Question

(R programming) The weight and systolic blood pressure of 26 randomly selected males in the age group 25-30 are found in the attached CSV file. Fit a simple linear regression model relating systolic blood pressure (y) to weight (2). (30 points) (a) Provide the R code which fits the model and outputs the summary of the fitting. (b) Report the model fitting summary (c) Write down the fitted linear model and its interpretation. (d) Write down the null and alternative hypotheses for testing the slope, B1. Draw a conclusion based on the observed test statistic and p-value.

Rstudio programming

Answer 1

a)

R code

® RStudio File Edit Code - OR View Plots Build Session O Debug Profile Tools Go to file/function Help 69 - Addins - H Project: (None) - O Untitled7* X Untitled6 Q test ® MSR.RX ® sample_size_calculation.RX ® study of acute effect of caffier >> Sp a H Source on Save a . I + Run + Source - 1 weight=c(165,167,180,155,212,175,190,210,200,149, 158, 169, 170,172,159,168, 174,183,215,195,180,143,240,235,192,187) 3 sys_bp=c(130,133,150,128,151,146,150,140,148,125,133,135,150,153,128,132, 149, 158, 150, 163,156,124, 170, 165, 160,159) fit=1m (sys_bp-weight) ;fit 6 summary (fit) Environment History Connections H E Import Dataset List - Global Environment - Data fit List of 12 Values sys_bp num [1:26] 130 133 150 128... weight num [1:26] 165 167 180 155. 6:13 (Top Level) R Script Files Plots Packages Zoom Help Viewer Export - 3 + Console - > weight=C(165,167,180,155,212,175,190,210,200,149,158,169,170,172,159,168, 174,183,215,195,180,143,240,235,192,187) > weight=c(165, 167, 180,155,212,175,190,210,200,149, 158, 169, 170,172,159,168, 174.183.215 195' 174,183,215,195,180,143,240,235,192,187) > sys_bp=C(130,133,150, 128, 151, 146,150,140,148,125,133,135,150,153,128,132, 149, 158, 150, 163,156,124,170, 165, 160, 159) > fit=1m (sys_bp-weight) > fit=1m (sys_bp-weight) ;fit + Call: Im(formula = sys_bp ~ weight) Coefficients: (Intercept) 69. 1044 weight 0.4194 > summary(fit) Activate Windows Go to Settings to activate Windows. Call: Im(formula = sys_bp ~ weight) A 4:04 PM ENG 27/01/2020

b)

R-output Model fitting summary

Version:1.0 StartHTML:0000000107 EndHTML:0000002858 StartFragment:0000000127 EndFragment:0000002840

> summary(fit) Call: lm(formula = sys_bp ~ weight) Residuals: Min 1Q Median 3Q Max -17.182 -6.485 -2.519 8.926 12.143 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 69.10437 12.91013 5.353 1.71e-05 *** weight 0.41942 0.07015 5.979 3.59e-06 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1 Residual standard error: 8.681 on 24 degrees of freedom Multiple R-squared: 0.5983, Adjusted R-squared: 0.5815 F-statistic: 35.74 on 1 and 24 DF, p-value: 3.591e-06

c)

Fitted linear regression model is

sys_bp =69.10437 + 0.41942 * weight

Interpretation :The Values of multiple R-squared: 0.5983 and Adjusted R-squared: 0.5815 are not closed to 1 so we can conclude that fitted model is not good linear fit between sys_bp and weight.

d)

Ho : B1 =0 ie. weight is not significant to predict sys_bp

VS

H1 : B1 $\neq$ 0 i.e. weight is statistically significant.

P-value : 3.59e-06 = 0.00000359 < 0.05

   so we reject null hypothesis (H0)

Conclusion : B1 $\neq$ 0 so weight is statistically significant with $\alpha$ =0.05

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