidence. A 95% confidence interval for the slope, \( \beta \), in the context of a regression line, suggests that we can be 95% sure that the interval calculated from the sample data contains the true slope of the population regression line.

The exercise indicates the formula for constructing a confidence interval for \( \beta \). It incorporates the slope estimated from the sample data, \( b_1 \), and accounts for the variability in this estimate given as the standard error, \( SE(b_1) \), and the critical value from the t-distribution, \( t^* \).

To determine whether Leonardo da Vinci鈥檚 assertion that armspan is roughly equal to height (implying a slope of 1) holds, one must check if the value of 1 falls within the constructed confidence interval. If it does, this provides support for da Vinci鈥檚 claim in the context of the sampled data. The exercise guides through these steps to explore the historical supposition by quantifying the uncertainty around the slope estimate.

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Most popular questions from this chapter

An experiment was conducted to observe the effect of an increase in temperature on the potency of an antibiotic. Three 1 -ounce portions of the antibiotic were stored for equal lengths of time at each of these temperatures: \(30^{\circ}, 50^{\circ}, 70^{\circ},\) and \(90^{\circ} .\) The potency readings observed at each temperature of the experimental period are listed here: $$ \begin{array}{l|l|l|l|l} \text { Potency Readings, } y & 38,43,29 & 32,26,33 & 19,27,23 & 14,19,21 \\ \hline \text { Temperature, } x & 30^{\circ} & 50^{\circ} & 70^{\circ} & 90^{\circ} \end{array} $$ Use an appropriate computer program to answer these questions: a. Find the least-squares line appropriate for these data. b. Plot the points and graph the line as a check on your calculations. c. Construct the ANOVA table for linear regression. d. If they are available, examine the diagnostic plots to check the validity of the regression assumptions. e. Estimate the change in potency for a 1 -unit change in temperature. Use a \(95 \%\) confidence interval. f. Estimate the average potency corresponding to a temperature of \(50^{\circ} .\) Use a \(95 \%\) confidence interval. g. Suppose that a batch of the antibiotic was stored at \(50^{\circ}\) for the same length of time as the experimental period. Predict the potency of the batch at the end of the storage period. Use a \(95 \%\) prediction interval.

How many weeks can a movie run and still make a reasonable profit? The data that follow show the number of weeks in release \((x)\) and the gross to date (y) for the top 10 movies during a recent week. \({ }^{17}\) $$ \begin{array}{lcc} & \text {Gross to Date (in } & \text { Weeks } \\ \text { Movie } & \text { millions) } & \text { in Release } \\ \hline \text { 1. The Prestige } & \$ 14.8 & 1 \\ \text { 2. The Departed } & \$ 77.1 & 3 \\ \text { 3. Flags of Our Fathers } & \$ 10.2 & 1 \\ \text { 4. } \text { Open Season } & \$ 69.6 & 4 \\ \text { 5. Flicka } & \$ 7.7 & 1 \\ \text { 6. } \text { The Grudge } 2 & \$ 31.4 & 2 \\ \text { 7. } \text { Man of the Year } & \$ 22.5 & 2 \\ \text { 8. } \text { Marie } \text { Antoinette } & \$ 5.3 & 1 \\ \text { 9. } \text { The Texas Chainsaw Massacre: } & \$ 36.0 & 3 \\ \text {The Beginning } \\ \text { 10. } \text { The Marine } & \$ 12.5 & 2 \\ \hline \end{array} $$ a. Plot the points in a scatterplot. Does it appear that the relationship between \(x\) and \(y\) is linear? How would you describe the direction and strength of the relationship? b. Calculate the value of \(r^{2}\). What percentage of the overall variation is explained by using the linear model rather than \(\bar{y}\) to predict the response variable \(y ?\) c. What is the regression equation? Do the data provide evidence to indicate that \(x\) and \(y\) are linearly related? Test using a \(5 \%\) significance level. d. Given the results of parts \(b\) and \(c,\) is it appropriate to use the regression line for estimation and prediction? Explain your answer.

An agricultural experimenter, investigating the effect of the amount of nitrogen \(x\) applied in 100 pounds per acre on the yield of oats \(y\) measured in bushels per acre, collected the following data: $$ \begin{array}{l|llll} x & 1 & 2 & 3 & 4 \\ \hline y & 22 & 38 & 57 & 68 \\ & 19 & 41 & 54 & 65 \end{array} $$ a. Find the least-squares line for the data. b. Construct the ANOVA table. c. Is there sufficient evidence to indicate that the yield of oats is linearly related to the amount of nitrogen applied? Use \(\alpha=.05 .\) d. Predict the expected yield of oats with \(95 \%\) confidence if 250 pounds of nitrogen per acre are applied.e. Estimate the average increase in yield for an increase of 100 pounds of nitrogen per acre with \(99 \%\) confidence. f. Calculate \(r^{2}\) and explain its significance in terms of predicting \(y\), the yield of oats.

Give the equation and graph for a line with y-intercept equal to -3 and slope equal to 1 .

Is there any relationship between these two variables? To find out, we randomly selected 12 people from a data set constructed by Allen Shoemaker (Journal of Statistics Education) and recorded their body temperature and heart rate. \({ }^{13}\) $$ \begin{array}{l|llllll} \text { Person } & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \begin{array}{c} \text { Temperature } \\ \text { (degrees) } \end{array} & 96.3 & 97.4 & 98.9 & 99.0 & 99.0 & 96.8 \\ \text { Heart Rate } & 70 & 68 & 80 & 75 & 79 & 75 \\ \text { (beats per minute) } & & & & & & \end{array} $$ $$ \begin{array}{c|cccccc} \text { Person } & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline \begin{array}{c} \text { Temperature } \\ \text { (degrees) } \end{array} & 98.4 & 98.4 & 98.8 & 98.8 & 99.2 & 99.3 \\ \text { Heart Rate } & 74 & 84 & 73 & 84 & 66 & 68 \\ \text { (beats per minute) } & & & & & & \end{array} $$ a. Find the correlation coefficient \(r\), relating body temperature to heart rate. b. Is there sufficient evidence to indicate that there is a correlation between these two variables? Test at the \(5 \%\) level of significance.

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