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:

\(\displaystyle{x}-{0}=\pm{1.96}{\left({22.92}\right)}\)

Add 5.3 to each side:

\(\displaystyle{x}={0}\pm{1.96}{\left({22.92}\right)}\)

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.