21 Jun statistics assignment
HW1_STAT206.pdf
Statistical Inference II: J. Lee Assignment 1
Problem 1. Suppose the day after the Drexel-Northeastern basketball game, a poll of 1000 Drexel students was conducted and it was determined that 850 out of the 1000 watched the game (live or on television). Assume that this was a simple random sample and that the Drexel undergraduate population is 20000.
(a) Generate an unbiased estimate of the true proportion of Drexel undergraduate students who watched the game.
(b) What is your estimated standard error for the proportion estimate in (a)?
(c) Give a 95% confidence interval for the true proportion of Drexel undergraduate students who watched the game.
Problem 2. (Exercise 18 in Chapter 7 of Rice) From independent surveys of two populations, 90% con- fidence intervals for the population means are conducted. What is the probability that neither interval contains the respective population mean? That both do?
Problem 3. (Exercise 23 in Chapter 7 of Rice)
(a) Show that the standard error of an estimated proportion is largest when p = 1/2.
(b) Use this result and Corollary B of Section 7.3.2 (also, on Page 17 of the lecture notes) to conclude that the quantity
1
2
√ N − n
N(n− 1)
is a conservative estimate of the standard error of p̂ no matter what the value of p may be.
(c) Use the central limit theorem to conclude that the interval
p̂±
√ N − n
N(n− 1)
contains p with probability at least .95.
HW2_STAT206.pdf
Statistical Inference II: J. Lee Assignment 2
Problem 1. The following data set represents the number of NBA games in January 2016, watched by 10 randomly selected student in STAT 206.
7, 0, 4, 2, 2, 1, 0, 1, 2, 3
(a) What is the sample mean?
(b) Calculate sample variance.
(c) Estimate the mean number of NBA games watched by a student in January 2016.
(d) Estimate the standard error of the estimated mean.
Problem 2. True or false? Tell me why for the false statements.
(a) The center of a 95% confidence interval for the population mean is a random variable.
(b) A 95% confidence interval for µ contains the sample mean with probability .95.
(c) A 95% confidence interval contains 95% of the population.
(d) Out of one hundred 95% confidence intervals for µ, 95 will contain µ.
Problem 3. An investigator quantifies her uncertainty about the estimate of a population mean by reporting X ± sX . What size confidence interval is?
Problem 4. For a random sample of size n from a population of size N , consider the following as an estimate of µ:
Xc =
n∑ i=1
ciXi,
where the ci are fixed numbers and X1, . . . , Xn are the sample. Find a condition on the ci such that the estimate is unbiased.
Problem 5. A sample of size 100 has the sample mean X = 10. Suppose the we know that the population standard deviation σ = 5. Find a 95% confidence interval for the population mean µ.
Problem 6. Suppose the we know that the population standard deviation σ = 5. Then how large should a sample be to estimate the population mean µ with a margin of error not exceeding 0.5?
Problem 7. You flip a fair coin n times and keep track of the sample mean, X (n)
(the fraction of heads
among the n flips). Of course, when n is very large, you expect that the random variable X (n)
will be very close to 0.5 (since the coin is fair).
(a) Use the Central Limit Theorem to estimate how large n must be in order for you to be 95% confident
that X (n)
is between 0.45 and 0.55.
(b) Use Chebyshev inequality to obtain a number K such that you can guarantee that if n is at least K,
then the probability that X (n)
is between 0.45 and 0.55 is at least 0.95.
Rice_HW1.pdf
Dr. Jinwook Lee !
Survey Sampling
(Ref: Ch 7.1-7.3.3 in Rice)
2!
Introduction Many applications of statistics is for inference on a fixed and finite population – estimation of population parameters and providing some sort of quantification of accuracy. Typically the estimates/ accuracies are generated via some sort of “random” sampling of the population. This Lecture will describe the appropriate probability (and hence statistical) models for results of random sampling.
3!
(a) A population is a class of things/elements and we denote its size by N. We assume that associated to each thing/element is a number xi which is the characteristic of interest. So a population is: (b) Population mean is: (c) Population total is: (d) Population variance is:
Population Parameters
4!
(e) An important special case is when all of the xi’s are 0 or 1. • The population consists of those having or not having a
particular characteristic. – – • Refer to the population mean as the population proportion and
denote it by p. • In this case the variance is:
Population Parameters
5!
Definition. For a population of size N, we say that a random sample of size n is a simple random sample (srs) if (i) Sampling is done without replacement. (ii) All ︎“N choose n” ︎(“N combination n”) subsets of size n in the population have an equally likely chance of being chosen. Remark. Actually carrying out a srs can be very hard to do in practice.
Sampling
6!
Example.
Sampling
7!
Suppose X1,X2,…,Xn are random variables representing a srs from a population. They are not independent (due to the sampling without replacement), but they do have common mean and variance. Proposition. Suppose X1, X2, . . . , Xn is a srs from a population of size N, mean µ, and variance of σ2. Then
Expectation and Variance for srs
8!
Definitions. (a) Suppose that X1,X2,…,Xn denote a srs of size n. Then the
sample mean is defined as: (b) In the case where population values are 0 or 1, then the sample proportion is: (c) For a srs of X1,X2,…,Xn from a population, a natural estimate of the population total τ is given by: Remarks. The sample mean, the sample proportion, and the population total estimate are natural estimates for the respective population parameters of mean µ, population proportion p, and population total τ.
Sample Mean as Estimate
9!
Remark. As in prediction, we want to quantify how good the estimates: are in estimating parameters µ, τ, p – natural to do this via mean-squared error (MSE) which can be defined for any estimator. Definition. Bias of the estimate is defined as: Remark.
MSE/Bias/Variance of Estimators
µ̂
10!
Corrollary. Suppose that X1,X2,…,Xn is a srs from a population with mean µ and total τ. Then (a) is an unbiased estimate for µ. (b) T is an unbiased estimate for τ. (c) if the population consists of 0 and 1, the sample proportion is an unbiased estimate for p.
Sample Means are Unbiased
p̂
X
11!
Remarks.
Sample Means are Unbiased
12!
Theorem. Suppose that X1,X2,…,Xn are random variables corresponding to a srs from a population of size N and which has a mean of µ and variance of σ2. Then Remarks.
Variance of sample mean from a srs
13!
Notation and Terminology. Corollary 1. Corollary 2.
Variance of sample mean from a srs
14!
Remark. Typically one does not know the mean or the variance of a population – that is why one is sampling and doing estimation. The standard errors of the and T depend on the underlying population standard deviation, and the standard error of depends on the population proportion p, the very parameter we are trying to estimate. Recall.
Estimation of Population Variance
X
p̂
15!
Theorem. Suppose X1,X2,…,Xn is a srs from a population of size N with a mean of µ and variance of σ2. Then we have:
Estimation of Population Variance
16!
We want to translate previous results into deriving unbiased estimates for the standard errors of our 3 estimators. Notation. Let denote the estimate of the standard error . Corollary A. Corollary A´.
Unbiased estimation of standard errors
sX �X
17!
Corollary B. Terminology. We refer to as estimated standard errors. For each, the squared values are unbiased estimates of the variance.
Unbiased estimation of standard errors
sX , sT , sp̂
18!
Population Parameters, Estimates, Std. Errors
19!
Summary
20!
• In case of sampling with replacement, could invoke the CLT to derive the approximate sampling distribution for reasonable sized n, i.e., n ≥ 25
• There is a generalization of the CLT which applies for the case where one has a srs – essentially says that CLT applies if – sample size n is large enough and – the sampling fraction, n, is small enough
CLT Approximation for srs
21!
CLT Result for srs
22!
Recall. Remark. It is desirable to have a more direct statement quantifying the accuracy of the estimate – one standard way for doing this is via confidence intervals.
Introduction to Confidence Intervals
23!
Definition. Suppose θ is a (general) population parameter and X1, X2, . . . , Xn is a srs. Then a 100(1−α)% confidence interval is (a) an random interval I to be computed/derived from the srs. (b) in advance we know that P(θ I) ≥ 1 − α.
Overview of Confidence Intervals
24!
Remarks. • Once the data is collected and a confidence interval is
computed, the parameter is either in or out of the confidence interval – there is no longer any probability.
! Hence the terminology of term confidence interval (as opposed to probability interval)!
• One potentially helpful interpretation is that if one were to collect srs’s and compute 95% confidence intervals over and over again (say a 1000 times), approximately 95% of those intervals would contain the true parameter
! We simply do not know which ones! • Typical values for α are .10, .05., 01, resulting in 90, 95, and
99 percent confidence intervals
Overview of Confidence Intervals
25!
Specifics of Confidence Intervals
26!
• Confidence intervals for µ
• Confidence intervals for p
Confidence Intervals for srs
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