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WA8. Estimating the Expectation

1.Estimating the Expectation

statement 1

  • A measurement follows the normal distribution with a standard deviation of 15 and an unknown expectation μ. You can consider that measurement to be the “original” distribution.
  • Two statisticians propose two distinct ways to estimate the unknown quantity μ with the aid of a sample of size 36.
  • They will do that by evaluating two different SAMPLING DISTRIBUTIONS to determine which method is better.
  • There are two statisticians involved in this task: statistician “A” and statistician “B.”
  • Statistician A proposes to use the sampling distribution of the sample average as an estimate.
  • Statistician B proposes to use the sampling distribution of the sample median instead.
  • In order to choose between the two options, they agree to prefer the statistic that has a smaller variance (with respect to the sampling distribution).
  • Tasks 1-9 refer to this problem of comparing the two statistics to each other.

Questions 1

  1. Assume that the actual expectation of the measurement is equal to 5 (μ=5). Then the expectation of the statistic that was proposed by Statistician A is equal to: ?
  2. Assume that the actual expectation of the measurement is equal to 5 (μ=5). Then the standard deviation of the statistic that was proposed by Statistician A is equal to: ?
  3. the measurement is equal to 5 (μ=5). Then the expectation of the statistic that was proposed by Statistician B is equal to: ?
  4. Assume that the actual expectation of the measurement is equal to 5 (μ=5). Then the standard deviation of the statistic that was proposed by Statistician B is equal to: ?
  5. Assume that the actual expectation of the measurement is equal to 2.3 (μ=2.3). Then the expectation of the statistic that was proposed by Statistician A is equal to: ?
  6. Assume that the actual expectation of the measurement is equal to 2.3 (μ=2.3). Then the standard deviation of the statistic that was proposed by Statistician A is equal to: ?
  7. Assume that the actual expectation of the measurement is equal to 2.3 (μ=2.3). Then the expectation of the statistic that was proposed by Statistician B is equal to: ?
  8. Assume that the actual expectation of the measurement is equal to 2.3 (μ=2.3). Then the standard deviation of the statistic that was proposed by Statistician B is equal to: ?
  9. Based on the information collected in Tasks 1-8, which of the two statistics produces values which tends to be more concentrated about the expectation of the measurement?

Solution 1

# N ~ (m, 15)
n = 36
N = 10^5 # population size (assumed)

# m = mu.X = ? # mean of the population
sigma.X = 15 # sd of the population
var.X = (15)^2 # variance of the population

m1 = 5
XA = rep(0, N)
XB = rep(0, N)

for (i in 1:N){
    X.samp = rnorm(n, m1, sd.X) # generate samples of size n=36
    XA[i] = mean(X.samp) # XA: the mean
    XB[i] = median(X.samp) # XB: the median
}

### 1
XA.mu = mean(XA) # mean of XA, or: XA.mu = mu.X = m1 = 5


### 2
XA.var = var(XA)  # variance
XA.sigma = sqrt(XA.var) # sd of XA

### 3
XB.mu = mean(XB) # mean of XB, or XB.mu = mu.X = m1 = 5

### 4
XB.var = var(XB)
XB.sigma = sqrt(XB.var)



m2 = 2.3
YA = rep(0, N)
YB = rep(0, N)

for (i in 1:N){
    Y.samp = rnorm(n, m2, sd.X) # generate samples of size n=36
    YA[i] = mean(Y.samp) # YA: the mean
    YB[i] = median(Y.samp) # YB: the median
}

### 5
YA.mu = mean(YA) # mean of YA


### 6
YA.var = var(YA)  # variance
YA.sigma = sqrt(YA.var) # sd of YA

### 7
YB.mu = mean(YB) # mean of YB,

### 8
YB.var = var(YB)
YB.sigma = sqrt(YB.var)

### 9
# we see that Statistician A has smaller variance in both cases, so he is more accurate.

2.Normal Approximation of the Sampling Distribution of a Sum

statement 2

  • Suppose that the expected number of phone calls that are handled by a switchboard in each second is 5.35.
  • Assume that the distribution of the number of phone calls per second follows the Poisson distribution.
  • Tasks 10-12 refer to this information.

Questions 2

  1. The number of phone calls that the switchboard is expected to handle in 1 minute is: ?
  2. The 0.80-percentile of the number of calls per minute is (approximately) equal to: ?
  3. The probability that the switchboard will need to handle no more than 300 call in a minute is: ?

Solution 2

# Poisson ~ (lambda = 5.35)
lambda = 5.35
mu.X.sec = lambda = 5.35
var.X.sec = 5.35
sigma.X.sec = sqrt(var.X)

mu.X.minute = lambda * 60
var.X.minute= lambda * 60
sigma.X.minute = sqrt(var.X.minute)

### 1
lambda * 60 # number of phone calls per minute

### 2
qnorm(0.80, mu.X.minute, sigma.X.minute)

### 3
# P (x <= 300)
pnorm(300, mu.X.minute, sigma.X.minute)

3. average height

statement 3

  • Recall that the population average of the heights in the file “pop1.csv” is μ = 170.035. Using simulations we found that the probability of the sample average of the height falling within 1 centimeter of the population average is approximately equal to 0.626.
  • From the simulations we also got that the standard deviation of the sample average is (approximately) equal to 1.122.
  • In the next 3 questions you are asked to apply the Normal approximation to the distribution of the sample average using this information. The answer may be rounded up to 3 decimal places of the actual value.

Questions 3

  1. Using the Normal approximation, the probability that sample average of the heights falls within 1 centimeter of the population average is
  2. Using the Normal approximation we get that the central region that contains 95% of the distribution of the sample average is of the form 170.035 ± z · 1.122. The value of z is
  3. Using the Normal approximation, the probability that sample average of the heights is less than 168 is

solution 3

### 1
mu <- 170.035
sig <- 1.122
pnorm(mu+1,mu,sig) - pnorm(mu-1,mu,sig)

### 2
qnorm(0.975)

### 3
pnorm(168,mu,sig)

4. Internal Revenue Service

statement 4

  • According to the Internal Revenue Service, the average length of time for an individual to complete (record keep, learn, prepare, copy, assemble and send) IRS Form 1040 is 10.53 hours (without any attached schedules).
  • The distribution is unknown. Let us assume that the standard deviation is 2 hours.
  • Suppose we randomly sample 36 taxpayers and compute their average time to completing the forms.

Questions 4

  1. Then the probability that the average is more than 11 hours is approximately equal to (The answer may be rounded up to 3 decimal places of the actual value.) ?

solution 4

sd = 2
n = 36
mu = 10.35
1 - pnorm(11, mu, sd/sqrt(n) )

5. class Runners

statement 5

  • Suppose that a category of world class runners are known to run a marathon (26 miles) in an expectation of 145 minutes with a standard deviation of 14 minutes.
  • Consider 49 of the races. In the next 3 questions you are asked to apply the Normal approximation to the distribution of the sample average using this information. The answer may be rounded up to 3 decimal places of the actual value.

Questions 5

  1. The probability that the runner will average between 142 and 146 minutes in these 49 marathons is
  2. The 0.80-percentile for the average of these 49 marathons is
  3. The median of the average running time is

solution 5

mu.X = 145
sd.X = 14
n=49

mu.bar = 145
sd.bar = sd.X / sqrt(n)


###1
pnorm(146, mu.bar, sd.bar) - pnorm(142, mu.bar, sd.bar)

### 2
qnorm(0.80, mu.bar, sd.bar)

### 3
qnorm(0.50, mu.bar, sd.bar)

6. Rural bus wait

statement 6

  • The time to wait for a particular rural bus is distributed uniformly from 0 to 75 minutes.
  • 100 riders are randomly sampled and their waiting times are measured.

Questions 6

  1. The 90th percentile of the average waiting time (in minutes) for a sample of 100 riders is (approximately):

solution 6

n =100
# Uniform (0, 75)
mu.X = 75/2
var.X = (75)^2 / 12
sd.X = sqrt(var.X)

mu.bar = mu.X
sd.bar = sd.X/sqrt(n)
qnorm(0.90, mu.bar, sd.bar )

7. switching board

statement 7

  • A switching board receives a random number of phone calls. The expected number of calls is 5.3 per minute. Assume that the distribution of the number of calls is Poisson.
  • The average number of calls per minute is recorded by counting the total number of calls received in one hour, divided by 60, the number of minutes in an hour.
  • In the next 4 questions you are asked to apply the Normal approximation to the distribution of the sample average using this information. The answer may be rounded up to 3 decimal places of the actual value.

questions 7

  1. The expectation of the average is
  2. The standard deviation of the average is
  3. The probability that the average is less than 5
  4. The probability that number of calls in a random minute is less than 5 is

solution 7

lambda = 5.3

### 1
mu.X = lambda

### 2
sd.X = sqrt(lambda) / sqrt (60)

### 3
pnorm(5, mu.X, sd.X)

### 4
ppois(5, lambda)

8. Computer time before reboot

statement 8

  • It is claimed that the expected length of time some computer part may work before requiring a reboot is 2 months. In order to examine this claim 80 identical parts are set to work.
  • Assume that the distribution of the length of time the part can work (in months) is Exponential.
  • In the next 4 questions you are asked to apply the Normal approximation to the distribution of the average of the 80 parts that are examined.
  • The answer may be rounded up to 3 decimal places of the actual value.

Questions 8

  1. The expectation of the average is
  2. The standard deviation of the average is
  3. The central region that contains 90% of the distribution of the average is of the form E(X) ± c, where E(X) is the expectation of the sample average. The value of c is
  4. The probability that the average is more than 2.5 months is

solution 8

lambda = 2
n = 80

### 1
mu.X = 2

### 2
sd.X = mu.X / sqrt(n)

### 3
# mu.X + qnorm(0.90) . sd.X =>
qnorm(0.95) * sd.X

### 4
1 - pnorm(2.5, mu.X, sd.X)