Question

Question: State casket sales restrictions. Refer to the Journal of Law and Economics (February 2008) study of the impact of lifting casket sales restrictions on the cost of a funeral, Exercise 12.123 (p. 803). Recall that data collected for a sample of 1,437 funerals were used to fit the model, $\mathit{E}\left(y\right)\mathbf{=}{\mathit{\beta }}_{\mathbf{0}}\mathbf{+}{\mathit{\beta }}_{\mathbf{1}}{\mathit{x}}_{\mathbf{1}}\mathbf{+}{\mathit{\beta }}_{\mathbf{2}}{\mathit{x}}_{\mathbf{2}}\mathbf{+}{\mathit{\beta }}_{\mathbf{3}}{\mathit{x}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{2}}$, where y is the price (in dollars) of a direct burial,{1 if funeral home is in a restricted state, 0 if not}, and{1 if price includes a basic wooden casket, 0 if no casket}.The estimated equation (with standard errors in parentheses) is:

$\hat{\mathbf{y}}\mathbf{=}\mathbf{1432}\mathbf{+}\mathbf{793}{\mathit{x}}_{\mathbf{1}}\mathbf{-}\mathbf{252}{\mathit{x}}_{\mathbf{2}}\mathbf{+}\mathbf{261}{\mathit{x}}_{\mathbf{1}}{\mathit{x}}_{\mathbf{2}}\mathbf{,}{\mathit{R}}^{\mathbf{2}}\mathbf{=}\mathbf{.}\mathbf{78}$

(70) (134) (109)

1. Interpret the reported value of R².
2. Use the value of R²to compute the F-statistic for testing the overall adequacy of the model. Test atrole="math" localid="1660811116911" $\mathbf{\alpha }\mathbf{=}\mathbf{.}\mathbf{05}$.
3. Compute the predicted price of a direct burial with a basic wooden casket for a funeral home in a restrictive state.
4. Estimate the difference between the mean price of a direct burial with a basic wooden casket and the mean price of a burial with no casket for a funeral home in a restrictive state.
5. Estimate the difference between the mean price of a direct burial with a basic wooden casket and the mean price of a burial with no casket for a funeral home in a non-restrictive state.
6. Is there sufficient evidence to indicate that the difference between the mean price of a direct burial with a basic wooden casket and the mean price of a burial with no casket depends on whether the funeral home is in a restrictive state? Test using$\mathbf{\alpha }\mathbf{=}\mathbf{.}\mathbf{05}$.

Short answer

Answers

1. The value R² of is 0.78 indicating that almost 78% of the variation in the data is explained by the model. The higher the value of the better the model is fit for the data. 78% is very high value indicating that the model is a good fit for the data.
2. At 95% significance level, at least one of the parameters${\beta }_1,{\beta }_2,{\beta }_3$ is non zero.
3. The prediction price of a direct burial with a basic wooden casket for a funeral home in a restrictive state is 2234.
4. The difference in mean prices of direct burial in a restrictive state is$\left(2234-2225\right)=9$ .
5. The difference in mean prices of direct burial in a non-restrictive state is$\left(1180-1432\right)=-252$ .
6. At 95% significance level,${\beta }_3\ne 0$ indicating that the difference between the mean price of a direct burial with a basic wooden casket and the mean price of a burial with no casket does not depends on whether the funeral home is in a restrictive state.

Step-by-step solution

Stating the given information  

The total number of samples are 1437 and the model is given as,$E\left(y\right)={\beta }_0+{\beta }_1{x}_1+{\beta }_2{x}_2+{\beta }_3{x}_1{x}_2$ where y is the price (in dollars) of a direct burial, x₁ ={1 if funeral home is in a restricted state, 0 if not}, and x₂ = {1 if price includes a basic wooden casket, 0 if no casket}. The estimated model is R-Squared and the standard errors are given as $\hat{y}=1432+793x_1-252x_2+261x_1{x}_2,R^2=.78$, 70,134,109 for $x_1{x}_{2}and{x}_1{x}_2$respectively.

Interpretation of 

a. A study was conducted to find the impact of casket sales restriction on the cost of funerals. A first order equation with interaction was fitted to the data collected from 1437 observations.

The regression equation with the standard errors of the coefficients and the R² value is mentioned below.

,$\hat{y}=1432+793x_1-252x_2+261x_1{x}_2$

(70) (134) (109)

The value of R² is 0.78 indicating that almost 78% of the variation in the data is explained by the model. The higher the value of R² the better the model is fit for the data. 78% is very high value indicating that the model is a good fit for the data.

Overall adequacy of the model

b.

$H_0\text{:}{\beta }_1={\beta }_2={\beta }_3=0$

H_a :At least one of the parameters${\beta }_1,{\beta }_2,{\beta }_3$ is non zero.

Here

$$\begin{gathered} F-\text{test} \text{statistic}=\frac{\frac{R^2}{k}}{\frac{\left(1-R^2\right)}{\left[n-\left(k+1\right)\right]}} \\ =\frac{\frac{0.78}{3}}{\frac{0.22}{\left[1437-\left(3+1\right)\right]}} \\ =1269.2727 \end{gathered}$$

Value of $F_{0.05,1436,1436}$is 1.517.

H₀ is rejected if $F-\text{statistic}>F_{0.05,28,28}$. For$\alpha =0.05$ , since $F>F_{0.05,1436,1436}$there is a sufficient evidence to reject H₀ at 95% confidence interval.

Therefore, at least one of the parameters${\mathbf{\beta }}_{\mathbf{1}}\mathbf{,}{\mathbf{\beta }}_{\mathbf{2}}\mathbf{,}{\mathbf{\beta }}_{\mathbf{3}}$ is non zero.

Prediction value

c. The regression equation is $\hat{y}=1432+793x_1-252x_2+261x_1{x}_2$.

The prediction price of a direct burial with a basic wooden casket for a funeral home in a restrictive state can be calculated when$x_1=1,x_2=1$

$$\begin{gathered} \hat{y}=1432+793\left(1\right)-252\left(1\right)+261\left(1\right)\left(1\right) \\ \hat{y}=2234 \end{gathered}$$

The prediction price of a direct burial with a basic wooden casket for a funeral home in a restrictive state is 2234.

Difference in mean prices 

d. The mean price of a direct burial with a basic wooden casket and the mean price of a burial with no casket for a funeral home in a restrictive state can be calculated when $x_1=1and{x}_2=1$or 0.

Mean price of a direct burial with a basic wooden casket in a restrictive state is

$$\begin{gathered} \hat{y}=1432+793\left(1\right)-252\left(1\right)+261\left(1\right)\left(1\right) \\ \hat{y}=2234 \end{gathered}$$

Mean price of a burial with no casket for a funeral home in a restrictive state is

$$\begin{gathered} \hat{y}=1432+793\left(1\right)-252\left(0\right)+261\left(1\right)\left(0\right) \\ \hat{y}=2225 \end{gathered}$$

The difference in mean prices is (2234-2225) = 9.

Difference in mean prices

e. The mean price of a direct burial with a basic wooden casket and the mean price of a burial with no casket for a funeral home in a non-restrictive state can be calculated when $x_1=1and{x}_2=1$or 0.

Mean price of a direct burial with a basic wooden casket in a restrictive state is

$$\begin{gathered} \hat{y}=1432+793\left(0\right)-252\left(1\right)+261\left(0\right)\left(1\right) \\ \hat{y}=1180 \end{gathered}$$

Mean price of a burial with no casket for a funeral home in a restrictive state is

$$\begin{gathered} \hat{y}=1432+793\left(0\right)-252\left(0\right)+261\left(0\right)\left(0\right) \\ \hat{y}=1432 \end{gathered}$$

The difference in mean prices is (1180-1432) = -252.

Significance of  β3

f.

$$\begin{gathered} H_0\text{:}{\beta }_3=0 \\ {H}_a\text{:}{\beta }_3\ne 0 \end{gathered}$$

$$\begin{gathered} \text{Here}, t-\text{test} \text{statistic}=\frac{{\hat{\beta }}_3}{s{\hat{\beta }}_3} \\ =\frac{261}{109} \\ =2.39 \end{gathered}$$

Value of ${t_0}_{.025,1436}$is 1.96.

H₀is rejected if $t-\text{statistic}>t_{0.05,1436}$.

For$\alpha =0.05$ , since$t>t_{0.05,31}$ sufficient evidence to reject H₀ at 95% confidence interval.

Therefore ,indicating that the difference between the mean price of a direct burial with a basic wooden casket and the mean price of a burial with no casket does not depends on whether the funeral home is in a restrictive state.