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研究所、轉學考(插大)◆統計學
> 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
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
科目:
研究所、轉學考(插大)◆統計學 |
年份:
113年 |
選擇題數:
0 |
申論題數:
8
試卷資訊
所屬科目:
研究所、轉學考(插大)◆統計學
選擇題 (0)
申論題 (8)
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
| Lot Number | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| --- | --- | --- | --- | --- | --- | --- | --- | --- | --- | --- |
| Company Test Results (X) | 0.50 | 0.58 | 0.90 | 1.17 | 1.14 | 1.25 | 0.75 | 1.22 | 0.74 | 0.80 |
| Vendor Test Results (Y) | 0.79 | 0.71 | 0.82 | 0.82 | 0.73 | 0.77 | 0.72 | 0.79 | 0.72 | 0.91 |
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: t0.95(9) = 1.833; t0.975(9) = 2.262; t0.95(10) = 1.812; t0.975(10) = 2.228, where P(t < t1-α(r)) = 1 - α for a t random variable with a freedom r)
2. (20%) Each of four persons fires one shot at a target. Let \( C_k \) denote the event that the target is hit by person \( k, k = 1, 2, 3, 4 \). If \( C_1, C_2, C_3, C_4 \) are independent and if \( P(C_1) = P(C_2) = 0.7 \), \( P(C_3) = 0.9 \), and \( P(C_4) = 0.4 \), compute the probability that (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.
3. (20%) A random sample \( X_{1}, X_{2}, \ldots, X_{n} \) of size \( n \) is taken from a Poisson distribution with a mean of \( \lambda \), \( 0 < \lambda < \infty \).
(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 \( \lambda \).
4、(20%) Some dentists were interested in studying the fusion of embryonic rat palates by using a standard transplantation technique. When no treatment (vitamin A) is used, the probability of fusion approximately equals 0.89. They would like precisely to estimate p, the probability of fusion, when vitamin A is lacking.
(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.)
5、(20%) Let the random variables X and Y have the joint p.d.f. f(x, y) = (2/θ²)exp(-(x+y)/θ), 0 < x < y < ∞, zero elsewhere.
(Hint: $\Gamma(z) = \int_0^\infty t^{z-1} e^{-t} dt$, and $\Gamma(n) = (n-1)!$ for $n = 1, 2, \ldots, \Gamma(1) = 0! = 1$)
(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 \(5\theta^2 / 4\).
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