According to a 2009 Lawyers.com survey, only 35% of adult Americans had a will. Suppose a recent survey of 100 adult Americans found 40 adults with wills. At the 5% significance level, is the proportion different?
Hint: \( \sqrt{0.2275} = 0.4770 \)
Null Hypothesis \( H_0: p = 0.35 \), Alternate Hypothesis \( H_a: p \neq 0.35 \)
Significance level \( \alpha = 0.05 \), Critical Values \( [-1.96, 1.96] \)
Test Statistics \( z = 1.0482 \)
Reject \( H_0 \)? No
在這項假設檢定中，我們的零假設 \( H_0 \) 為成年美國人有遺囑的比例為 35%。對立假設 \( H_a \) 為該比例不等於 35%。標準誤差是樣本比例標準差的估計，計算公式為 \( SE = \sqrt{\frac{p_0 \cdot (1 - p_0)}{n}} \)，其中 \( p_0 \) 是零假設下的比例，\( n \) 是樣本大小。
使用給定的公式和數據，我們得到標準誤差 \( SE = 0.0477 \)。接著，我們計算了測試統計量 \( z \) 值，這是樣本比例 \( p_1 \) 與零假設比例 \( p_0 \) 差異的度量，其計算公式為 \( z = \frac{p_1 - p_0}{SE} \)。
在 \( \alpha = 0.05 \) 的顯著性水平下，我們的臨界值為 \( \pm 1.96 \)。這表示，如果 \( z \) 值介於 \( -1.96 \) 和 \( 1.96 \) 之間，我們不能拒絕零假設。在本例中，計算得出的 \( z \) 值為 1.0482，未超出臨界值，因此我們不能拒絕零假設，即沒有足夠的證據表明成年美國人有遺囑的比例有所改變。
Suppose taxi fare from Logan Airport to downtown Boston is known to be normally distributed with a standard deviation of $2.40. The last 9 times John has taken a taxi from Logan to downtown Boston, the fares have been $22, $26, $25, $27, $25, $25, $30, $26, $28. What is a 95% confidence interval for the population mean taxi fare?
The 95% confidence interval for the population mean taxi fare is B.[$24.432, $27.568].
在計算信心區間時，我們首先確定了樣本的平均值為 $26.00，然後計算了標準誤差為 $0.80。接著，我們根據 95% 的信心水平找到 \( z \) 分數為 1.96。計算誤差幅度後為 $1.568，因此得出 95% 的信心區間為 [$24.432, $27.568]。
首先，計算 9 次車費的平均值（樣本均值），得到 $26.00。由於已知標準差為 $2.40 和樣本大小為 9，我們可以計算出標準誤差為標準差除以樣本大小的平方根，即 $0.80。根據 95% 的信心水平，相對應的 \( z \) 值約為 1.96。信心區間的計算公式為 \( \text{樣本均值} \pm z \times \text{標準誤差} \)，這裡即為 \( \$26.00 \pm 1.96 \times \$0.80 \)，結果是 $24.432 到 $27.568。因此，我們對總體平均車費的 95% 信心區間是 $24.432 至 $27.568。
A marketing firm wants to estimate how much root beer the average teenager drinks per year. A previous study found a standard deviation of 1 liters. How many teenagers \( n \) = ＿ must the firm interview in order to have a margin of error of at most 0.1 liter when constructing a 91.08% confidence interval?
To construct a 91.08% confidence interval with a margin of error of at most 0.1 liter, the firm must interview at least 289 teenagers.
為了計算所需的樣本大小，我們首先找到了對應於 91.08% 信心水平的 \( z \) 分數為 1.6996。根據誤差幅度公式 \( E = z \frac{\sigma}{\sqrt{n}} \)，我們重組解為 \( n \) 並計算得出結果為 288.875，取整後為 289。
為了計算所需的樣本大小，我們使用信心區間和誤差範圍的公式 \( n = \left(\frac{z \times \sigma}{E}\right)^2 \)，其中 \( z \) 是對應於信心水平的 \( z \) 分數，\( \sigma \) 是標準差，\( E \) 是希望達到的誤差範圍。對於 91.08% 的信心水平，\( z \) 分數約為 1.6996。將這些值代入公式，我們得到 \( n = \left(\frac{1.6996 \times 1}{0.1}\right)^2 = 288.875 \)。由於我們不能訪問部分人，所以我們需要向上取整，即至少需要訪問 289 名青少年。
Professor Elderman has given the same multiple choice final exam in his Principles of Microeconomics class for many years. After examining his records from the past 10 years, he finds that the scores have a mean of 76 and a standard deviation of 12. Professor Elderman offers his class of 36 a pizza party if the class average is above 80. What is the probability that he will have to deliver on his promise?
The probability that Professor Elderman will have to deliver on his promise is A. 0.0228.
我們首先計算班級平均分超過 80 分的標準誤差，得到 \( SE = \frac{12}{\sqrt{36}} = 2 \)。接著計算 \( z \) 分數 \( z = \frac{80 - 76}{2} = 2 \)。使用標準正態分佈表或計算，我們找到 \( z \) 分數大於 2 的概率為 0.0228，這表示 Elderman 教授不太可能需要兌現承諾，因為這一事件發生的機率非常小。
A university administrator expects that 25% of students in a core course will receive an A. He looks at the grades assigned to 60 students. The probability that the proportion of students who receive an A is between 0.20 and 0.35 is _________.
The probability that the proportion of students who receive an A is between 0.20 and 0.35 is B. 0.7766.
大學管理員預計 25% 的學生將在一門核心課程中獲得 A 等級。給定標準差 \( \sigma = 1 \) 並考慮到樣本大小為 60，我們計算得到標準誤差為 \( SE = \sqrt{\frac{0.25 \times 0.75}{60}} \)。然後計算得到 20% 和 35% 對應的 \( z \) 分數，並找到這兩個分數之間的概率，這就是學生獲得 A 等級的比例在 0.20 和 0.35 之間的概率，這個概率是 0.7766。
The time to complete the construction of a soapbox derby car is normally distributed with a mean of three hours and a standard deviation of one hour. Find the probability that it would take between 2.5 and 3.5 hours to construct a soapbox derby car.
The probability that the construction time is between 2.5 and 3.5 hours is B. 0.3830.
建造一輛肥皂箱賽車的時間服從均值為 3 小時、標準差為 1 小時的正態分佈。首先計算 2.5 小時和 3.5 小時對應的 \( z \) 分數：
\( z_{\text{low}} = \frac{2.5 - 3}{1} = -0.5 \)
\( z_{\text{high}} = \frac{3.5 - 3}{1} = 0.5 \)
然後查找這兩個 \( z \) 分數之間的標準正態分佈面積，得到建造時間介於 2.5 小時至 3.5 小時之間的概率。這個概率約為 0.3830，表示有大約 38.30% 的機率建造時間會落在這個範圍內。
The salaries of teachers in a particular school district are normally distributed with a mean of 50,000 and a standard deviation of 2,500. Due to budget limitations, it has been decided that the teachers who are in the top 2.5% of the salaries would not get a raise. What is the salary level that divides the teachers into one group that gets a raise and one that doesn't?
The salary level that divides the teachers is D. $54,900.
某校區教師的薪資服從均值為 50,000 美元、標準差為 2,500 美元的正態分佈。由於預算限制，決定不給排在薪資前 2.5% 的教師加薪。這意味著我們需要找到對應於前 97.5% 的 \( z \) 分數，這可以通過逆標準正態分佈函數來計算：
\( z = \text{invNorm}(1 - 0.025) = 1.96 \)
接著，我們將 \( z \) 分數轉換為薪資水平：
\( \text{薪資水平} = 50,000 + (1.96 \times 2,500) = 54,900 \)
因此，薪資大於等於 54,900 美元的教師將不會獲得加薪。
It is known that 10% of the calculators shipped from a particular factory are defective. What is the probability that at least one in a random sample of four calculators is defective?
The probability that at least one calculator is defective is C. 0.3439.
已知某工廠出貨的計算器中有 10% 是有缺陷的。在隨機抽取的四台計算器中，至少有一台是有缺陷的概率可通過計算全部沒有缺陷的計算器的概率，然後取其餘數得出。完全沒有缺陷的計算器的概率為 \( (1 - 0.10)^4 \)。
計算得 \( (1 - 0.10)^4 = 0.6561 \)，這是四台計算器都沒有缺陷的概率。
因此，至少有一台計算器有缺陷的概率為 \( 1 - 0.6561 = 0.3439 \)，即大約 34.39%。
Romi, a production manager, is trying to improve the efficiency of his assembly line. He knows that the machine is set up correctly only 60% of the time. He also knows that if the machine is set up correctly, it will produce good parts 80% of the time, but if set up incorrectly, it will produce good parts only 20% of the time. Romi starts the machine and produces one part before he begins the production run. He finds the first part to be bad. What is the revised probability that the machine was set up correctly?
The revised probability that the machine was set up correctly after finding the first part to be bad is approximately 0.2727.
根據貝葉斯定理，當我們觀察到第一個零件不良時，機器設置正確的概率需要修正。首先，我們計算出生產不良零件的總概率，然後根據貝葉斯定理，我們可以修正機器設置正確的概率。
生產不良零件的概率是由兩部分組成的：設置正確時生產不良零件的概率，以及設置不正確時生產不良零件的概率。計算如下：
\( P(\text{Bad Part}) = P(\text{Correct}) \times P(\text{Bad | Correct}) + P(\text{Incorrect}) \times P(\text{Bad | Incorrect}) \)
\( P(\text{Bad Part}) = 0.60 \times 0.20 + 0.40 \times 0.80 = 0.44 \)
然後應用貝葉斯定理計算機器設置正確給出不良零件後的修正概率：
\( P(\text{Correct | Bad}) = \frac{P(\text{Correct}) \times P(\text{Bad | Correct})}{P(\text{Bad Part})} \)
\( P(\text{Correct | Bad}) = \frac{0.60 \times 0.20}{0.44} \approx 0.2727 \)
這意味著在第一個零件是不良的情況下，機器設置正確的修正概率大約為 27.27%。
The table below gives statistics relating to a hypothetical 10-year record of two portfolios. Assume other statistics relating to these portfolios are the same and the risk-free rate is 3.5%. Using the coefficient of variation and the Sharpe ratio, the fund that is preferred in terms of relative risk and return per unit of risk is:
The preferred fund in terms of relative risk and return per unit of risk is B. Portfolio A since it has a lower coefficient of variation and a higher Sharpe ratio.
投資組合 A 的變異系數為 1.4902，表示相對於其平均回報，其回報的波動性較大。然而，其 Sharpe 比率為 0.5175，顯示每單位風險的預期超額回報較高。相比之下，投資組合 B 的變異系數為 1.5588，Sharpe 比率為 0.4214，顯示它有較高的風險和較低的每單位風險回報。
根據這些指標，投資組合 A 是較佳的選擇，因為它在承擔相同風險的情況下提供了較高的調整後回報。