i am using minitab to solve the problem.
steps:-
copy the data in minitab
stat
regression
regression
fit regression model
in responses select fall festival, in continuous predictors select
4 th of july
results
tick coefficients,analysis of variance
ok
ok.
to round off your answer to 3 decimal places right click on the
value you want to round off
decimal places
3
your minitab output be:-
Regression Analysis: fall festival versus 4 th of july
Analysis of Variance
| Source | DF | Adj SS | Adj MS | F-Value | P-Value |
| Regression | 1 | 47.751 | 47.751 | 29.03 | 0.000 |
| 4 th of july | 1 | 47.751 | 47.751 | 29.03 | 0.000 |
| Error | 13 | 21.386 | 1.645 | ||
| Lack-of-Fit | 11 | 19.521 | 1.775 | 1.90 | 0.395 |
| Pure Error | 2 | 1.865 | 0.933 | ||
| Total | 14 | 69.137 |
Coefficients
| Term | Coef | SE Coef | T-Value | P-Value | VIF |
| Constant | 1.276 | 1.498 | 0.85 | 0.410 | |
| 4 th of july | 0.846 | 0.157 | 5.39 | 0.000 | 1.00 |
SOLUTION TO THE PROBLEM:-
a).the regression equation be:-

b).the completed table be:-
| predictor | coefficient | SE coefficient | t | p value |
| constant | 1.276 | 1.498 | 0.85 | 0.410 |
| july | 0.846 | 0.157 | 5.39 | 0.000 |
c).the ANOVA table be:-
| source | DF | SS | MS | F | p value |
| regression | 1 | 47.751 | 47.751 | 29.03 | 0.000 |
| residual error | 13 | 21.386 | 1.545 | 0 | 0 |
| total | 14 | 69.137 | 0 | 0 | 0 |
d).there is significant relationship between attendance at the fall festival and the fireworks.
[ p value of F statistic = 0.000 < 0.05 (alpha).
so, we reject the null hypothesis and conclude that your sample data provide sufficient evidence to conclude that the regression model fits the data better than the model with no independent variables.]
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