A statistics student is asked to estimate y8081 she...

Name: a statistics student is asked to estimate y8081 she calculates the following values 1 220 2440 568 385 9 1120 n 11 which of the following is the value of y if x equals 2 multiple choice 6338 74561
Uploaded: 2022-12-06T18:30:07-08:00
Duration: 2 min 3 s
Channel: Sri K
Description: a statistics student is asked to estimate y8081 she calculates the following values 1 220 2440 568 385 9 1120 n 11 which of the following is the value of y if x equals 2 multiple choice 6338 74561

A statistics student is asked to estimate y=80+81 + ε. She calculates the following values: Στ₁ = 220, Σ(-)²=440. (-) (-) 568, Σ. – 385, Σ(μ. – 9) = 1,120, n= 11. Which of the following is the value of y if x equals 2? Multiple Choice 63.38 6.62 11.78 58.22

(a) Based on the following statistics, give the equation for the least squares regression line. n = 78, x̄ = 68.35, ȳ = 65.22, s_x^2 = 87.51, s_y^2 = 84.39, Σ(x_i - x̄)(y_i - ȳ) = 4863.78, Σ(y_i - ŷ)^2 = 3757.657. (b) It turns out that the estimator for β_1, β̂_1, has the following sampling distribution: β̂_1 ~ N(β, σ^2 / Σ(x_i - x̄)^2) Where an estimate of σ^2 that we can calculate from data is σ̂^2 = 1/(n-2) Σ(y_i - ŷ)^2. In order to test our claim of "ESP", we could do the following hypothesis test: H_0 : β_1 = 0 H_A : β_1 ≠ 0 In other words, showing that β_1 ≠ 0 would give us evidence that students have some ability to predict their exam scores. Based on all the information provided, we can conduct a normal based Hypothesis test in the same fashion that we have done before. Give the value of the test statistic, calculate the p-value, and draw a conclusion.

Sri K.

Question

A statistics student is asked to estimate y=80+81 + ε. She calculates the following values: Στ₁ = 220, Σ(-)²=440. (-) (-) 568, Σ. – 385, Σ(μ. – 9) = 1,120, n= 11. Which of the following is the value of y if x equals 2? Multiple Choice 63.38 6.62 11.78 58.22

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