# The manager of the store in the preceding exercise calculated the residual for each point in the scatterplot and made a dotplot of the residuals. The distribution of residuals is roughly Normal with a mean of $0 and standard deviation of$22.92. The middle 95% of residuals should be between which two values? Use this information to give an interval of plausible values for the weekly sales revenue if 5 linear feet are allocated to the store's brand of men's grooming products.

Question
Normal distributions
The manager of the store in the preceding exercise calculated the residual for each point in the scatterplot and made a dotplot of the residuals.
The distribution of residuals is roughly Normal with a mean of $0 and standard deviation of$22.92.
The middle 95% of residuals should be between which two values? Use this information to give an interval of plausible values for the weekly sales revenue if 5 linear feet are allocated to the store's brand of men's grooming products.

2021-02-28

Step 1
Given:
$$\displaystyle\mu={M}{e}{a}{n}={0}$$
$$\displaystyle\sigma=Standard\ deviation={22.92}$$
Middle 95% of residuals
The lower cutoff value of the middle 95% has $$\displaystyle{\frac{{{100}\%-{95}\%}}{{{2}}}}={\frac{{{5}\%}}{{{2}}}}={2.5}\%$$ of the data values below it.
Similarly, the upper cutoff value of the middle 95% has 2.5% of the data values above it and thus $$100\% - 2.5\% = 97.5\%$$ of the data values below it.
Let us determine the z-score that corresponds with a probability of 2.5% or 0.025 in the normal probability table of the appendix.
We note that the closest probability is 0.0250 which lies in the row -1.9 and in the column .06 of the normal probability table and thus the corresponding z-score is then $$1.9 + 06 = -1.96$$.
Similarly, we obtain that the z-score 1.96 corresponds with a probability of 0.9750.
$$\displaystyle{z}=\pm{1.96}$$
The z-score is the value decreased by the mean, divided by the standard deviation.
$$\displaystyle{z}={\frac{{{x}-\mu}}{{\sigma}}}={\frac{{{x}-{0}}}{{{22.92}}}}$$
The two found expressions of the z-score then have to be equal:
$$\displaystyle{\frac{{{x}-{5.3}}}{{{22.92}}}}=\pm{1.96}$$
Multiply each side by 22.92:
$$\displaystyle{x}-{0}=\pm{1.96}{\left({22.92}\right)}$$
$$\displaystyle{x}={0}\pm{1.96}{\left({22.92}\right)}$$
Evaluate:
$$\displaystyle{x}=\pm{44.9232}$$
Thus the middle 95% of the residuals should be between -$44.9232 and$44.9232.
Step 2
Interval sales
$$x=5$$
$$\displaystyle\hat{{{y}}}={317.940}+{152.680}{x}$$ (result previous exercise)
Let us first determine the predicted value, by evaluating the equation of the
least-squares regression line at $$x = 5$$.
$$\displaystyle\hat{{{y}}}={317.940}+{152.680}{\left({5}\right)}={317.940}+{763.40}={1081.340}$$
The residual is the difference between the actual y-value (sales) and the predicted y-value:
Residual $$\displaystyle={y}-\hat{{{y}}}$$
We are interested in the actual sales, thus let us add hat{y} to each side of the previous equation:
y=Residual $$\displaystyle+\hat{{{y}}}$$
We know that the residuals are between -$44.9232 and$44.9232.
$$y=$$Residual $$\displaystyle+\hat{{{y}}}=-\{44.9232}+\{1081.340}=\{1036.4168}$$
$$y=$$Residual $$\displaystyle+\hat{{{y}}}=-\{44.9232}+\{1081.340}=\{1126.2632}$$
Thus the plausible values for the weekly sales revenue are then between
$1036.4168 and between$1126.2632.
Result:
Residuals: Between -$44.9232 and$44.9232.
Sales:Between $1036.4168 and between$1126.2632.

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