---
title: "Answers"
output: pdf_document
---

## Question 1:

i. In this data we note that the number of hours per month is less variable for first few months. As month increases, the number of hours becomes for variable across samples. The scatter plot of the data shows that there is a quadratic relationship between the number of hours and months. Therefore, we can use $x, x^2$ as the independent variable where $x$ is tje number of months.

ii. a) The fitted model is $y = 20.7 - 23.23\times x + 8.113 \times x^2$.

b) We note from the above model that the coefficients of $x$ and $x^2$ are significant at $5\%$ level of significance. This is simply because of the small p-value for the respective variables in the Coefficients table. Secondly the adjusted R-square value of $95.5\%$ indicate that the model fit is very good. Hence nearly $95.5\%$ of the variability in the number of months can be explained by months.

c) We first find the $90\%$ confidence interval for the coefficient of $x$. This is given by

```{r}
# Lower confidence level
Lower = c(-23.23 - qt(0.95, df = 15)* 9.159)
# Upper confidence level
Upper = c(-23.23 + qt(0.95, df = 15)* 9.159)
# combine them together
confidence_x = c(Lower = Lower, Upper = Upper)
confidence_x
```

Next, we find the $90\%$ confidence interval for the coefficient of $x^2$.

```{r}
# Lower confidence level
Lower = c(8.113 - qt(0.95, df = 15)* 1.281)
# Upper confidence level
Upper = c(8.113 + qt(0.95, df = 15)* 1.281)
# combine them together
confidence_x_square = c(Lower = Lower, Upper = Upper)
confidence_x_square
```

d) We need to compute the $90\%$ predictive interval for the future observation $y$. We know that such interval is given by

$$
\left(\hat{y} - t_{\alpha/2, n-p} \sqrt{\hat{\sigma}^2(1 + \mathbf{x}'(X'X)^{-1}\mathbf{x})}, \quad \hat{y} + t_{\alpha/2, n-p} \sqrt{\hat{\sigma}^2(1 + \mathbf{x}'(X'X)^{-1}\mathbf{x})} \right).
$$
We have been given the matrix $(X'X)^{-1}$
```{r}
XX_inverse = matrix(c(1.066, -0.650, 0.083, -0.650, 0.456, -0.063, 0.083,-0.063,0.009),
nrow = 3, byrow = T)
XX_inverse
```

We first find

PROBLEM 2 Suppose Y follows a bivariate normal distribution with p...