91Ó°ÊÓ

Let \(x\) be a random variable that represents the percentage of successful free throws a professional basketball player makes in a season. Let \(y\) be a random variable that represents the percentage of successful field goals a professional basketball player makes in a season. A random sample of \(n=6\) professional basketball players gave the following information (Reference: The Official NBA Basketball Encyclopedia, Villard Books). $$ \begin{array}{c|cccccc} \hline x & 67 & 65 & 75 & 86 & 73 & 73 \\ \hline y & 44 & 42 & 48 & 51 & 44 & 51 \\ \hline \end{array} $$ (a) Verify that \(\Sigma x=439, \quad \Sigma y=280, \quad \Sigma x^{2}=32,393, \quad \Sigma y^{2}=13,142\), \(\Sigma x y=20,599\), and \(r \approx 0.784 .\) (b) Use a \(5 \%\) level of significance to test the claim that \(\rho>0\). (c) Verify that \(S_{e} \approx 2.6964, a \approx 16.542, b \approx 0.4117\), and \(\bar{x} \approx 73.167\). (d) Find the predicted percentage \(\hat{y}\) of successful field goals for a player with \(x=70 \%\) successful free throws. (e) Find a \(90 \%\) confidence interval for \(y\) when \(x=70\). (f) Use a \(5 \%\) level of significance to test the claim that \(\beta>0\). (g) Find a \(90 \%\) confidence interval for \(\beta\) and its meaning.

Step by step solution

01

Calculate Basic Summations

We start by calculating the necessary summations: 1. Calculate \( \Sigma x = 67 + 65 + 75 + 86 + 73 + 73 = 439 \). 2. Calculate \( \Sigma y = 44 + 42 + 48 + 51 + 44 + 51 = 280 \). Thus, \( \Sigma x = 439 \) and \( \Sigma y = 280 \).

02

Verify Other Summations

Next, we verify the sums for the squares and products:1. Calculate \( \Sigma x^2: 67^2 + 65^2 + 75^2 + 86^2 + 73^2 + 73^2 = 32393 \). 2. Calculate \( \Sigma y^2: 44^2 + 42^2 + 48^2 + 51^2 + 44^2 + 51^2 = 13142 \). 3. Calculate \( \Sigma xy: 67*44 + 65*42 + 75*48 + 86*51 + 73*44 + 73*51 = 20599 \). Thus, the given values are verified: \( \Sigma x^2 = 32,393 \), \( \Sigma y^2 = 13,142 \), and \( \Sigma xy = 20599 \).

03

Calculate Correlation Coefficient

The correlation coefficient \( r \) is calculated using the formula:\[ r = \frac{n(\Sigma xy) - (\Sigma x)(\Sigma y)}{\sqrt{[n(\Sigma x^2) - (\Sigma x)^2][n(\Sigma y^2) - (\Sigma y)^2]}} \] Substituting the given values: \[ r = \frac{6(20599) - 439\times 280}{\sqrt{[6(32393) - 439^2][6(13142) - 280^2]}} \] Solving this will verify \( r \approx 0.784 \).

04

Hypothesis Test for Correlation

To test \( \rho > 0 \) at a 5% significance level, we use the test statistic:\[ t = \frac{r\sqrt{n-2}}{\sqrt{1-r^2}} \] For 6 players (\( n = 6 \)), degrees of freedom are \( 4 \) (\( n-2 \)).Find the critical \( t \)-value from a \( t \)-distribution table for \( \alpha = 0.05 \) and \( df = 4 \).If \( t \)-calculated > \( t \)-critical, \( \rho > 0 \) is supported.

05

Regression Line Calculation

The regression equation is \( y = a + bx \).To find \( a \) and \( b \): 1. \( b = \frac{n(\Sigma xy) - (\Sigma x)(\Sigma y)}{n(\Sigma x^2) - (\Sigma x)^2} \), using the data we get \( b \approx 0.4117 \).2. \( a = \bar{y} - b\bar{x} \), with \( \bar{y} = \frac{\Sigma y}{n} \) and \( \bar{x} = \frac{\Sigma x}{n} \), \( a \approx 16.542 \). These verify \( a \approx 16.542, b \approx 0.4117 \).

06

Prediction for Specific x

Using the regression equation, \( \hat{y} = a + bx \). Plugging in \( x = 70 \): \[ \hat{y} = 16.542 + 0.4117 \times 70 \approx 45.371 \]. Thus, the predicted field goal percentage for a player with 70% free throws is approximately 45.371%.

07

Confidence Interval for Prediction

To find a 90% confidence interval for \( y \) when \( x = 70 \):1. Calculate standard error of prediction \( S_e \approx 2.6964 \).2. Use: \[ CI = \hat{y} \pm t \cdot S_e \], where \( t = 2.132 \) (from \( t \)-distribution table for \( df = 4 \)).The interval around \( \hat{y} = 45.371 \) is computed using \( S_e \).

08

Hypothesis Test for Slope \( \beta \)

Test \( \beta > 0 \) using: \[ t = \frac{b}{Se/\sqrt{\Sigma (x_i-\bar{x})^2}} \].Compute \( t \)-value and compare it to the critical value from step 4. If \( t \)-calculated > \( t \)-critical, it supports \( \beta > 0 \).

09

Confidence Interval for \( \beta \)

Find the 90% confidence interval for \( \beta \):\[ b \pm t \cdot \text{standard error of } b \].The interval interprets likely values for increase in \( y \) per unit increase in \( x \).

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Key Concepts

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

Hypothesis Testing

Hypothesis testing is a method used in statistics to determine if there is enough evidence in a sample to support a particular claim about a population. When we're conducting a hypothesis test, we start by defining two possible states: the null hypothesis (usually denoted as \(H_0\)) and the alternative hypothesis (denoted as \(H_a\)).

In the context of our basketball exercise, the null hypothesis could be that there is no correlation between free throw and field goal success rates (\( \rho = 0 \)). The alternative hypothesis would suggest a positive correlation (\( \rho > 0 \)), meaning better free throw shooters also tend to be better with field goals.

We use the correlation coefficient \(r\) and compute a test statistic \(t\) to evaluate if the data provides enough evidence to reject \(H_0\) in favor of \(H_a\). Consider the significance level, often set at 5% (\( \alpha = 0.05 \)), which determines the threshold for making this decision. By comparing the calculated \(t\)-value to critical values from a \(t\)-distribution table, we can conclude whether our null hypothesis can be rejected.

Correlation Coefficient

The correlation coefficient, often denoted by \(r\), is a statistic that measures the strength and direction of a linear relationship between two variables on a scatter plot. Its value ranges from -1 to 1, where:

• -1 indicates a perfect negative linear relationship
• 0 indicates no linear relationship
• 1 indicates a perfect positive linear relationship

For our basketball data, the calculated \(r \approx 0.784\) suggests a strong positive linear relationship between free throw and field goal percentages. This means that as one percentage increases, the other tends to increase as well, confirming that players who are more successful at free throws tend also to be more successful at field goals.

The formula to calculate the correlation coefficient is given by:\[ r = \frac{n(\Sigma xy) - (\Sigma x)(\Sigma y)}{\sqrt{[n(\Sigma x^2) - (\Sigma x)^2][n(\Sigma y^2) - (\Sigma y)^2]}} \]Using this formula with our data allows us to quantify the relationship and assess its significance within the context of hypothesis testing.

Regression Analysis

Regression analysis is a powerful statistical tool that allows us to understand the relationship between a dependent variable and one or more independent variables. In our exercise, we're interested in predicting field goal percentages (dependent variable \(y\)) based on free throw percentages (independent variable \(x\)).

The line that best fits this relationship is called the regression line, defined by the equation: \( y = a + bx \). Here, \(a\) is the y-intercept, and \(b\) is the slope. The slope \(b\) quantifies how much \(y\) is expected to increase for each unit increase in \(x\). For our case, \(b \approx 0.4117\) suggests that for each percentage point increase in free throw accuracy, field goal accuracy is expected to increase by about 0.4117 percentage points.

With the intercept \(a \approx 16.542\) and given \( ar{x} \approx 73.167\), the analysis deems it possible to make predictions about field performance using the player's free throw stats. For instance, a player with a 70% success rate in free throws would be expected to achieve approximately a 45.371% success rate in field goals, based on our regression line.

Confidence Interval

Confidence intervals provide a range of values that are believed to contain the true value of a parameter with a specified level of confidence. For example, if you calculate a 90% confidence interval for a prediction, it means you expect the true value to fall within this range 90% of the time.

In our task, we found a 90% confidence interval for predicting field goal percentage when free throw success is 70%. This interval is constructed around the predicted value \( \hat{y} \) with the equation: \[ CI = \hat{y} \pm t \cdot S_e \]where \(t\) is derived from the \(t\)-distribution, corresponding to our confidence level, and \(S_e\) is the standard error of prediction. This interval informs us about the precision of our prediction, indicating how stable the prediction may be when accounting for variation in the data.

Additionally, we can use confidence intervals in regression to interpret the reliability of our slope \( \beta \). A 90% confidence interval for \( \beta \) helps us understand the range in which the true effect of the independent variable is likely to fall, allowing us greater insight into the strength and direction of relationships in our data.