113年 - 113 國立臺北大學_日間學士班轉學生招生考試試題_金融與合作經營學日間學士班2年級：統計學#147591

1、(20%) A vendor of milk products produces and sells low-fat dry milk to a company that uses it to produce baby formula. In order to determine the fat content of the milk, both the company and the vendor take a sample from each lot and test it for fat content in percent. Ten sets of paired test results

Let $\mu_D = \mu_X - \mu_Y$ denote the mean of the differences between X and Y. Test H0: $\mu_D = \mu_X - \mu_Y = 0$ against H1: $\mu_D = \mu_X - \mu_Y > 0$ using t-test for the differences. Let $\alpha = 0.05$. (All values are rounded to the third decimal place.) (Hints: t_0.95(9) = 1.833; t_0.975(9) = 2.262; t_0.95(10) = 1.812; t_0.975(10) = 2.228, where P(t < t1-α(r)) = 1 - α for a t random variable with a freedom r)

(a) all of them hit the target; 

(b) exactly one hits the target;

(c) no one hits the target;

(d) at least one hits the target.

(a) \( (15\%) \) Find the maximum likelihood estimator for \( \lambda \), and prove it.

(b) (5%) Let \( X \) equal the number of flaws per 100 feet of a used computer tape. Assume that \( X \) has a Poisson distribution with a mean of \( \lambda \). If 40 observations of \( X \) yielded 5 zeros, 7 ones, 12 twos, 9 threes, 5 fours, 1 five and 1 six, find the maximum likelihood estimate of \( \lambda \).

(a) \( (10\%) \) Let \( y \) denote the number of fusions in the sample. How large a sample \( n \) of rat embryos is needed for \( y / n \pm 0.10 \) to be a \( 95\% \) confidence interval for \( p \)?

(b) \( (10\%) \) If \( y = 44 \) out of \( n = 60 \) palates showed fusion, give a \( 95\% \) confidence interval for \( p \). (All values are rounded to the third decimal place.)

(a) \( (10\%) \) Show that the mean and variance of Y are, respectively, \( 3\theta /2 \) and \( 5\theta^2 /4 \)

(b) \( (10\%) \) Show that \( \operatorname{E}(Y|\mathbf{x}) = x + \theta \). In accordance with the theory, the expected value of \( X + \theta \) is that of Y, namely, \( 3\theta / 2 \), and the variance of \( X + \theta \) is less than that of Y. Show the variance of \( X \) is in face \( 5\theta^2 / 4 \).