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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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    I am still confused because I am not sure what $\epsilon$ stands for here? Or where I should apply the standard deviation and mean to the problem.2012-11-29
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    The $\epsilon$ is a normally distributed random variable with mean $0$, it is the "error." You might want to give it a more random variable type of name, such as $W$. Then our profit in the new shop is the random variable $7+W$.2012-11-29
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    Thanks a lot! I am going to reattempt the problem with your given suggestion.2012-11-29
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    Andre, you think you can provide me a hint on how to do part a with the graph and profit? I know how to do part b now with your guidelines given above.2012-11-29
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    OK. In the post, I changed $X$ to $Y$ to be more consistent with your notation, and with what follows. In (almost) the notation of the problem, the profit $Y$ is related to the size $x$ by $Y=2x+1+W$, where $W$ is a random variable with mean $0$ that represents the inevitable variation there will be. So $E(Y)=2x+1+E(W)$. But $E(W)=0$, so $E(Y)=2x+1$. The line $y=2x+1$ represents mean profit $y$ as a function of store size $x$.2012-11-29
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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