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a.) Draw the graph of the straight line with equation $y = 2x + 1$. Now assume that a company has several shops. Let $Y_i$ be the profit the shop number $i$ makes in the coming year. Let $x_i$ be the size of the shop number $i$. We assume that for all these shops the following relationship holds. $Y_i = 2x_i + 1 + \epsilon_i$ where $\epsilon_i$ is a random term for which $E[\epsilon_i] = 0$ and such that $\epsilon_1,\epsilon_2,...$ are $i.i.d$. So, if $\alpha = 2$ and $\beta = 1$, we can write $Y_i = \alpha + \beta x_i +\epsilon_i$.

Now, the company plans to open a new shop with size 3. What is the expected profit for that shop? Also, write it in terms of $\alpha$ and $\beta$. What does the straight line $y = \alpha + \beta x$ represent?

b.) Assume that the errors are normal and that it is known that the standard deviation of $\epsilon = 2$ for all shops. Give a $95 \%$-confidence interval for the profit of that shop with size 3.

For part a, I drew the graph which is the easy part , but I do not know what it represents or how to do the problem.

For part b, I know that the profit of the shop is going to be a normal with known expectation and standard deviation, so the confidence interval is simple expected value plus/minus standard deviation times $c$, where $c$ is the constant so that $P(-c \le N(0,1) \le c) =0.95$

but then I get lost on what I have to do next.

1 Answers 1

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The profit for the shop of size $3$ is a normally distributed random variable $Y$ with mean $7$ and standard deviation $2$.

From a table of the standard normal, you can see that with probability $0.95$, a normally distributed random variable will be within $1.96$ standard deviation units of the mean.

More formally, we want the $d\gt 0$ such that $\Pr(-d\le Y-7\le d)=0.95.$ Divide by the standard deviation $2$, and recall that $\dfrac{Y-7}{2}$ is standard normal. So we want $d$ such that $\Pr\left( -\frac{d}{2} \le Z \le \frac{d}{2}\right)=0.95.$

Or else let's give the random variable called $\epsilon$ a more random variable type of name, like $W$. Then $Y=7+W$. Since $W$ has mean $0$ and standard deviation $2$, it follows that $Y$ has mean $7$ and standard deviation $2$. We want to find a $95\%$ (symmetric) confidence interval for $Y$. So let's find a $95\%$ confidence interval for $W$. This is the interval $[-(2)(1.96),(2)(1.96)]$. From this we can readily find the confidence interval for $Y$.

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    All the subscripts, like $i$ in $x_i$, are probably there in preparation for a question about the company's **total** profit.2012-11-29