---
title: "Linear regression"
author: "Alizeh Azhar"
date: "9/20/2021"
output: pdf_document
---

```{r setup, include=FALSE}
knitr::opts_chunk$set(echo = TRUE)
```

## Problem 1: OLS Exercise 8.4
a.	The plot does not form a linear pattern, the pattern is nonlinear, linear model is not reasonable in this case.
b.	The plot also form a nonlinear pattern, linear model is not reasonable in this case.
c.	The plot suggests presence of strong linear relationship, because it forms a perfect linear pattern. Application of linear model is reasonable
d.	The pattern formed by the plot suggests weak linear relationship. Linear model is a reasonable 
e.	The pattern linear pattern formed by the plot suggests presence of weak linear relationship. Linear model can be applied
f.	He pattern formed suggests moderate linear relationship. Linear model is reasonable because it forms linear pattern.

## Problem 2: OLS Exercise 8.5
a.Exam 2 has the strongest correlation with the final exam. This is because the plot forms a better linear pattern as compared to Exam 1. Exam 1 has many outliers while majority of the points in Exam 2 linear on or closer to the fitted line as compared to exam 1.
b.	Exam 2 is close to final exam, the student is more prepare for Exam2 than Exam 1 because it’s the predictor of final performance.

## Problem 3:OLS Exercise 8.5
a)	There is strong linear relationship between husband and wife age. As husband’s age increases woman’s age rises.
b)	There is weak relationship between husband’s and wife’s height.
c)	Plot of husband and wife’s age shows stronger correlation. There is a clear linear pattern with few outliers.
d)	Conversion of units for heights from centimeters to inches does not influence the correlation between husband’s and wife’s height.

## Problem 4: OLS Exercise 8.20
0.5 shows under estimation of the incidences of skin cancer. This is because the residual is positive.

## Problem 5: OLS Exercise 8.24
	a) Equation of the regression line 
```{r, echo=FALSE}
#Y_bar=b0+b1 x-BAR
#b1=(Cov(xy))/sX^2 
b1=0.67/(10.37*10.37)
b1
#b0=Y_bar-b1X_bar
b0=171.4-b1*107.20
b0
# Equation
#height=170.73+0.00624 girth
```
  Equation
height=170.73+0.00624 girth
  b)	Interpretation
The model states that a change in girth by 1 cm increases height by 0.00623 cm.

c)	Predicted height 

```{r, echo=FALSE}
height=170.73+0.00624*100
height
```

d)	Residual 
```{r, echo=FALSE}
Residual=160-171.5 
Residual 
```
The residual means that the height is over predicted by 11.5 cm.

## Problem 6
a. Using lm(), fit a linear regression model predicting birthweight from gestation
```{r, echo=FALSE}
babies <-read.csv("https://www.openintro.org/data/csv/babies.csv")
attach(babies)
m1=lm(bwt~gestation)
summary(m1)
```
b.	 Write the equation of the line using mathematical notation (2 pts)

y=-10.06418+0.46426 x, Where y represent birth weight while x represent gestation
`