Body weight Bench press 126 130 202 165 176 154 154 130 209 174 133 117 128 110
ID: 3361736 • Letter: B
Question
Body weight Bench press
126 130
202 165
176 154
154 130
209 174
133 117
128 110
168 149
152 114
157 138
188 157
219 184
213 161
205 170
218 161
219 179
137 124
124 135
133 131
212 157
186 155
125 121
138 122
193 152
162 144
215 173
142 137
196 168
149 159
219 163
3. 115 points I Previous Answers My Note An exercise science major wants to try to use body weight to predict how much someone can bench press. He collects the data shown below on 30 male students. Both quantities are measured in pounds Assignment 10q3 data a) What type of association does there appear to be between these two variables? O positive association O no association O can not be determined negative association b) Compute a 95% confidence interval for the average bench press of 150 pound males, what is the lower limit? Give your answer to two decimal places. c) compute a 95% confidence interval for the average bench press of 150 pound males, what is the upper limit? Give your answer to two decimal places. d) Compute a 95% prediction interval for the bench press of a 15 pound male what is the lower limit? Give your answer to two decima places e) compute a 95% prediction interval for the bench press of a 150 pound male, what is the upper limit? Give your answer to two decimal places.Explanation / Answer
we can do this in R as shown below , The R snippet is as follows
# read the data into R dataframe
data.df<- read.csv("C:\Users\586645\Downloads\Chegg\bench.csv",header=TRUE)
str(data.df)
# perform anova analysis
a<- lm(Bench.press~ Body.weight,data=data.df)
#summarise the results
summary(a)
##
newdat <- data.frame(Body.weight= 150)
# 95% Confidence interval
predict(a,newdat,interval = "confidence",level=0.95)
# 95% Prediction interval
predict(a,newdat,interval = "prediction",level=0.95)
The result is
> summary(a)
Call:
lm(formula = Bench.press ~ Body.weight, data = data.df)
Residuals:
Min 1Q Median 3Q Max
-22.479 -7.414 1.631 5.710 24.118
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 55.56362 9.08100 6.119 1.33e-06 ***
Body.weight 0.53234 0.05142 10.353 4.45e-11 *** as the coefficient is positve hence there is a positive relation
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1
Residual standard error: 9.64 on 28 degrees of freedom
Multiple R-squared: 0.7929, Adjusted R-squared: 0.7855
F-statistic: 107.2 on 1 and 28 DF, p-value: 4.447e-11
The intervals are
> # 95% Confidence interval
> predict(a,newdat,interval = "confidence",level=0.95)
fit lwr upr
1 135.4143 131.0551 139.7734
> # 95% Prediction interval
> predict(a,newdat,interval = "prediction",level=0.95)
fit lwr upr
1 135.4143 115.1926 155.6359