(2). Central Limit Theorem. (6%)

7. An equipment manufacturer has developed a new electrical product. Heclaims that the product has a mean life length of 10,000 hours, with a standarddeviation of 150 hours. A consumer tested 16 random samples of the productand obtained a mean life length of 9,775 hours. Evaluate at the 0.05 level ofsignificance whether the life length announcement for the new product isreliable ? (12%)

(1) Assume that the error terms εᵢ are independent N(0, σ²) and that σ² is known. State the likelihood function for the n sample observations on Y and obtain the maximum likelihood estimator of β₁. Is it the same as the least squares estimator of β₁? (10%)

(2) Show that the maximum likelihood estimator of β₁ is unbiased. (5%)

(1). two random variables are independent and identically distributed. (6%)

2. Given the joint probability distribution of two random variables X and Y as following, (4%)

3. 普松隨機變數(Poisson random variable)X之機率函數為P(x=k) = , 試証其符合機率總和等於1的特性。(8%)

 (1).對一個隨機變數X而言,其平均值(mean)與變異數(variance)之意義各為何? (6%)

(2). 試証 Var(X) = E(X²) - [E(X)]² (8%)

5. 一隨機變數X之機率密度函數(probability density function)為試繪製此隨機變數之機率密度函數與累積機率函數(cumulative distribution function)的圖形。(6%)

6. The manager of a convenient store wants to know whether the customer arrival in every ten-minute period follows Poisson distribution. The alternatives are as follows:H₀: The population distribution is Poisson.H₁: The population distribution is not Poisson.One day he observes from 8 a.m. to 4 p.m. The data on the number of customers (X) arriving at his store in each ten-minute period are as follows:Hence the sample mean = 6.0.Using the Chi-square procedure, for α = 0.05, test whether the probability distribution of the ten-minute customer arrival is Poisson. (20%)