Please show all parts. This problem is to be solved by hand. The weight and syst

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Please show all parts.

This problem is to be solved by hand. The weight and systolic blood pressure of 12 randomly selected males in the age group 25-30 are shown in the table below 3. Subject 12 Weight (ibs), x 165 167 180 165 212 175 190 212 200 | 149 158 169 y130 133 150 128 151146150 140 148 125 133 135 Systolic BP, (a) Plot the points on a scatter diagram. Does there appear to be a linear trend? Estimate the least- squares regression line (b) Estimate the average systolic BP for a male that weighs 195. (c) Construct a 95% confidence interval for -Interpret the interval. (d) Construct a 95% confidence interval and a 95% prediction interval for the average systolic BP if x- 195 lbs. Explain the difference between these two intervals (e) Test for significance of regression coefficient using the two-sided t-test and = 0.05 (f) Construct an analysis of variance table and use it to perform the hypothesis test of part (e) Restate the null and alternative hypotheses and interpret your results. (g) Perform a residual analysis on the data. (h) Summarize your analysis of parts (a)-(g) in paragraph form to determine the appropriateness of the model.

Explanation / Answer

x=c(165,167,180,165,212,175,190,212,200,149,158,169)
y=c(130,133,150,128,151,146,150,140,148,125,133,135)
#(a):
#Part 1: Scatter diagram
plot(x,y)
#Comment: From the scatter diagram we can conclude that there is linear trend is
#present and there is positive correlation between them.
#part 2: Estimation of the Regression equation
lm(y~x)
#Hence the regression eqn is: y=76.43289+0.35098 x
#(b) For weight 195 we have to estimate Systolic Bp .
y=76.43289+0.35098*195
#hence the estimated systolic Bp is 144.874
#(c)
#Part 1:95% Confidence interval for B1
summary(lm(y~x))
#here estimate of B1 is b1=0.35098
#standard error of B1 0.09423
qt(0.975,10)
#Critical value is upper 97.5% point of t distribution with (n-2)=10 df,2.228139
#hence upper confidence limit b1(estimated B1)+standard error*critical value=0.5609375
#hence lower confidence limit b1(estimated B1)-standard error*critical value=0.1410225

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