A corporation owns several companies. The strategic planner for the corporation believes dollars spent on advertising can to some extent be a predictor of total sales dollars. As an aid in long-term planning, she gathers the following sales and advertising information from several of the companies for 2017 (in millions).

Advertising (x): 12.5, 3.7, 21.6, 60.0, 37.6, 6.1, 16.8, 41.2
Sales (Y): 148, 55, 338, 994, 541, 89, 126, 379

Consider the following linear model:

Yi=α+β(xi−
ˉ
x
)+ϵi,ϵi∼N(0,σ2).

Supose that we know that Y given x is N(α+β(xi−
ˉ
x
),σ2), where σ2=1092 is known. Use a N(mean=0,variance=102) prior for β. Find a 95% credible interval for β and provide the numerical value of the Upper Credible Limit (UCL).

Round your final answer to four decimal places.

Question

A corporation owns several companies. The strategic planner for the corporation believes dollars spent on advertising can to some extent be a predictor of total sales dollars. As an aid in long-term planning, she gathers the following sales and advertising information from several of the companies for 2017 (in millions).

Advertising (x):	12.5, 3.7, 21.6, 60.0, 37.6, 6.1, 16.8, 41.2
Sales (Y): 	148, 55, 338, 994, 541, 89, 126, 379

Consider the following linear model:

Yi=α+β(xi−
ˉ
x
)+ϵi,ϵi∼N(0,σ2).

Supose that we know that Y given x is N(α+β(xi−
ˉ
x
),σ2), where σ2=1092 is known. Use a N(mean=0,variance=102) prior for β.  Find a 95% credible interval for β and provide the numerical value of the Upper Credible Limit (UCL).

Round your final answer to four decimal places.

BlackTom AI · Accepted Answer

We start by restating the given setup and data to keep the problem clear. The model is Yi = α + β(xi − x̄) + εi with εi ∼ N(0, σ^2) and σ^2 = 1092 is known. A prior for β is β ∼ N(0, τ^2) with τ^2 = 102. The data provided are the eight pairs (xi, Yi) where xi are the advertising figures and Yi are the sales figures. First, compute the centered predictor values xi − x̄ and the centered response values Yi − Ȳ, since centering makes the intercept α orthogonal to the slope β and simplifies the regression algebra.

Step 1: Compute x̄ and the centered x’s (xi − x̄).
- The sum of x is 199.5, so x̄ = 199.5/8 = 24.9375.
- Centered x values, xi − x̄, are: −12.4375, −21.2375, −3.3375, 35.0625, 12.6625, −18.8375, −8.1375, 16.2625.

Step 2: Compute Sxx, the sum of squared centered x values (∑(xi − x̄)^2).
- Using the centered x’s, Sxx ≈ 2691.89.

Step 3: Compute Ȳ and the centered y values (Yi − Ȳ).
- The sum of Y is 2670, so Ȳ = 2670/8 = 333.75.
- Centered Y values, Yi − Ȳ, are: −185.75, −278.75, 4.25, 660.25, 207.25, −244.75, −207.75, 45.25.

Step 4: Compute β̂ (the OLS estimator for β with the intercept included, but using centered x so the intercept is orthogonal).
- The cross-product ∑(xi − x̄)(Yi − Ȳ) ≈ 41027.00 (summed from the eight products).
- Therefore β̂ = ∑[(xi − x̄)(Yi − Ȳ)] / ∑[(xi − x̄)^2] ≈ 41027.00 / 2691.89 ≈ 15.24.

Step 5: Combine prior and likelihood to obtain the posterior for β.
- The likelihood for β, given known σ^2 and x-centered, has variance σ^2/Sxx for β around β̂, i.e., β | data ∼ N(β̂, σ^2/Sxx) when α is integrated out with an appropriate (flat) prior for α, and the intercept has been accounted for via centering.
- The prior is β ∼ N(0, τ^2) with τ^2 = 102, so the prior precision is 1/τ^2 = 1/102 ≈ 0.009804.
- The likelihood precision for β is Sxx/σ^2 ≈ 2691.89 / 1092 ≈ 2.465.
- The posterior precision for β is the sum of these: 2.465 + 0.009804 ≈ 2.474804.
- The posterior mean for β when combining a prior 0 and data β̂ with precision Sxx/σ^2 is μ_post = β̂ × (Sxx/σ^2) / (Sxx/σ^2 + 1/τ^2).
- Substituting numbers: μ_post ≈ 15.24 × 2.465 / 2.474804 ≈ 15.18.
- The posterior variance for β is Var(β|data) = 1 / (Sxx/σ^2 + 1/τ^2) ≈ 1 / 2.474804 ≈ 0.4044, so the posterior standard deviation is sd_post ≈ √0.4044 ≈ 0.6358.

Step 6: Form the 95% credible interval for β.
- A 95% interval for a Normal posterior is μ_post ± 1.96 × sd_post.
- Compute the limits: 15.18 ± 1.96 × 0.6358 ≈ 15.18 ± 1.246.
- This gives approximately (13.934, 16.426).

Step 7: Present the result with four-decimal precision as requested.
- Lower bound ≈ 13.9340, Upper bound ≈ 16.4260.

Thus, the 95% credible interval for β is approximately (13.9340, 16.4260).

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