TABLE 16-12
the Manager of a Health Club Has Recorded

question 89

True/False

TABLE 16-12
The manager of a health club has recorded average attendance in newly introduced step classes over the last 15 months: 32.1, 39.5, 40.3, 46.0, 65.2, 73.1, 83.7, 106.8, 118.0, 133.1, 163.3, 182.8, 205.6, 249.1, and 263.5. She then used Microsoft Excel to
obtain the following partial output for both a first- and second-order autoregressive model.
SUMMARY OUTPUT - 2nd Order Model

$\begin{array} { l r }\text { Regression Statistics}\\ \text { Multiple R } & 0.993 \\ \text { R Square } & 0.987 \\ \text { Adjusted R Square } & 0.985 \\ \text { Standard Error } & 9.276 \\ \text { Observations } & 15 \end{array}$

$\begin{array}{ll} & \text { Coefficients } \\\text { Intercept } & 5.86 \\\text { X Variable 1 } & 0.37 \\\text { XVariable 2 } & 0.85\end{array}$
SUMMARY
OUTPUT - 1st Order
Model
Regression Statistics
$\begin{array} { l r } \text { Multiple R } & 0.993 \\ \text { R Square } & 0.987 \\ \text { Adjusted R Square } & 0.985 \\ \text { Standard Error } & 9.150 \\ \text { Observations } & 15 \end{array}$

$\begin{array}{ll} & \text { Coefficients } \\\text { Intercept } & 5.66\end{array}$
$\underline{\text {X Variable 1 \quad 1.10}}$
-Referring to Table 16-12, based on the parsimony principle, the second-order model is the better model for making forecasts.

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