Question

In: Statistics and Probability

Listed below are paired data consisting of movie budget amounts and the amounts that the movies...

Listed below are paired data consisting of movie budget amounts and the amounts that the movies grossed. Find the regression​ equation, letting the budget be the predictor​ (x) variable. Find the best predicted amount that a movie will gross if its budget is

​$125

million. Use a significance level of

alpha equals 0.05α=0.05.

Budget left parenthesis $ right parenthesisBudget ($)in Millionsin Millions

3939

2121

114114

6969

7575

5454

115115

6868

99

6565

126126

1616

1111

148148

88

Gross left parenthesis $ right parenthesisGross ($)

in Millionsin Millions

123123

1616

104104

7474

123123

108108

9494

107107

4747

105105

205205

3434

1212

293293

4040

Solutions

Expert Solution

Here, x = Budget (in Millions), y = Gross (in Millions).
The values are: x = (39, 21, 114, 69, 75, 54, 115, 68, 9, 65, 126, 16, 11, 14), y = (123, 16, 104, 74, 123, 108, 94,107, 47, 105, 205, 34, 12, 29). We input this data set in R programming software and use the summary(.) and lm(.) functions to answer the given question. The R code and output is given below.

> x = c(39, 21, 114, 69, 75, 54, 115, 68, 9, 65, 126, 16, 11, 14)
> y = c(123, 16, 104, 74, 123, 108, 94,107, 47, 105, 205, 34, 12, 29)
> mb=lm(y~x)
> summary(lm(y~x))

Call:
lm(formula = y ~ x)

Residuals:
Min 1Q Median 3Q Max
-50.233 -24.565 1.443 18.034 57.032

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 25.8053 15.8375 1.629 0.129184
x 1.0298 0.2292 4.494 0.000735 ***
---
Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 33.69 on 12 degrees of freedom
Multiple R-squared: 0.6272, Adjusted R-squared: 0.5962
F-statistic: 20.19 on 1 and 12 DF, p-value: 0.0007347


The regression equation is given as: = 25.8053 + (1.0298) (Please round the coefficients to the required number of decimal places).
Now, for = 125, predicted value of Gross = 25.8053 + (1.0298 * 125) = 154.53 millions (Please round the coefficients to the required number of decimal places).


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