| xi | 1.0 | 2.0 | 2.5 | 3.0 | 3.5 | 4.5 |
| yi | 2.0 | 3.0 | 3.5 | 4.2 | 5.0 | 5.4 |
First, we create a table to calculate the necessary sums: ∑xi, ∑yi, ∑xi2, and ∑xiyi.
| i | xi | yi | xi2 | xiyi |
|---|---|---|---|---|
| 1 | 1.0 | 2.0 | 1.00 | 2.00 |
| 2 | 2.0 | 3.0 | 4.00 | 6.00 |
| 3 | 2.5 | 3.5 | 6.25 | 8.75 |
| 4 | 3.0 | 4.2 | 9.00 | 12.60 |
| 5 | 3.5 | 5.0 | 12.25 | 17.50 |
| 6 | 4.5 | 5.4 | 20.25 | 24.30 |
| Sum | 16.5 | 23.1 | 52.75 | 71.15 |
Number of observations, n=6.
xˉ=n∑xi=616.5=2.75
yˉ=n∑yi=623.1=3.85
The formula for the slope is:
β^1=∑xi2−nxˉ2∑xiyi−nxˉyˉ
Numerator (Sxy):
71.15−6(2.75)(3.85)=71.15−63.525=7.625
Denominator (Sxx):
52.75−6(2.75)2=52.75−6(7.5625)=52.75−45.375=7.375
Slope:
β^1=7.3757.625≈1.0339
The formula for the intercept is:
β^0=yˉ−β^1xˉ
Substitute the values:
β^0=3.85−(1.0339)(2.75)
β^0=3.85−2.8432
β^0≈1.0068
β^0+β^1=1.0068+1.0339
β^0+β^1=2.0407
Rounding to two decimal places:
Answer: 2.04
Let \( X_1, X_2, \dots, X_7 \) be a random sample from a population having the probability density function \[ f(x) = \frac{1}{2} \lambda^3 x^2 e^{-\lambda x}, \quad x>0, \] where \( \lambda>0 \) is an unknown parameter. Let \( \hat{\lambda} \) be the maximum likelihood estimator of \( \lambda \), and \( E(\hat{\lambda} - \lambda) = \alpha \lambda \) be the corresponding bias, where \( \alpha \) is a real constant. Then the value of \( \frac{1}{\alpha} \) equals __________ (answer in integer).