QuestionA chemist studies how carbon atoms in hydrocarbons affect energy release during combustion. Estimate $\theta_{0}$ and $\theta_{1}$ for $y = \theta_{0} + \theta_{1} x$ . Options: $\theta_{0}=-569.6, \theta_{1}=530.9$ ; $\theta_{0}=-1780.0, \theta_{1}=530.9$ ; $\theta_{0}=-569.6, \theta_{1}=-530.9$ ; $\theta_{0}=-1780.0, \theta_{1}=-530.9$ .

Studdy Solution

STEP 1

Assumptions1. The number of hydrocarbons in a molecule (x) and the heat release when burned (y) have a linear relationship. . The relationship can be represented by the linear regression model $h_{\theta}(x)=\theta_{0}+\theta_{1} x$ .
3. The heat release when burned (y) is a dependent variable and the number of hydrocarbons in a molecule (x) is an independent variable.
4. The values for $\theta_{0}$ and $\theta_{1}$ can be estimated without actually implementing linear regression.

STEP 2

First, let's understand the meaning of $\theta_{0}$ and $\theta_{1}$ in the linear regression model. $\theta_{0}$ is the y-intercept, which is the value of y when x is0. $\theta_{1}$ is the slope of the line, which represents the change in y for each unit change in x.

STEP 3

From the given dataset, we can observe that as the number of hydrocarbons in a molecule increases, the heat release when burned also increases. This means that the slope of the line ( $\theta_{1}$ ) should be positive.

STEP 4

Therefore, we can eliminate the options where $\theta_{1}$ is negative. The options left are $\theta_{0}=-569.6, \theta_{1}=530.9$ and $\theta_{0}=-1780.0, \theta_{1}=530.9$ .

STEP 5

Next, let's consider the y-intercept ( $\theta_{0}$ ). In this context, it represents the heat release when there are no carbon atoms in the molecule. Since it's not possible to have a molecule with no carbon atoms in this context, the y-intercept is not physically meaningful and can be any value.

STEP 6

However, in the context of fitting the best line to the data, we can consider the range of y values in the dataset. The y values range from -890 to -5471. A y-intercept of -569.6 seems too small compared to the range of y values, while -1780.0 seems more reasonable.

STEP 7

Therefore, based on the observations and reasoning, the best estimate for the values of $\theta_{0}$ and $\theta_{1}$ would be $\theta_{0}=-1780.0, \theta_{1}=530.9$ .
The estimated values for $\theta_{0}$ and $\theta_{1}$ are -1780.0 and530.9 respectively.

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Studdy Solution

STEP 1

STEP 2

First, let's understand the meaning of $\theta_{0}$ and $\theta_{1}$ in the linear regression model. $\theta_{0}$ is the y-intercept, which is the value of y when x is0. $\theta_{1}$ is the slope of the line, which represents the change in y for each unit change in x.

STEP 3

From the given dataset, we can observe that as the number of hydrocarbons in a molecule increases, the heat release when burned also increases. This means that the slope of the line ( $\theta_{1}$ ) should be positive.

STEP 4

Therefore, we can eliminate the options where $\theta_{1}$ is negative. The options left are $\theta_{0}=-569.6, \theta_{1}=530.9$ and $\theta_{0}=-1780.0, \theta_{1}=530.9$ .

STEP 5

Next, let's consider the y-intercept ( $\theta_{0}$ ). In this context, it represents the heat release when there are no carbon atoms in the molecule. Since it's not possible to have a molecule with no carbon atoms in this context, the y-intercept is not physically meaningful and can be any value.

STEP 6

However, in the context of fitting the best line to the data, we can consider the range of y values in the dataset. The y values range from -890 to -5471. A y-intercept of -569.6 seems too small compared to the range of y values, while -1780.0 seems more reasonable.

STEP 7

Therefore, based on the observations and reasoning, the best estimate for the values of $\theta_{0}$ and $\theta_{1}$ would be $\theta_{0}=-1780.0, \theta_{1}=530.9$ .
The estimated values for $\theta_{0}$ and $\theta_{1}$ are -1780.0 and530.9 respectively.

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Studdy Solution

STEP 1

STEP 2

First, let's understand the meaning of θ0\theta_{0}θ0​ and θ1\theta_{1}θ1​ in the linear regression model. θ0\theta_{0}θ0​ is the y-intercept, which is the value of y when x is0. θ1\theta_{1}θ1​ is the slope of the line, which represents the change in y for each unit change in x.

STEP 3

From the given dataset, we can observe that as the number of hydrocarbons in a molecule increases, the heat release when burned also increases. This means that the slope of the line (θ1\theta_{1}θ1​) should be positive.

STEP 4

Therefore, we can eliminate the options where θ1\theta_{1}θ1​ is negative. The options left are θ0=−569.6,θ1=530.9\theta_{0}=-569.6, \theta_{1}=530.9θ0​=−569.6,θ1​=530.9 and θ0=−1780.0,θ1=530.9\theta_{0}=-1780.0, \theta_{1}=530.9θ0​=−1780.0,θ1​=530.9.

STEP 5

STEP 6

However, in the context of fitting the best line to the data, we can consider the range of y values in the dataset. The y values range from -890 to -5471. A y-intercept of -569.6 seems too small compared to the range of y values, while -1780.0 seems more reasonable.

STEP 7

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First, let's understand the meaning of $\theta_{0}$ and $\theta_{1}$ in the linear regression model. $\theta_{0}$ is the y-intercept, which is the value of y when x is0. $\theta_{1}$ is the slope of the line, which represents the change in y for each unit change in x.

From the given dataset, we can observe that as the number of hydrocarbons in a molecule increases, the heat release when burned also increases. This means that the slope of the line ( $\theta_{1}$ ) should be positive.

Therefore, we can eliminate the options where $\theta_{1}$ is negative. The options left are $\theta_{0}=-569.6, \theta_{1}=530.9$ and $\theta_{0}=-1780.0, \theta_{1}=530.9$ .