I asked this question yesterday and the answer was incorrect. Please see previous responses below that were incorrect for your reference. Thanks again! β = 0.4091 α = 3.1455 The lineal regression equation is y = 3.1455 +0.4091 x (second screenshot is the second step to the question that I received yeste
Correlation
Correlation defines a relationship between two independent variables. It tells the degree to which variables move in relation to each other. When two sets of data are related to each other, there is a correlation between them.
Linear Correlation
A correlation is used to determine the relationships between numerical and categorical variables. In other words, it is an indicator of how things are connected to one another. The correlation analysis is the study of how variables are related.
Regression Analysis
Regression analysis is a statistical method in which it estimates the relationship between a dependent variable and one or more independent variable. In simple terms dependent variable is called as outcome variable and independent variable is called as predictors. Regression analysis is one of the methods to find the trends in data. The independent variable used in Regression analysis is named Predictor variable. It offers data of an associated dependent variable regarding a particular outcome.
I asked this question yesterday and the answer was incorrect. Please see previous responses below that were incorrect for your reference. Thanks again!
β = 0.4091
α = 3.1455
The lineal regression equation is
y = 3.1455 +0.4091 x
(second screenshot is the second step to the question that I received yesterday as well)
![### Step 2
**Now we have fitted a linear equation**
When \( x = 5 \), then \( y = 5.191 \).
#### Data Table
| No | x | y | \( x - \bar{x} \) | \( (x - \bar{x})^2 \) |
|----|---|---|---------------|------------------|
| 1 | 5 | 5 | 3 - 1 | 1 |
| 2 | 5 | 4 | 6 - 1 | 1 |
| 3 | 4 | 5 | 5 - 2 | 4 |
| 4 | 5 | 6 | 7 - 0 | 0 |
| 5 | 10 | 8 | 7-4 | 16 |
| **Sum** | **30** | **28** | | **4.4** |
The standard error is \( (s_e) = 1.5831 = \sqrt{\text{MS residual}} \).
\( t_{\text{crit}, 5-2} = 3.1824 \) (Table value for 0.05 level of significance)
- **The 0.95 confidence interval for the mean predicted when \( x = 5 \):**
\[
\text{C.I.} = \left( \hat{y} - t_{b, (n-2)} \cdot s_e \cdot \sqrt{\frac{1}{n} + \frac{(x - \bar{x})^2}{\sum (x - \bar{x})^2}}, \hat{y} + t_{b, (n-2)} \cdot s_e \cdot \sqrt{\frac{1}{n} + \frac{(x - \bar{x})^2}{\sum (x - \bar{x})^2}} \right)
\]
Substituting values:
\[
\text{C.I.} = (5.191 - 3.1824 \cdot 1.5831 \cdot \sqrt{\frac{1}{5} + \frac{(5 - 5.6)^2}{4.4}}, 5.191 + 3.1824 \cdot](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb299a4ab-95e2-40fd-810f-07b7412657e7%2Ff351d913-e180-4aa8-9644-7b2632e0dc5b%2Fzwv9cc8_processed.jpeg&w=3840&q=75)
![The following sample observations were randomly selected. (Do not round the intermediate values. Negative amounts should be indicated by a minus sign. Round your answers to 3 decimal places.)
| **X:** | 5 | 5 | 4 | 6 | 10 |
| ------ |---|---|---|---|----|
| **Y:** | 3 | 6 | 5 | 7 | 7 |
**a.** Determine the 0.95 confidence interval for the mean predicted when x = 5.
[ , ]
**b.** Determine the 0.95 prediction interval for an individual predicted when x = 5.
[ , ]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb299a4ab-95e2-40fd-810f-07b7412657e7%2Ff351d913-e180-4aa8-9644-7b2632e0dc5b%2Fd4ieeza_processed.png&w=3840&q=75)
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