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

Short Answer

Expert verified

Answer

  1. 0is negative, indicating that the y-intercept is negative for the regression equation, while 1,2,and3 are all positive, indicating that the variables are all positively related to the dependent variable.
  2. At 95% significance level,1230 .
  3. At 95% significance level, 1=0.
  4. The predicted attendance on a sunny weekday at a temperature of 95 0F is 900.
  5. 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.

Step by step solution

01

(a) Interpretation of beta coefficient

,0is negative, indicating that the y-intercept is negative for the regression equation, while 1,2,and3are all positive, indicating that the variables are all positively related to the dependent variable.

Here, 1is positive, which implies that for every 1-unit increase in the predictor variable, the outcome variable will increase by the 1value (here,25), similarly ,2and3 are also positive; thus, for every 1 unit increase in the predictor variable, the outcome variable will increase by the 2value (here,100) and 3value (here,10).

02

(b) Overall model adequacy

H0:1=2=3=0

Ha : At least one of the parameters 1,2,3is non-zero.

Here, the F test statistic=R2k1-R2n-k+1=0.6531-0.6530-3+1=16.661

Value of F0.05,28,28is 1.901

H0

is rejected if F statistic > F0.05,28,28. For=0.05 , sinceF>F0.05,28,28 sufficient to reject H0 at 95% confidence interval.

Therefore,1230which means we accept the alternative hypothesis.

03

(c) Significance of  β1

H0:1=0Ha:10

Here, the t-test statistic

Value of t0.05,24,24is 1.699

H0 is rejected if t statistic >t0.05,24,24. For , since t>t0.05,24,24sufficient evidence to reject at a 95% confidence interval.

Therefore,localid="1660801540094" 1=0

04

(d) Prediction value

The regression equation is y^=-105+25x1+100x2+10x3.

The prediction value of daily attendance on sunny weekdays at950F can be calculated when x1=0,x2=1andx3=95

y^=-105+250+1001+1095=900

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

05

(e) Prediction interval

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.

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