Stats-101A-HW-5

Stats 101A HW 5 Ian Zhang UID: 205702810 2023-05-05 Question 4 rse <- 2.418 df <- 33 r2 <- . 7254 fstat <- 87.17 RSS <- df * rseˆ 2 RSS ## [1] 192.9419 SSreg <- fstat * (RSS/df) SSreg ## [1] 509.6589 meanSSreg <- SSreg / df meanSSreg ## [1] 15.44421 totalSS <- SSreg + RSS totalSS ## [1] 702.6008 r <- sqrt(SSreg / totalSS) r ## [1] 0.8516977 Question 1 arm <- read.csv( "armspans2022_gender.csv" ) mean(arm$is.female) ## [1] 0.3478261 m1 <- lm(armspan ~ is.female, data = arm) summary(m1) ## ## Call: ## lm(formula = armspan ~ is.female, data = arm) ## ## Residuals: 1

## Min 1Q Median 3Q Max ## -9.7586 -2.0248 0.2414 2.2414 8.2414 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 69.7586 0.7399 94.284 < 2e-16 *** ## is.female -7.7338 1.2408 -6.233 1.68e-07 *** ## --- ## Signif. codes: 0 ' *** ' 0.001 ' ** ' 0.01 ' * ' 0.05 ' . ' 0.1 ' ' 1 ## ## Residual standard error: 3.984 on 43 degrees of freedom ## (1 observation deleted due to missingness) ## Multiple R-squared: 0.4746, Adjusted R-squared: 0.4624 ## F-statistic: 38.85 on 1 and 43 DF, p-value: 1.676e-07 plot(armspan ~ is.female, data = arm) 0.0 0.2 0.4 0.6 0.8 1.0 55 60 65 70 75 is.female armspan b) the intercept is 69.758, which is the estimated mean armspan of males. c) the slope is -7.7338, which means that the difference in the estimated mean armspan of males and females is 7.7338. this means that females armspans are 7.7338 shorter than male armspans on average d) the t statistic and p value is testing if there is a difference in mean armspans between males and females. The null hypothesis would be that the slope = 0, which means that there is no difference. The alternative would be that the slope != 0. The p-value for the slope is 1.68e-7, which means that we can reject the null hypothesis, meaning that there is a significant difference between the mean armspans between males and females Question 2 iowa <- read.delim( "iowatest.txt" ) temp <- ifelse(iowa$City== "Iowa City" , 1 , 0 ) iowa$is.iowa <- temp 2

Question 3 m3 <- lm(Test ~ Poverty, data= iowa) plot(Test ~ Poverty, data = iowa) abline(m3) 0 20 40 60 80 100 20 30 40 50 60 70 80 90 Poverty Test summary(m3) ## ## Call: ## lm(formula = Test ~ Poverty, data = iowa) ## ## Residuals: ## Min 1Q Median 3Q Max ## -27.2812 -6.2097 0.5058 4.8252 22.3610 ## ## Coefficients: ## Estimate Std. Error t value Pr(>|t|) ## (Intercept) 74.60578 1.61325 46.25 <2e-16 *** ## Poverty -0.53578 0.03262 -16.43 <2e-16 *** ## --- ## Signif. codes: 0 ' *** ' 0.001 ' ** ' 0.01 ' * ' 0.05 ' . ' 0.1 ' ' 1 ## ## Residual standard error: 8.766 on 131 degrees of freedom ## Multiple R-squared: 0.6731, Adjusted R-squared: 0.6707 ## F-statistic: 269.8 on 1 and 131 DF, p-value: < 2.2e-16 From the regression line, we can see that there is a strong linear association between test scores and poverty — as poverty increases, test scores decrease. The line appears to be a good fit to the data, and as we look at the summary, the p value for the slope of the line is <2e-16, which means that we are able to reject the null hypothesis of 0 slope, and we can assume that the slope is not 0, meaning that there is a linear association. 4

Question 4 plot(m3) 20 30 40 50 60 70 -30 -20 -10 0 10 20 Fitted values Residuals lm(Test ~ Poverty) Residuals vs Fitted 70 43 47 -2 -1 0 1 2 -3 -2 -1 0 1 2 3 Theoretical Quantiles Standardized residuals lm(Test ~ Poverty) Normal Q-Q 70 47 43 5

condition. Since the points are randomly scattered and there is no clear trend and the red line is basically horizontal and the values are equally spread around the line, this proves that the model is valid. Ultimately, based off of the 3 residual plots, we can determine that neither the constant variance or normal distribution conditions were violated and the model is valid. Question 5 leverage <- hatvalues(m3) leverage[leverage == max(leverage)] ## 27 ## 0.04997855 #row 27 highlev <- leverage > ( 4 / 133 ) standardRes <- rstandard(m3) standardRes[highlev] < - 2 ## 7 27 46 64 67 89 109 120 126 ## FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE standardRes[highlev] > 2 ## 7 27 46 64 67 89 109 120 126 ## FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE FALSE Since there are no high leverage points that are < -2 or > 2, so there are no bad leverage points in this data. Question 6 The f test in question 3 tests whether there is a significant relationship between poverty and test scores. The null hypothesis is that the slope of the regression line is 0, and the alternate hypothesis is that it != 0. If the slope of the line is 0, that means there is not a linear association, whereas if it isn’t 0, then there would be a relationship between the 2. The p-value for this test is 2.2e-16, which is less than 0.05, thus we have significant evidence to reject the null hypothesis. This means that we can conclude there is a linear association between poverty and test scores. 7