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A 99% confidence interval estimate can be interpreted to mean that (A)lf all possible samples are taken and confidence interval estimates are developed, 99% of them would include the true population mean somewhere within their interval (B)We have 99% confidence that we have selected a sample whose interval does include the population mean (C)Both of the above (D)None of the above 99% 的信賴區間估計可以解釋為 (A) 如果採集了所有可能的樣本並得出了信賴區間估計值,那麼 99% 的樣本將包括其區間內某處的真實母體平均值 (B) 我們有 99% 的信賴度認為我們選擇了一個樣本,其區間確實包含母體平均值 (C) 以上兩者 (D) 以上都不是 (台大商研) 答:(C) For the following statements about a confidence interval (CI), which is (are) true? (a) The narrower the CI, the better, for the same level of confidence. (b) The narrower the CI, the higher the confidence level. (c) A 95% CI for the mean of population implies a .95 probability that the mean of the population lies in that interval. (A) (a)(b)(c) (B) (a)(c) (C) (a) (E) none (成大財金) 對於以下關於置信區間 (CI) 的陳述,哪個是(是)真的? (a) 對於相同的信賴水平,CI 越窄越好。 (b) CI 越窄,信賴水平越高。 (c) 母體平均數的 95% CI 意味著 0.95 的機率 母體平均數位於該區間內。 答:(C) (a) (b) CI 越窄,信賴水平越低 (c) 母體平均數的 95% CI 並非意味著 0.95 的機率母體平均數位於該區間內。相反,我們不知道母體平均數有多少機率會落在信賴區間。譬如,我們預測人類哪時候滅亡,五十年後、一百年後、一千年後甚至一億年後,當然區間估計範圍拉的越大,譬如估計一億年後人類滅亡,估計準確度就降低,所以並非區間拉的越大機率越大。The 95% confidence interval is a range of values that you can be 95% confident contains the true mean of the population. Which of the following statements about confidence interval is true? (A)For a certain level of confidence, we would prefer a wide confidence interval over a narrow confidence interval because the wide confidence interval is more likely to contain the true value of the population parameter (B)A good confidence interval has a large width and a high level of confidence (C)Suppose we construct a two-sided and a one-sided confidence intervals for the population mean. In each case, the level of confidence is 1-a. The lower confidence limit for the two-sided interval exceeds the lower confidence limit for the one-sided interval (D)Other things being equal, we could always increase sample size to reduce the width of confidence interval (E) None of the above answer is correct 下列關於信賴區間的說法正確的是? (A) 對於一定的信賴水準,我們更喜歡寬信賴區間而不是窄信賴區間,因為寬信賴區間更有可能包含母體平均數的真實值 (B) 一個好的信賴區間具有較大的寬度和較高的信賴水準 (C) 假設我們為母體平均值構建了一個雙尾和一個雙尾信賴區間。 在每種情況下,信賴水準都是 1-a。 兩側區間的信賴下限超過單側區間的信賴下限 (D) 在其他條件相同的情況下,我們總是可以增加樣本量以減少置信區間的寬度 (E) 以上答案都不正確 (中正企管) 答:(D) (A)錯,信賴區間越窄估計越精確 (B)錯,區間越窄越精確,雖然信賴水準越低區間越窄,但可靠度也會降低,估計範圍變小,並非好方法。應該增加樣本數使區間變窄,樣本越大母體參數容易穩定集中在較小範圍。 (C)兩側區間的信賴下限低於單側區間的信賴下限 一家雜誌社欲知其讀者的平均年齡,以作為雜誌內容走向的參考,根據其 對訂戶抽查所得的讀者平均年齡為 35 歲。當樣本數為 25,母體分配為常 態,樣本標準差為6歲,求該雜誌讀者平均年齡的95%信賴區間。 〈台北合經》
A sample of the reading scores of 36 fifth-graders has a mean of 85. The standard deviation of the sample is 12. (1) Find the point estimate of the mean reading scores of all fifth-graders. (2) Find the 95% confidence interval of the mean reading score of all fifth-graders. (淡江國貿) 36 名五年級學生的閱讀分數樣本的平均值為 85。 樣本的標準差為 12。 (1) 求所有五年級學生平均閱讀分數的點估計。 (2) 求所有五年級學生平均閱讀分數的 95% 信賴區間。 假設依據過去的經驗,池上米在分配大包裝的白米時,每包重量的標準差 為9公斤。今隨機抽樣選取 100 包白米稱重,得平均每包 105 公斤。 (1) 請問每包白米重量之 95%平均重量之信賴區間為何? (2) 又假設過去在分裝白米時並未加以紀錄,以致於未能得知每包白米重 量的標準差。今由隨機抽取之 100 包白米稱得知標準差為 10公斤,估 計每包白米之母體平均數之 98%信賴區間為何? 〈元智企管》 設由一常態母體抽出一組樣本數為 13之隨機樣本,求得其母體平均數 之95%的信賴區間為 (8.30, 13.72) ,試求: (1) 其樣本平均數與樣本標準差。 (2) 母體平均數μ之 95%的信賴區間的誤差界限 。 (3) μ之 90%的信賴區間。 《淡江保險 A group of 12 security analysts provided estimates of the year 2003 estimates of the year 2003 earnings per share for T company.
1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 |
1.40 | 1.40 | 1.45 | 1.49 | 1.37 | 1.27 | 1.40 | 1.55 | 1.40 | 1.42 | 1.48 | 1.63 |
(1) Compute the sample variance for the earnings per share estimate.
(2) Compute the sample standard deviation for the earnings per share estimate.
(3) Provide 95% confidence interval estimates of the population variance and
the population standard deviation.
(大同資訊經營)
一組 12 位證券分析師提供了 2003 年對 T 公司 2003 年每股收益的估計。
(1) 計算每股收益估計的樣本變異數。
(2) 計算每股收益估計的樣本標準差。
(3) 提供母體變異數和母體標準差的 95% 信賴區間估計值。
A random sample of 200 semiconductor devices results in six defects.
(1) The 90% C.I. of the fraction defective=________.
(2) How large a sample is needed to be a 95% confident that the error in
estimating the true value of the fraction defective is less than 0.02?
(朝陽工管)
200 個半導體元件的隨機樣本導致 6 個缺陷。
(1) 90% C.I. 有缺陷的部分=
(2) 需要多大的樣本才能有 95% 的信賴區間
估計缺陷部分的真實值小於 0.02?
The Dong Hwa Business Management are interested in the average annual salary difference for
employees in the manufacturing industry in the regions of Taipei and
Kaohsiung. An independent random sample in each region was taken. The
results are as follows: (in thousand dollars)
Region x̅ S n
Taipei 24.8 3.1 45
Kaohsiung 21.9 3.7 55
Construct a 90% confidence interval for the true population difference in the
annual salary of manufacturing employees of the two regions.
東華企業管理系專長在人事管理的學生對台北和高雄地區製造業員工的平均年薪差異感興趣。 在每個區域抽取一個獨立的隨機樣本。 結果如下:(單位:千元)
構建兩個地區製造業員工年薪真實母體差異的 90%信賴區間(東華企研)
Suppose that a certain drug A was administered to eight patients at random;
and after a fixed time period, the concentration of the drug in certain body
cells of each patient was measured in appropriate units. Suppose that these
concentrations for the eight patients were found to be as follows:
1.23, 1.42, 1.41, 1.62, 1.55, 1.51, 1.60, and 1.76
suppose also that a second drug B was administered to six different patients a
similar way for six patients, the results were as follows:
1.76, 1.41, 1.87, 1.49, 1.67, and 1.81
Let μ, denote the mean of each observation for drug A, and let μ, denote
the mean of each observation for drug B. Also suppose that all observations
have a common unknown variance. Please find a confidence interval for
μA - μB with confidence coefficient of 0.90.
假設將某種藥物 A 隨機給予 8 名患者; 並在固定時間段後,以適當的單位衡量每個患者某些體細胞中藥物的濃度。 假設 8 名患者的這些濃度如下:
1.23、1.42、1.41、1.62、1.55、1.51、1.60和1.76
還假設將第二種藥物 B 以類似的方式施用於六名不同的患者,結果如下:
1.76、1.41、1.87、1.49、1.67 和 1.81
讓μ表示藥物 A 的每個觀察值的平均值,並讓 μ 表示藥物 B 的每個觀察值的平均值。還假設所有觀察值都有一個共同的未知變異數。 請找出信賴係數為 0.90 的 μA - μB 的信賴區間。
(暨南經研)
A certain change in a manufacturing procedure for component parts is being
considered. Samples are taken using both the existing and the new procedure
in order to determine if the new procedure results in an improvement. If 75 of
1500 items from the existing procedure were found to be defective and 80 of
2000 items from the new procedure were found to be defective, find a 90%
confidence interval for the true difference in the fraction of defectives between
the existing and the new process.
正在考慮對零部件的製造程序進行某種更改。 使用現有程序和新程序進行抽樣,以確定新程序是否會帶來改進。 如果發現現有程序的 1500 件商品中有 75 件有缺陷,而新程序的 2000 件商品中有 80 件有缺陷,請找出現有程序和新程序之間缺陷品比例的真實差異的 90% 信賴區間
(台科大企管)
If the standard deviation of the lifetime of washing machines is estimated to
be 800 hours, how large a sample must be taken in order to be 97% confident
that the margin of error will not exceed 50 hours? (Z0.03 =1.88; Z0.015 = 2.17)〈中興企管》
如果洗衣機壽命的標準偏差估計為 800 小時,那麼必須抽取多大的樣本才能 97% 信賴誤差界限不會超過 50 小時? (Z0.03 =1.88;Z0.015 = 2.17)
經濟部商業司想要了解物流公司推動e化的效果,調查發現 500 家的公司
中有400 家的成效良好,試求:
(1) e化成效良好公司所有接受調查公司比例之估計p的 95%信賴區間為
何?
(2) 若使p的95%信賴區間之區間長度不超過 0.03,其所需之樣本數為何?
〈高第一運籌》
A proportion, p, that many public opinion polls estimate is the number of
citizens who would say yes to the question, "Do you think that it is necessary
for government to fund elementary computer courses in preschools?" In one
such random sample of 300 people, 75 say yes.
(1) Find an approximate 95% confidence interval for p.
(2) If a margin of error ±4 percent is required, how many people should be
questioned?〈暨南資管〉
許多民意調查估計的一個比例 p 是對這個問題說“是”的公民數量,“你認為有必要嗎?
讓政府資助學前班的基礎電腦課程?”在這樣一個隨機抽樣的 300 人中,有 75 人表示同意。
(1) 找到 p 的近似 95% 信賴區間。
(2) 如果要求有±4%的誤差,應該問多少人?〈暨南資管〉
It is desired to estimate the proportion of college students who feel a sudden
relief once their statistics class is over. How many students must be sampled
in order to estimate the true proportion to within 2% at the 90% confidence
level?〈台大國企〉
需要估計統計課結束後突然感到輕鬆的大學生的比例。 在90%的信賴水準下估計真實比例在2%以內,必須抽樣多少學生?〈台大國企〉
Suppose that a city of N = 100000 citizens is interested in assessing the
support for a new policy of waste recycling. To estimate the proportion p in
favor of the new policy, how large a sample is required so that with 95%
confidence the maximum error of the estimate of p is 0.03? 〈暨南資管〉
假設一個有 N = 100000 名市民的城市有興趣評估
支持垃圾回收新政策。 估計比例 p
贊成新政策,需要多大的樣本才能使顯著水準 95%
p 估計的最大誤差為 0.03?
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