Since Y depends linearly on X and Y is normally distributed, this is a perfect scenario for simple linear regression.
1) there are formulas for this, namely $\hat{\beta_1} = \frac{\sum_{i}(x_i-\bar{x})y_i }{\sum_{i} (x_i-\bar{x})^2} = 0.667$ and $\hat{\beta_0} = \bar{y} - \hat{\beta_1}\bar{X} = 1.859$. The hat on the betas means that they are estimated instead of the true population value. So the analyst believes that the expected profit in year X is $E(Y|X) =1.859+0.667X$
2) This is a one sided hypothesis test.
$H_0: \beta_1 \leq 0.5$
$H_1: \beta_1 > 0.5$
for the critical value we use a t table like this one: https://www.dummies.com/wp-content/uploads/451675.image0.jpg. We have 10 observations and 2 estimated betas so there are 10-2 = 8 degrees of freedom. The significance level is 0.005 so according to the table the critical value is 3.35539. Now we need the standard error of $\hat{\beta_1}$, the formula for which is $SE(\hat{\beta_1}) = \sqrt{\frac{\sigma^2}{\sum_{i} (x_i - \bar{x}) } } = 0.0293$.
The t value of our sample is $\frac{\hat{\beta_1} - \beta_1}{SE(\hat{\beta_1})} = 5.7$.
Since 5.7 is larger than the critical value of 3.35539, we reject the null hypothesis at the 0.5% significance level. There is enough evidence to conclude that the true beta is larger than 0.5.
3) We can first use the regression equation to get a point estimate of Y at X = 5. $E(Y|X = 5) =1.859+0.667*5 = 5.194$. The confidence interval will be of the form $5.194\pm t_{0.005;8}*s_{\bar{y}}$, notice that this is the same t value that we have already used in the last question, so 3.33539. $s_{\bar{y}}$ is the standard error of the mean of Y at a given X.
There also exists a formula for this, namely $s_{\bar{y}} = \sigma\sqrt{\frac{1}{10} + \frac{(x - \bar{x})^2}{\sum_{i}(x_i -\bar{x})^2 } } = 0.04823$ so the confidence interval is $5.194\pm 3.33539*0.04823 = [5.033, 5.35]$.
So we can be 95% confident that when X=5, the mean of Y is between 5.033 and 5.35.
4) We can again use the regression equation to find a point estimate $E(Y|X = 1) =1.859+0.667*1 = 2.526$. The prediction interval takes a similar form to the confidence interval, namely $2.526 \pm t_{\alpha/2}*s_{pred,y}$. Since the significance level has changed to 5%, we are now looking for the t at 0.05/2 = 0.025 and with 8 degrees of freedom, according to the table this is 2.306. $s_{pred,y}^2$ is the variance in the prediction of y, which is composed of the variance of Y and the variance of the mean of Y, both of which we have calculated before. $s_{pred,y}^2 = s^2 + s_{\bar{y}}^2 = 0.08^2 + 0.04823^2 = 0.008726$, so that the square root of that is $s_{pred,y} = 0.0934$
Putting in all the numbers we get $2.526 \pm 2.306*0.0934 = [2.311, 2.741]$. So we can be 95% confident that a new observation of Y at X = 1 is between 2.311 and 2.741
I hope that answers everything!
