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question:In the 2016 US presidential election, was New Hampshire the only state where more people identified as liberal than conservative and still voted more Republican than the national average?
answer:Yes, New Hampshire was the sole state with a higher percentage of self-identified liberals than conservatives that voted more Republican than the national average by a margin of 2.22 percentage points. According to Gallup, the six states with more liberals than conservatives are New Hampshire, New York, Washington, Vermont, Hawaii, and Massachusetts. While New Hampshire had a slim 0.37-point difference in favor of Clinton (47.62% to Trump's 47.25%), the national average difference was 2.1 percentage points. Here is a summary of the data for these states: State | Conservative (%) | Moderate (%) | Liberal (%) | Conservative Advantage | Clinton Vote (%) | Trump Vote (%) | Clinton Lead (pp) | National Advantage Difference (pp) New Hampshire | 28 | 36 | 30 | -2 | 47.62 | 47.25 | 0.37 | -1.85 New York | 27 | 35 | 30 | -3 | 59.01 | 36.52 | 22.49 | 20.27 Washington | 28 | 37 | 31 | -3 | 54.3 | 38.07 | 16.23 | 14.01 Vermont | 28 | 36 | 32 | -4 | 55.72 | 29.76 | 25.96 | 23.74 Hawaii | 22 | 45 | 28 | -6 | 62.88 | 30.36 | 32.52 | 30.3 Massachusetts | 21 | 38 | 35 | -14 | 60.98 | 33.34 | 27.64 | 25.42 Note that New Hampshire's difference from the national average is -1.85 percentage points, indicating it leaned more Republican compared to the overall country.
question:What is the name of a random variable that takes on values of either 0 or i with probabilities p and 1-p, respectively? It's similar to a Bernoulli random variable, but I can't find its specific name.
answer:This random variable is a type of indicator variable. Let 1_A be the random variable for which 1_A(omega)=1 if omegain A and 0 otherwise, where A is an event. Then your random variable can be described as i1_A. It's equal to i if omegain A, which happens with probability P(A), and 0 otherwise, which happens with probability 1-P(A).
question:Mercer Manufacturing's production data for the year is as follows: - Actual raw materials used: 29,000 lbs. at 5.10 per lb - Actual factory payroll: 8,100 hours for a total of 164,430 - Actual units produced: 48,000 The budgeted standards for each unit are: - 0.50 lbs. of raw material at 5.05 per lb - 10 minutes of direct labor at 22.00 per hour Calculate the following variances: 1. Direct materials price and quantity variances. 2. Direct labor rate and efficiency variances.
answer:1. Direct Materials Variances: - Standard quantity for actual production: 24,000 lbs (48,000 units * 0.50 lbs/unit) Price Variance: [ text{Price Variance} = (text{Actual Price} - text{Standard Price}) times text{Actual Quantity Purchased} ] [ text{Price Variance} = (5.10 - 5.05) times 29,000 ] [ text{Price Variance} = 0.05 times 29,000 ] [ text{Price Variance} = 1,450 text{ (Unfavorable)} ] Quantity Variance: [ text{Quantity Variance} = text{Standard Price} times (text{Actual Quantity Used} - text{Standard Quantity Required}) ] [ text{Quantity Variance} = 5.05 times (29,000 - 24,000) ] [ text{Quantity Variance} = 5.05 times 5,000 ] [ text{Quantity Variance} = 25,250 text{ (Unfavorable)} ] 2. Direct Labor Variances: - Standard labor hours for actual production: 8,000 hours (48,000 units * 10 min/unit / 60 min/hour) Rate Variance: [ text{Rate Variance} = (text{Actual Rate} - text{Standard Rate}) times text{Actual Labor Hours} ] [ text{Actual Rate} = frac{text{Actual Factory Payroll}}{text{Actual Labor Hours}} ] [ text{Actual Rate} = frac{164,430}{8,100} ] [ text{Actual Rate} = 20.30 ] [ text{Rate Variance} = (20.30 - 22.00) times 8,100 ] [ text{Rate Variance} = (-1.70) times 8,100 ] [ text{Rate Variance} = -13,770 text{ ( Favorable)} ] Efficiency Variance: [ text{Efficiency Variance} = text{Standard Price per Labor Hour} times (text{Actual Labor Hours} - text{Standard Labor Hours}) ] [ text{Efficiency Variance} = 22.00 times (8,100 - 8,000) ] [ text{Efficiency Variance} = 22.00 times 100 ] [ text{Efficiency Variance} = 2,200 text{ (Unfavorable)} ]
question:Consider a sequence of independent and identically distributed (i.i.d.) random variables X_1, X_2, cdots, X_n, each following a Bernoulli distribution with success probability p. Show that the sample mean bar{x} is an unbiased estimator of p.
answer:To demonstrate that bar{x} is an unbiased estimator of p, we first define the sample mean as follows: bar{x} = frac{1}{n}sum_{i=1}^n X_i Now, take the expectation of both sides: E(bar{x}) = Eleft(frac{1}{n}sum_{i=1}^n X_iright) Since the expectation operator is linear and the X_i are i.i.d., we can write: E(bar{x}) = frac{1}{n}sum_{i=1}^n E(X_i) For a Bernoulli random variable, the expectation of X_i is p: E(X_i) = p Substituting this back into the equation: E(bar{x}) = frac{1}{n}sum_{i=1}^n p As p is a constant, it can be factored out of the sum: E(bar{x}) = frac{p}{n}(n) Simplifying: E(bar{x}) = p Since p is the true mean of the Bernoulli distribution, we conclude that bar{x} is an unbiased estimator of p, as it yields the true parameter value on average: E(bar{x}) = p = mu where mu is the mean of the Bernoulli distribution. The key steps in this proof are the linearity of expectation and the identical distribution of the random variables.