Exam 6: Sampling Distributions

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 The sample mean, xˉ, is a statistic. \text { The sample mean, } \bar { x } \text {, is a statistic. }

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The probability distribution shown below describes a population of measurements. x 0 2 4 p(x) 1/3 1/3 1/3 Suppose that we took repeated random samples of n = 2 observations from the population described above. Find the expected value of the sampling distribution of the sample mean.

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Consider the population described by the probability distribution below. x 0 2 4 p(x) a. Find μ\mu . b. Find the sampling distribution of the sample mean for a random sample of n=3n = 3 measurements from this distribution. c. Find the sampling distribution of the sample median for a random sample of n=3n = 3 observations from this population. d. Show that both the mean and the median are unbiased estimators of μ\mu for this population. e. Find the variances of the sampling distributions of the sample mean and the sample median. f. Which estimator would you use to estimate μ\mu ? Why?

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a. μ=E(x)=13(0)+13(2)+13(4)=2\quad \mu = E ( x ) = \frac { 1 } { 3 } ( 0 ) + \frac { 1 } { 3 } ( 2 ) + \frac { 1 } { 3 } ( 4 ) = 2
b.
xˉ023432831034p(xˉ)12719297272919127\begin{array}{c|c|c|c|c|c|c|c}\bar{x} & 0 & \frac{2}{3} & \frac{4}{3} & 2 & \frac{8}{3} & \frac{10}{3} & 4 \\\hline p(\bar{x}) & \frac{1}{27} & \frac{1}{9} & \frac{2}{9} & \frac{7}{27} & \frac{2}{9} & \frac{1}{9} & \frac{1}{27}\end{array}
c.
M024p(M)7271327727\begin{array}{c|c|c|c}M & 0 & 2 & 4 \\\hline p(M) & \frac{7}{27} & \frac{13}{27} & \frac{7}{27}\end{array}

d. E(xˉ)=127(0)+19(23)+29(43)+727(2)+29(83)+19(103)+127(4)=2E ( \bar { x } ) = \frac { 1 } { 27 } ( 0 ) + \frac { 1 } { 9 } \left( \frac { 2 } { 3 } \right) + \frac { 2 } { 9 } \left( \frac { 4 } { 3 } \right) + \frac { 7 } { 27 } ( 2 ) + \frac { 2 } { 9 } \left( \frac { 8 } { 3 } \right) + \frac { 1 } { 9 } \left( \frac { 10 } { 3 } \right) + \frac { 1 } { 27 } ( 4 ) = 2 ; Since E(xˉ)=μE ( \bar { x } ) = \mu , the sample
mean is an unbiased estimator of μ\mu .
E(M)=727(0)+1327(2)+727(4)=2E ( M ) = \frac { 7 } { 27 } ( 0 ) + \frac { 13 } { 27 } ( 2 ) + \frac { 7 } { 27 } ( 4 ) = 2 ; Since E(M)=μE ( M ) = \mu , the sample median is an unbiased estimator
of μ\mu .
e. σ2x=127(02)2+19(232)2+29(432)2+727(22)2+29(832)2\quad \sigma \frac { 2 } { x } = \frac { 1 } { 27 } ( 0 - 2 ) ^ { 2 } + \frac { 1 } { 9 } \left( \frac { 2 } { 3 } - 2 \right) ^ { 2 } + \frac { 2 } { 9 } \left( \frac { 4 } { 3 } - 2 \right) ^ { 2 } + \frac { 7 } { 27 } ( 2 - 2 ) ^ { 2 } + \frac { 2 } { 9 } \left( \frac { 8 } { 3 } - 2 \right) ^ { 2 } +19(1032)2+127(42)2=89+ \frac { 1 } { 9 } \left( \frac { 10 } { 3 } - 2 \right) ^ { 2 } + \frac { 1 } { 27 } ( 4 - 2 ) ^ { 2 } = \frac { 8 } { 9 }
σM2=727(02)2+1327(22)2+727(42)2=5627\sigma _ { M } ^ { 2 } = \frac { 7 } { 27 } ( 0 - 2 ) ^ { 2 } + \frac { 13 } { 27 } ( 2 - 2 ) ^ { 2 } + \frac { 7 } { 27 } ( 4 - 2 ) ^ { 2 } = \frac { 56 } { 27 }
f. sample mean; The variance is smaller.

The weight of corn chips dispensed into a 16-ounce bag by the dispensing machine has been identified as possessing a normal distribution with a mean of 16.5 ounces and a standard deviation Of 0.2 ounce. Suppose 100 bags of chips are randomly selected. Find the probability that the mean Weight of these 100 bags exceeds 16.6 ounces.

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A point estimator of a population parameter is a rule or formula which tells us how to use sample data to calculate a single number that can be used as an estimate of the population parameter.

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The number of cars running a red light in a day, at a given intersection, possesses a distribution with a mean of 2.4 cars and a standard deviation of 4. The number of cars running the red light Was observed on 100 randomly chosen days and the mean number of cars calculated. Describe the Sampling distribution of the sample mean.

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The probability of success, pp , in a binomial experiment is a parameter, while the mean and standard deviation, μ\mu and σ\sigma , are statistics.

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Suppose studentsʹ ages follow a skewed right distribution with a mean of 24 years old and a standard deviation of 3 years. If we randomly sample 350 students, which of the following Statements about the sampling distribution of the sample mean age is incorrect?

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A random sample of n=600\mathrm { n } = 600 measurements is drawn from a binomial population with probability of success .08. Give the mean and the standard deviation of the sampling distribution of the sample proportion, p.

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In most situations, the true mean and standard deviation are unknown quantities that have to be estimated.

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The Central Limit Theorem is important in statistics because _____.

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The length of time a traffic signal stays green (nicknamed the ʺgreen timeʺ)at a particular intersection follows a normal probability distribution with a mean of 200 seconds and the standard Deviation of 10 seconds. Use this information to answer the following questions. Which of the Following describes the derivation of the sampling distribution of the sample mean?

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A statistic is biased if the mean of the sampling distribution is equal to the parameter it is intended to estimate.

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The sampling distribution of the sample mean is shown below. 4 5 6 7 8 () 1/9 2/9 3/9 2/9 1/9 Find the expected value of the sampling distribution of the sample mean.

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A random sample of size nn is to be drawn from a population with μ=1200\mu = 1200 and σ=200\sigma = 200 . What size sample would be necessary in order to reduce the standard error to 25 ?

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The probability distribution shown below describes a population of measurements that can assume values of 1, 5, 9, and 13, each of which occurs with the same frequency: x 1 5 9 13 p(x) Consider taking samples of n=2n = 2 measurements and calculating xˉ\bar { x } for each sample. Construct the probability histogram for the sampling distribution of xˉ\bar { x } .

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Which of the following does the Central Limit Theorem allow us to disregard when working with the sampling distribution of the sample mean?

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Suppose a random sample of n\mathrm { n } measurements is selected from a binomial population with probability of success p=.35p = .35 . Given n=200n = 200 , describe the shape, and find the mean and the standard deviation of the sampling distribution of the sample proportion, p^\hat { p } .

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One year, the distribution of salaries for professional sports players had mean $1.6 million and standard deviation $0.8 million. Suppose a sample of 400 major league players was taken. Find the Approximate probability that the average salary of the 400 players that year exceeded $1.1 million.

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The probability distribution shown below describes a population of measurements that can assume values of 5, 10, 15, and 20, each of which occurs with the same frequency: x 5 10 15 20 p(x) Find E(x)=μE ( x ) = \mu . Then consider taking samples of n=2n = 2 measurements and calculating xˉ\bar { x } for each sample. Find the expected value, E(xˉ)E ( \bar { x } ) , of xˉ\bar { x } .

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