### Solution:
First, we will address each part of the question step-by-step.
#### a) Finding the Regression Equation
To find the regression equation of the form [tex]\(\hat{y} = b_0 + b_1x\)[/tex]:
- The intercept ([tex]\(b_0\)[/tex]), given as approximately 0.8962.
- The slope ([tex]\(b_1\)[/tex]), given as approximately -0.003146.
Thus, the regression equation is:
[tex]\[
\hat{y} = 0.8962 - 0.0031x
\][/tex]
#### b) Predicting Bone Density for 21 Colas per Week
Using the regression equation [tex]\(\hat{y} = 0.8962 - 0.0031x\)[/tex], we substitute [tex]\(x = 21\)[/tex] to predict the bone density:
[tex]\[
\hat{y} = 0.8962 - 0.0031 \times 21
\][/tex]
[tex]\[
\hat{y} = 0.8962 - 0.0651
\][/tex]
[tex]\[
\hat{y} = 0.8311 \, \text{grams per cubic centimeter}
\][/tex]
#### c) Finding the Correlation Coefficient
The correlation coefficient measures the strength and direction of the linear relationship between two variables. The given correlation coefficient is:
[tex]\[
r = -0.9150
\][/tex]
#### d) Determining the Correct Alternative Hypothesis
The alternative hypothesis in this context examines whether the slope of the regression line is different from zero (indicating some relationship between the number of colas and bone mineral density). The correct alternative hypothesis is:
[tex]\[
H_a: \mu \neq 0
\][/tex]
#### e) Finding the [tex]\(p\)[/tex]-value
The [tex]\(p\)[/tex]-value helps in determining the statistical significance of our results. The given [tex]\(p\)[/tex]-value is:
[tex]\[
p \text{-value} = 0.0
\][/tex]
#### f) Determining Statistical Significance at the 5% Level
To determine if the correlation is statistically significant at the 5% significance level, we compare the [tex]\(p\)[/tex]-value to 0.05. Since the [tex]\(p\)[/tex]-value given is 0.0, which is less than 0.05, we conclude that the correlation is statistically significant at the 5% level:
[tex]\[
\text{Statistically significant} = \text{True}
\][/tex]
### Summary of Answers:
Let's now update each component based on the given values and results.
a) Regression equation: [tex]\(\hat{y} = 0.8962 - 0.0031x\)[/tex]
b) Bone density prediction for 21 colas per week: [tex]\(\hat{y} = 0.8311 \, \text{grams per cubic centimeter}\)[/tex]
c) Correlation coefficient: [tex]\(r = -0.9150\)[/tex]
d) Correct alternative hypothesis: [tex]\(H_a: \mu \neq 0\)[/tex]
e) [tex]\(p\)[/tex]-value: [tex]\(0.0\)[/tex]
f) Statistically significant at 5% level: Yes (True)
We can complete the solution for the problem with exactly these values.
