Loading the libraries

library(tidyverse)

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## ✔ lubridate 1.9.3     ✔ tidyr     1.3.1
## ✔ purrr     1.0.2     
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## ℹ Use the conflicted package (<http://conflicted.r-lib.org/>) to force all conflicts to become errors

loading the data set

setwd("C:/Users/eyong/Downloads/New folder - Copy")
nba <- read_csv("NBAStandings2019.csv")

## Rows: 30 Columns: 6
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## chr (1): Team
## dbl (5): Wins, Losses, WinPct, PtsFor, PtsAgainst
## 
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.

clean the colmun names

model_1 <-  lm(WinPct~  PtsFor+PtsAgainst ,data = nba)
summary(model_1)

## 
## Call:
## lm(formula = WinPct ~ PtsFor + PtsAgainst, data = nba)
## 
## Residuals:
##       Min        1Q    Median        3Q       Max 
## -0.065838 -0.017353  0.000247  0.019217  0.055061 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept)  0.302999   0.198357   1.528    0.138    
## PtsFor       0.030244   0.001405  21.522   <2e-16 ***
## PtsAgainst  -0.028470   0.001494 -19.050   <2e-16 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.03003 on 27 degrees of freedom
## Multiple R-squared:  0.961,  Adjusted R-squared:  0.9581 
## F-statistic: 332.5 on 2 and 27 DF,  p-value: < 2.2e-16

Equation: Winpct =0.303 +0.0302 * ptsFor -0.0285 * PtsAgainst

The eventual NBA playoff champion Toronto Raptors won 58 games in the regular season while losing only 24 games (WinPCT = 0.707). They scored an average of 114.4 points per game while giving up an average of 108.4 points against. Find the predicted winning percentage for the Raptors using this model and compute the residual.

Comment on the effectiveness of each predictor in this model. (p-values) both p-values are significant .therfore deffnese and offesnes stategies are improatnt for winning games

Do we do much better by including both predictors in a model? Choose some measure to compare the effectiveness of a simple linear model based on either PtsFor or PtsAgainst to this two-predictor model ## the model with the first predictor is better

prop.test(c(84200, 102598  ), c(144790 , 211693), alternative = "greater")

## 
##  2-sample test for equality of proportions with continuity correction
## 
## data:  c(84200, 102598) out of c(144790, 211693)
## X-squared = 3234.9, df = 1, p-value < 2.2e-16
## alternative hypothesis: greater
## 95 percent confidence interval:
##  0.09408942 1.00000000
## sample estimates:
##    prop 1    prop 2 
## 0.5815319 0.4846547

WinPct = 0.303 + 0.0302PTSfor - 0.0285 PysAgainst WinPct= 0.303 + 0.0302* 114 - 0.0285* 108.4 = 0.676 = 67.6%

Residual :0.707-0676 = 0.031

C/ Effectiveness of each predictor in this model :

info<- summary(model_1)
info$coefficients

##                Estimate  Std. Error    t value     Pr(>|t|)
## (Intercept)  0.30299945 0.198357432   1.527543 1.382574e-01
## PtsFor       0.03024363 0.001405245  21.521964 1.574358e-18
## PtsAgainst  -0.02846984 0.001494489 -19.049885 3.485649e-17

Both P_values are significant. Consequentlty, both defense and offense strategies are important for winning games.

Both Predictors , r_squared = 0.961 with one :

Model 1

model_1 <- lm(WinPct ~ PtsFor, data = nba)
summary(model_1)

## 
## Call:
## lm(formula = WinPct ~ PtsFor, data = nba)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.19736 -0.08943  0.02367  0.08390  0.17095 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -2.133765   0.565723  -3.772 0.000772 ***
## PtsFor       0.023684   0.005084   4.659 7.05e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.1121 on 28 degrees of freedom
## Multiple R-squared:  0.4366, Adjusted R-squared:  0.4165 
## F-statistic:  21.7 on 1 and 28 DF,  p-value: 7.051e-05

model_2 <- lm(WinPct ~ PtsFor, data = nba)
summary(model_2)

## 
## Call:
## lm(formula = WinPct ~ PtsFor, data = nba)
## 
## Residuals:
##      Min       1Q   Median       3Q      Max 
## -0.19736 -0.08943  0.02367  0.08390  0.17095 
## 
## Coefficients:
##              Estimate Std. Error t value Pr(>|t|)    
## (Intercept) -2.133765   0.565723  -3.772 0.000772 ***
## PtsFor       0.023684   0.005084   4.659 7.05e-05 ***
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 
## Residual standard error: 0.1121 on 28 degrees of freedom
## Multiple R-squared:  0.4366, Adjusted R-squared:  0.4165 
## F-statistic:  21.7 on 1 and 28 DF,  p-value: 7.051e-05

R_ squred = 0.42 when considering points forthe team and R-squared = 0.27 when only considering points against . therefore , including both predictors in the model is mich better .

Difference in Means Example: Use StudentSurvey dataset to test whether the data provide

evidence that there is a difference in the mean number of hours spent watching television per week between males and females.

library(tidyverse)
setwd("C:/Users/eyong/Downloads/New folder - Copy")
student_survey <- read_csv("StudentSurvey.csv")

## Rows: 79 Columns: 17
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## chr  (5): Year, Sex, Smoke, Award, HigherSAT
## dbl (12): Exercise, TV, Height, Weight, Siblings, BirthOrder, VerbalSAT, Mat...
## 
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.

Perform t-test for TV watching hours by gender

Hypotheses

H0: μ_male = μ_female No difference in mean TV watching hours

H1: μ_male is not equal to μ_female There is a difference in mean TV watching hours

tv_hours_test <- t.test(TV ~ Sex, data = student_survey)

tv_hours_test$p.value

## [1] 0.001271132

Conclusion:

p.value is 0.001271132 which is less than 0.05, Reject the null hypothesis. There is significant evidence of a difference in TV watching hours between males and females.

ICU Infections

Hypotheses

H0: Null Hypothesis There is no difference in the proportion of ICU patients with an infection between males and females

H1: There is a difference in the proportion of ICU patients with an infection between males and females.

setwd("C:/Users/eyong/Downloads/New folder - Copy")
icu_data <- read_csv("ICUAdmissions.csv")

## Rows: 200 Columns: 21
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## dbl (21): ID, Status, Age, Sex, Race, Service, Cancer, Renal, Infection, CPR...
## 
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.

Count the number of infected males and females

infection_prop_test <- prop.test(
  x = table(icu_data$Sex, icu_data$Infection)[,2],
  n = table(icu_data$Sex)
)
infection_prop_test$p.value

## [1] 0.8640718

conclusion

the p value is 0.8640718 which is greater than so Fail to reject the null hypothesis There is not enough evidence to conclude a difference in infection proportions between males and females

3 Chi-Squared Test for Black Friday Shopping

H0: No association between sex and Black Friday shopping plans H1: There is an association between sex and Black Friday shopping plans

shopping_data <- data.frame(
  Male = c(82, 100),
  Female = c(433, 400),
  row.names = c("Shopping", "Not Shopping")
)


# Perform Chi-squared test
chi_squared_test <- chisq.test(shopping_data)
chi_squared_test$p.value

## [1] 0.1071273

Conclusion

p.value is 0.1071273 which is greater 0.05

Fail to reject the null hypothesis. There is not enough evidence to conclude an association between sex and Black Friday shopping plans

4 ANOVA Test for School Pressure and Hang Hours

Hypotheses

H0: Mean hang hours are the same across all school pressure levels H1: At least one school pressure level has a different mean of hang hours

setwd("C:/Users/eyong/Downloads/New folder - Copy")
pa_seniors <- read_csv("PASeniors.csv")

## Rows: 457 Columns: 36
## ── Column specification ────────────────────────────────────────────────────────
## Delimiter: ","
## chr (16): Gender, Hand, GetToSchool, Activity, Music, BirthMonth, Season, Al...
## dbl (20): Year, Age, Height, Foot, Armspan, Languages, TravelTime, ReactionT...
## 
## ℹ Use `spec()` to retrieve the full column specification for this data.
## ℹ Specify the column types or set `show_col_types = FALSE` to quiet this message.

anova model

anova_model <- aov(HangHours ~ SchoolPressure, data = pa_seniors)
anova_model

## Call:
##    aov(formula = HangHours ~ SchoolPressure, data = pa_seniors)
## 
## Terms:
##                 SchoolPressure Residuals
## Sum of Squares         1563.48  48557.41
## Deg. of Freedom              3       443
## 
## Residual standard error: 10.4695
## Estimated effects may be unbalanced
## 10 observations deleted due to missingness

summary(anova_model)

##                 Df Sum Sq Mean Sq F value  Pr(>F)   
## SchoolPressure   3   1563   521.2   4.755 0.00283 **
## Residuals      443  48557   109.6                   
## ---
## Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
## 10 observations deleted due to missingness

conclusion

the p value is 0.00283 which is less than 0.05 so we fail to reject the null hypothesis
There is significant evidence that the amount of school pressure is related to time spent hanging out with friends

Extra credit

Eyong

2024-11-19

Loading the libraries

loading the data set

clean the colmun names

Both P_values are significant. Consequentlty, both defense and offense strategies are important for winning games.

Model 1

R_ squred = 0.42 when considering points forthe team and R-squared = 0.27 when only considering points against . therefore , including both predictors in the model is mich better .

Difference in Means Example: Use StudentSurvey dataset to test whether the data provide

Perform t-test for TV watching hours by gender

Hypotheses

H0: μ_male = μ_female No difference in mean TV watching hours

H1: μ_male is not equal to μ_female There is a difference in mean TV watching hours

Conclusion:

ICU Infections

Hypotheses

H0: Null Hypothesis There is no difference in the proportion of ICU patients with an infection between males and females

H1: There is a difference in the proportion of ICU patients with an infection between males and females.

Count the number of infected males and females

conclusion

3 Chi-Squared Test for Black Friday Shopping

Conclusion

4 ANOVA Test for School Pressure and Hang Hours

Hypotheses

anova model

conclusion