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

EunoR 2021-02-27 Answered
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 stores
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svartmaleJ
Answered 2021-02-28 Author has 92 answers

Step 1
Given:
μ=Mean=0
σ=Standard deviation=22.92
Middle 95% of residuals
The lower cutoff value of the middle 95% has 100%95%2=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.
z=±1.96
The z-score is the value decreased by the mean, divided by the standard deviation.
z=xμσ=x022.92
The two found expressions of the z-score then have to be equal:
x5.322.92=±1.96
Multiply each side by 22.92:
x0=±1.96(22.92)
Add 5.3 to each side:
x=0±1.96(22.92)
Evaluate:
x=±44.9232
Thus the middle 95% of the residuals should be between -$44.9232 and $44.9232.
Step 2
Interval sales
x=5
y^=317.940+152.680x (result previous exercise)
Let us first determine the predicted value, by evaluating the equation of the
least-squares regression line at x=5.
y^=317.940+152.680(5)=317.940+763.40=1081.340
The residual is the difference between the actual y-value (sales) and the predicted y-value:
Residual =yy^
We are interested in the actual sales, thus let us add hat{y} to each side of the previous equation:
y=Residual +y^
We know that the residuals are between -$44.9232 and $44.9232.
y=Residual +y^=$44.9232+$1081.340=$1036.4168
y=Residual +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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