Regression Statirties Muiple R R Square Practice Problem 3 0.9645 0.9303 Adjusted RSquare 0.9283 Standard Error Observations 230.5 37 The owner of an ice cream shop records the daily high temperature and daily sales for 37 days during the summer. Daily high temperatures ranged fro 100 degrees. Sales ranged from $1,824 to $4,783. Type your answers in this text box. ANOVA a. Interpret bo in the context of this problem. SgniticanceF 1 24821242.3 24821242.3 467. 1819406 7.90202E-22 AMS Regression Residual 35 1859539.95 53129.7128 Total 36 26680782.3 Ob. Interpret bi in the context of this problem. Conlicient. Standard Enor -3817.8 342.707286 88.011 4.07186316 Lower S bw 0:oww Kerper Star -11. 140038 4.67134E-13 -4513.50494 -3122 -4513.5 -3122 21614392 7.90202E-22 79.74452478 96.277 79.745 96.277 Pvale ntercept Temperature c. What percent of total variation in sales is explained by the regression model? Provide evidence to support your answer. d. Can the daily high temperature explairsales? Provide evidence to support your answer.

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The owner of an ice cream shop records the daily high temperature and daily sales for 37 days during the summer.  Daily high temperatures ranged from 63 to 100 degrees.  Sales ranged from $1,824 to $4,783.

Attached file.

**Regression Analysis of Ice Cream Sales**

The analysis examines the relationship between daily high temperature and ice cream sales over 37 days. The daily temperatures ranged from 63 to 100 degrees, with sales varying from $1,824 to $4,783.

### Regression Output Summary

- **Multiple R**: 0.9345
- **R Square**: 0.8733
- **Adjusted R Square**: 0.8683
- **Standard Error**: 230.5
- **Observations**: 37

### ANOVA

- **Regression**
  - Degrees of Freedom (df): 1
  - Sum of Squares (SS): 2482142.3
  - Mean Square (MS): 2482142.3
  - F Statistic: 467.1819406
  - Significance F: 7.90202E-22

- **Residual**
  - df: 35
  - SS: 315239.5
  - MS: 5312.7128

- **Total**
  - df: 36
  - SS: 2687382.7

### Regression Coefficients

- **Intercept**
  - Coefficient: -3817.8
  - Standard Error: 342.702706
  - t Stat: -11.130038
  - P-value: 4.67134E-13
  - Lower 95%: -4513.50434
  - Upper 95%: -3122.09566
  - Lower 95.0%: -4513.504
  - Upper 95.0%: -3122.096

- **Temperature (b1)**
  - Coefficient: 88.011
  - Standard Error: 4.07186316
  - t Stat: 21.64392
  - P-value: 7.90022E-22
  - Lower 95%: 79.7454278
  - Upper 95%: 96.2765722
  - Lower 95.0%: 79.745
  - Upper 95.0%: 96.277

### Practice Problem #3

1. **Interpret b0 in the context of this problem.**
2. **Interpret b1 in the context of this
Transcribed Image Text:**Regression Analysis of Ice Cream Sales** The analysis examines the relationship between daily high temperature and ice cream sales over 37 days. The daily temperatures ranged from 63 to 100 degrees, with sales varying from $1,824 to $4,783. ### Regression Output Summary - **Multiple R**: 0.9345 - **R Square**: 0.8733 - **Adjusted R Square**: 0.8683 - **Standard Error**: 230.5 - **Observations**: 37 ### ANOVA - **Regression** - Degrees of Freedom (df): 1 - Sum of Squares (SS): 2482142.3 - Mean Square (MS): 2482142.3 - F Statistic: 467.1819406 - Significance F: 7.90202E-22 - **Residual** - df: 35 - SS: 315239.5 - MS: 5312.7128 - **Total** - df: 36 - SS: 2687382.7 ### Regression Coefficients - **Intercept** - Coefficient: -3817.8 - Standard Error: 342.702706 - t Stat: -11.130038 - P-value: 4.67134E-13 - Lower 95%: -4513.50434 - Upper 95%: -3122.09566 - Lower 95.0%: -4513.504 - Upper 95.0%: -3122.096 - **Temperature (b1)** - Coefficient: 88.011 - Standard Error: 4.07186316 - t Stat: 21.64392 - P-value: 7.90022E-22 - Lower 95%: 79.7454278 - Upper 95%: 96.2765722 - Lower 95.0%: 79.745 - Upper 95.0%: 96.277 ### Practice Problem #3 1. **Interpret b0 in the context of this problem.** 2. **Interpret b1 in the context of this
Expert Solution
Step 1

Hi! Thank you for the question, As per the honor code, we are allowed to answer three sub-parts at a time so we are answering the first three as you have not mentioned which of these you are looking for. Please re-submit the question separately for the remaining sub-parts.

 

From the given excel output, 

 

Coefficient

Standard error

P-value

Intercept

-3817.8

342.707286

0.000

Temperature

88.011

4.07186316

0.000

 

The R-square is 0.9303.

Step 2

By using the given information, the regression model is:

y^=b0+b1×xSales=-3817.8+88.011×Tempertaure

 

 

From the given information,

the p-value is less than 0.05 level of significance for both the overall regression model and the independent variable (temperature). Thus, the model and variable both are statistically significant.

Step 3

3).

a).

Interpretation of b0 of a regression line,

The y-intercept of the least-squares regression means that the expected mean value of when all values of X are 0.

Thus, it can be interpreted as:

If the temperature is 0, then the predicted value of daily sales will be -3817.8.

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