Question: Forecasting daily admission of a water park. To determine whether extra personnel are needed for the day, the owners of a water adventure park would like to find a model that would allow them to predict the day’s attendance each morning before opening based on the day of the week and weather conditions. The model is of the form

where,

y = Daily admission

x₁ = 1 if weekend

0 otherwise

X₂ = 1 if sunny

0 if overcast

X₃ = predicted daily high temperature (°F)

These data were recorded for a random sample of 30 days, and a regression model was fitted to the data.

The least squares analysis produced the following results:

with

1. Interpret the estimated model coefficients.
2. Is there sufficient evidence to conclude that this model is useful for predicting daily attendance? Use α = .05.
3. Is there sufficient evidence to conclude that the mean attendance increases on weekends? Use α = .10.
4. Use the model to predict the attendance on a sunny weekday with a predicted high temperature of 95°F.
5. Suppose the 90% prediction interval for part d is (645, 1,245). Interpret this interval.

_{$,{\beta }_0$}is negative, indicating that the y-intercept is negative for the regression equation, while ${\beta }_1,{\beta }_2,and{\beta }_3$are all positive, indicating that the variables are all positively related to the dependent variable.

Here, ${\beta }_1$is positive, which implies that for every 1-unit increase in the predictor variable, the outcome variable will increase by the _{${\beta }_1$}value (here,25), similarly _,${\beta }_{2}and{\beta }_3$ are also positive; thus, for every 1 unit increase in the predictor variable, the outcome variable will increase by the ${\beta }_2$value (here,100) and ${\beta }_3$value (here,10).

$H_0\text{:}{\beta }_1={\beta }_2={\beta }_3=0$

Ha : At least one of the parameters ${\beta }_1,{\beta }_2,{\beta }_3$is non-zero.

Here, the F test statistic$=\frac{\frac{R^2}{k}}{\frac{\left(1-R^2\right)}{\left[n-\left(k+1\right)\right]}}=\frac{\frac{0.65}{3}}{\frac{1-0.65}{30-\left(3+1\right)}}=16.661$

Value of $F_{0.05,28,28}$is 1.901

H₀

is rejected if F statistic > $F_{0.05,28,28}$. For$\alpha =0.05$ , since$F>F_{0.05,28,28}$ sufficient to reject H₀ at 95% confidence interval.

Therefore,${\mathbf{\beta }}_{\mathbf{1}}\ne {\mathbf{\beta }}_{\mathbf{2}}\ne {\mathbf{\beta }}_{\mathbf{3}}\ne \mathbf{0}$which means we accept the alternative hypothesis.

$$\begin{gathered} H_0\text{:}{\beta }_1=0 \\ {H}_a\text{:}{\beta }_1\ne 0 \end{gathered}$$

Here, the t-test statistic

Value of $t_{0.05,24,24}$is 1.699

H₀ is rejected if t statistic >$t_{0.05,24,24}$. For , since $t>t_{0.05,24,24}$sufficient evidence to reject at a 95% confidence interval.

Therefore,localid="1660801540094" ${\mathit{\beta }}_{\mathbf{1}}\mathbf{=}\mathbf{0}$

The regression equation is $\hat{y}=-105+25x_1+100x_2+10x_3$.

The prediction value of daily attendance on sunny weekdays at95⁰F can be calculated when $x_1=0,x_2=1and{x}_3=95$

$$\begin{gathered} \hat{y}=-105+25\left(0\right)+100\left(1\right)+10\left(95\right) \\ =900 \end{gathered}$$

Therefore, the predicted attendance on a sunny weekday at a temperature of 95⁰Fis 900.

The 90% prediction interval for daily attendance is (645, 1245), indicating that the future values of the dependent variables will fall between the interval. From part d, the value of 900 is also falling into the prediction interval.