STAT 3090 Section 10 Lab Activity-1

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Jan 9, 2024

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STAT 3090 L AB A CTIVITY S ECTION 10 F ALL 2023 S IMPLE L INEAR R EGRESSION N AME : O BJECTIVES : Upon completion of this project, you should be able to: Complete a least squares regression line for a given data set on JMP Determine important quantities from the output and interpret them Evaluate the fit of a simple linear regression model Calculate confidence intervals for regression slopes Perform residual analysis Use the simple linear regression model Identify symbols and important terminology used in linear regression It is a fun fact that you can calculate the temperature based on the frequency of cricket chirping. This is sometimes called Dolbears law, as a formula for temperature based on the number of times crickets chirp in 15 seconds was published by Amos Dolbear in 1897. The formula states that the temperature in degrees Fahrenheit should be equal to the number of times crickets chirp in 15 seconds plus 40 degrees. In this assignment we want to see if this is true, and find the formula for ourselves. In the JMP file “Dolbear’s_Law.JMP” we have various recordings of cricket chirps and temperature, provided in the GLOBE article that can be found here ( https://www.globe.gov/explore-science/scientists-blog/archived- posts/sciblog/index.html_p=45.html ). D IRECTIONS : Answer the following questions using complete sentences where appropriate as though you were presenting your analysis to an academic journal where Dolbear’s law is researched. Please provide any appropriate output and/or screenshots from JMP. You will need the file Dolbear’s_Law.jmp Place your answers and any copies of output into this document. This lab is worth 100 points. Day 1: We will find the LSRL for cricket chirps per 15 seconds and temperature in Fahrenheit. In this application we will have the number of cricket chirps per 15 seconds be our independent variable and the temperature in Fahrenheit be our dependent variable. We know that the rate that crickets are chirping is in response to the temperature not the other way around. Please put yourself in the shoes of Mr. Dolbear in 1897. He is thinking of this in reverse and that is how we will proceed for this example. 1. Use JMP to plot temperature (y-axis) vs cricket chirps per 15 seconds (x-axis) and find the LSRL for the data. The instructions on how to use JMP for linear regression are in section 10.4 of your lecture guide. Provide a copy of the scatterplot with the least squares 1
STAT 3090 L AB A CTIVITY S ECTION 10 F ALL 2023 S IMPLE L INEAR R EGRESSION regression line drawn on it, and the main output consisting of the Linear Fit, Summary of Fit, Analysis of Variance and Parameter Estimates. (10 pts) 2. What is the value of the estimated y-intercept? If there is an interpretation for the y- intercept, please state it in the context of the problem. If there is no interpretation for the y-intercept explain why that is the case. (5 pts) For a night where there are no chirps from crickets the estimated average temperature is 40.025248 We do not have any data points for 0 chirps per 15 seconds. Since 0 is outside the scope of our model, we do not have an interpretation for the y-intercept. 2
STAT 3090 L AB A CTIVITY S ECTION 10 F ALL 2023 S IMPLE L INEAR R EGRESSION 3. What is the value of the estimated slope? Provide an interpretation for the slope in the context of the problem. (5 pts) The value of the slope b1 is 0. 89 ∘? . For every additional chirp per 15 seconds, the temperature increases on average by 0.89 ∘? . 4. Using this model, what is the predicted average temperature when crickets are chirping at a rate of 31 chirps per 15 seconds? Show your work by filling in the formula. (5 pts) Y^ = 40.025248 + .8917975 * 31 = 67.6709705 5. The observed temperature when the crickets were chirping at a rate of 31 chirps per 15 seconds was 69 degrees Fahrenheit. Calculate the residual. (5 pts) e i = y i γ i = 69 67.67097 05 = 1.3290295 6. Find and interpret the standard error found by your regression analysis. (5 pts) S = 1.6097 For any given value for the number of chirps in 15 seconds , the observed temperature should vary from the prediction on a n average by 1.6097 F. WebAssign Section 10, Q1-6 Day 2: 7. Find and interpret the coefficient of determination. (5 pts) ? 2 = 0.960901 96% of the variation in the temperatures of our data set is explained by the number of chirps per 15 seconds. 8. Find and interpret the correlation coefficient. (5 pts) ? = √ ? 2 = √0.960901 = 0.980255 Since the slope is positive, the value for r is positive. There is a strong positive linear relationship between the number of cricket chirps in 15 3
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STAT 3090 L AB A CTIVITY S ECTION 10 F ALL 2023 S IMPLE L INEAR R EGRESSION seconds and temperature 9. Identify the SSE for the least squares regression line in your analysis. (5 pts) SSE = 137.3382 10. Add to your linear regression model the plots of the residuals. Go to the red triangle next to Linear Fit and select Plot Residuals. Place a copy of the Residual Normal Quantile Plot and Residual by Predicted Plot here. (10 pts) 4
STAT 3090 L AB A CTIVITY S ECTION 10 F ALL 2023 S IMPLE L INEAR R EGRESSION 11. Write the 5 conditions that should be met for the estimated regression model to be valid. For each condition, include the name of the plot that demonstrates the condition has been met for this data set and state how the condition is met. (10 pts) 1. Linear Relationship: Based on the scatterplot, we can easily see a linear relationship between the number of chirps per 15 seconds and the temperature. 2. The error terms, 𝜀? , are normally distributed: To determine that the residuals follow a normal distribution, we create a normal probability plot of the residuals. Since the residuals in this plot seem to follow a linear pattern, it appears that this assumption is met. 3. The error terms, 𝜀? , are independent: As seen in the residuals vs predicted plot, the residuals are randomly scattered about the line y=0 4. The error terms, 𝜀? , have a mean of 0: As seen in the residuals vs predicted plot and in the histogram, the residuals have a mean of 0. 5. The error terms, 𝜀? , have constant variance: As seen in residuals vs predicted plot, the residuals are randomly scattered about the line y=0 12. Calculate and interpret a 95% confidence interval for the population slope β 1 . (10 pts) 𝒃? ± ( 𝒕? ? ⁄ , ? ? )( 𝒔𝒃? ) 𝒃? = ? . 𝟖𝟗?𝟕𝟗𝟕? , 𝒕 . ??? , ?? = ? . ???𝟖 , 𝒔𝒃? = ? . ??𝟒𝟕? 𝑪? : [ ? . 𝟖𝟒?? , ? . 𝟗𝟒?𝟒 ] We are 95% confident that the true population slope of the regression line using chirps per 15 seconds to predict temperature is between ? . 𝟖𝟒?? and ? . 𝟗𝟒?𝟒 . 13. Based on your confidence interval, would you say that the slope is significantly different than zero, indicating a linear relationship? Justify your answer. (10 pts) Yes, since 0 does not fall within our 95% confidence interval. 5
STAT 3090 L AB A CTIVITY S ECTION 10 F ALL 2023 S IMPLE L INEAR R EGRESSION 14. WebAssign Section 10 Linear Regression (10 pts) 6
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