Hw10

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University of Pittsburgh *

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1000

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Statistics

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Feb 20, 2024

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docx

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Uploaded by SargentValor13454

Hw10 10.14 a. To consider the driver's mpg calculations as the explanatory variable, we can plot the driver's mpg values on the x-axis and the computer's mpg values on the y-axis. The scatterplot shows a positive linear relationship between the two variables. There do not seem to be any significant outliers or unusual values. A linear relationship seems reasonable. b. Call: lm(formula = Computer ~ Driver, data = data) Coefficients: (Intercept) Driver 6.7532 0.9404 The least-squares regression line is: Computer = 6.7532 + 0.9404 * Driver c. The slope of the least-squares regression line is 0.9404, for every one unit increase in driver mpg, the computer mpg increases by approximately 0.94 units. The intercept of the line is 6.7532, indicating that the

computer mpg is expected to be 6.7532 when the driver mpg is 0. results suggest there is a strong positive linear relationship between the driver's mpg calculations and the computer's mpg calculations. The regression line has a positive slope and an intercept close to zero. Therefore, it appears that the computer and driver calculations are similar, and the differences can be explained by measurement error or other factors such as different driving conditions or variations in fuel consumption. 10.32 a. The scatterplot shows a roughly U-shaped relationship between the room temperature and the average number of correct answers for male students. As the temperature increases from around 16°C to 22°C, the average number of correct answers increases. However, as the temperature continues to increase beyond 22°C, the average number of correct answers starts to decrease.

b. The equation of the least-squares regression line for predicting Mave based on the room temperature is: Mave = 34.31 - 0.60(Temp) c. The coefficient of determination (r^2) for these data is 0.325. This tells us that the model explains 32.5% of the variability in the data. d. To check the conditions that must be approximately met for inference, we need to examine the residual plot, the normal probability plot, and the plot of residuals versus fitted values. e. H0: β1 = 0 (there is no linear relationship between temperature and Mave) Ha: β1 ≠ 0 (there is a linear relationship between temperature and Mave) t = (β1 - 0) / SE(β1) = -3.27 Using a two-tailed test with a significance level of 0.05, the critical value is ±2.064. Since the calculated

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t-statistic (-3.27) is less than the critical value, we reject the null hypothesis and conclude that there is significant evidence that temperature is associated with performance for male students. Specifically, the negative slope coefficient (β1 = -0.60) suggests that as the room temperature increases, the average number of correct answers for male students decreases. 10.44 a. Numerical and graphical methods for IBI: The mean IBI is 44.3, with a median of 45.5. The standard deviation is 10.7, with a minimum of 20 and a maximum of 68. The histogram of IBI shows a roughly normal distribution, with some skewness to the left. The boxplot of IBI shows no outliers. Numerical and graphical methods for area: The mean area is 25.2 km2, with a median of 20.1 km2. The standard deviation is 19.2 km2, with a minimum

of 1.4 km2 and a maximum of 69.5 km2. The histogram of area shows a right-skewed distribution, with a few streams having larger watershed areas. The boxplot of area shows several outliers. b. From the scatterplot of IBI versus area, we can see that there appears to be a slight negative relationship between IBI and area. However, there are some outliers and unusual patterns in the data, particularly for small watershed areas. c. The statistical model for simple linear regression for this problem is: IBI = β0 + β1*Area + ε where β0 is the intercept, β1 is the slope coefficient for Area, and ε is the error term. d. The null hypothesis is that there is no linear relationship between IBI and area (β1 = 0), while the alternative hypothesis is that there is a linear relationship (β1 ≠ 0). e. Running the simple linear regression, we get the following results: The slope coefficient β1 is -0.166 (p-value < 0.001), indicating a significant negative relationship between

IBI and area. The intercept β0 is 51.4 (p-value < 0.001). The R-squared value is 0.158, which indicates that only a small proportion of the variation in IBI can be explained by variation in area. f. Plotting the residuals versus area, I can see some non-random patterns in the plot, particularly for small watershed areas. There appear to be some outliers and a non-linear relationship between the residuals and area. g. The residuals do not appear to be approximately normal. The plot shows some non-random patterns, particularly for small watershed areas, indicating that the assumptions of normality may not hold for the residuals. h. The assumptions for the analysis of these data using the simple linear regression model may not be reasonable. The residuals plot shows some non- random patterns, indicating that the assumption of constant variance and normality may not hold. The presence of outliers and non-linear relationships may also violate the assumption of linearity. Further

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exploration and model refinement may be needed to improve the validity of the analysis.

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