Statistics for Business and Economics

Transcription

1 Statistics for Business and Economics Chapter 6 Sampling and Sampling Distributions Ch. 6-1

2 6.1 Tools of Business Statistics n Descriptive statistics n Collecting, presenting, and describing data n Inferential statistics n Drawing conclusions and/or making decisions concerning a population based only on sample data Ch. 6-2

3 Populations and Samples n A Population is the set of all items or individuals of interest n Examples: All likely voters in the next election All parts produced today All sales receipts for November n A Sample is a subset of the population n Examples: 1000 voters selected at random for interview A few parts selected for destructive testing Random receipts selected for audit Ch. 6-3

4 Population vs. Sample Population a b c d ef gh i jk l m n o p q rs t u v w x y z Sample b c g i n o r u y Ch. 6-4

5 Why Sample? n Less time consuming than a census n Less costly to administer than a census n It is possible to obtain statistical results of a sufficiently high precision based on samples. Ch. 6-5

6 Simple Random Samples n n n Every object in the population has an equal chance of being selected Objects are selected independently Samples can be obtained from a table of random numbers or computer random number generators n A simple random sample is the ideal against which other sample methods are compared Ch. 6-6

7 Inferential Statistics n Making statements about a population by examining sample results Sample statistics Population parameters (known) Inference (unknown, but can be estimated from sample evidence) Sample Population Ch. 6-7

8 Inferential Statistics Drawing conclusions and/or making decisions concerning a population based on sample results. n Estimation n e.g., Estimate the population mean weight using the sample mean weight n Hypothesis Testing n e.g., Use sample evidence to test the claim that the population mean weight is 120 pounds Ch. 6-8

9 6.2 Sampling Distributions n A sampling distribution is a distribution of all of the possible values of a statistic for a given size sample selected from a population Ch. 6-9

10 Chapter Outline Sampling Distributions Sampling Distribution of Sample Mean Sampling Distribution of Sample Proportion Sampling Distribution of Sample Variance Ch. 6-10

11 Sampling Distributions of Sample Means Sampling Distributions Sampling Distribution of Sample Mean Sampling Distribution of Sample Proportion Sampling Distribution of Sample Variance Ch. 6-11

12 Developing a Sampling Distribution n Assume there is a population n Four types of people A B C D n Random variable, X, is age of individuals n Possible Values of X: 18, 20, 22, 24 (years) Ch. 6-12

13 Developing a Sampling Distribution (continued) Summary Measures for the Population Distribution: P(x) µ = = σ 2 = (X i µ) 2 4 = A B C D x Uniform Distribution Ch. 6-13

14 Developing a Sampling Distribution Now consider all possible samples of size n = 2 1 st 2 nd Observation Obs ,18 18,20 18,22 18, ,18 20,20 20,22 20, ,18 22,20 22,22 22, ,18 24,20 24,22 24,24 16 possible samples (sampling with replacement) (continued) 16 Sample Means 1st 2nd Observation Obs Ch. 6-14

15 Developing a Sampling Distribution Sampling Distribution of All Sample Means (continued) 16 Sample Means 1st 2nd Observation Obs P(X) _ Sample Means Distribution (no longer uniform) _ X Ch. 6-15

16 Developing a Sampling Distribution Summary Measures of this Sampling Distribution: (continued) µ X = ! = 21 σ X = (18-21)2 + 2 (19-21) 2 +!+ (24-21) 2 16 =1.58 Ch. 6-16

17 Comparing the Population with its Sampling Distribution P(X).3.2 Population µ = 21 σ = Sample Means Distribution n = 2 P(X).3.2 µ X = 21 σ X =1.58 _ A B C D X _ X Ch. 6-17

18 Expected Value of Sample Mean n Let X 1, X 2,... X n represent a random sample from a population n The sample mean value of these observations is defined as X = 1 n n i= 1 X i Ch. 6-18

19 Standard Error of the Mean n n Different samples of the same size from the same population will yield different sample means A measure of the variability in the mean from sample to sample is given by the Standard Error of the Mean: σ X = σ n n n Note that the standard error of the mean decreases as the sample size increases If n=2, then σ /σ X = 2 = Ch. 6-19

20 If the Population is Normal n If a population is normal with mean µ and standard deviation σ, the sampling distribution of is also normally distributed with X µ X = µ and σ X = σ n Ch. 6-20

21 Z-value for Sampling Distribution of the Mean n Z-value for the sampling distribution of : X Z = (X µ) σ X where: X µ σ x = sample mean = population mean = standard error of the mean Ch. 6-21

22 Sampling Distribution Properties µ x = µ Normal Population Distribution x (i.e. is unbiased ) Normal Sampling Distribution (has the same mean) µ x µ x x Ch. 6-22

23 Sampling Distribution Properties (continued) n For sampling with replacement: As n increases, σ x decreases Larger sample size Smaller sample size µ x Ch. 6-23

24 If the Population is not Normal n We can apply the Central Limit Theorem: n Even if the population is not normal, n sample means from the population will be approximately normal as long as the sample size is large enough. Properties of the sampling distribution: µ x = µ and σ x = σ n Ch. 6-24

25 Central Limit Theorem As the sample size gets large enough n the sampling distribution becomes almost normal regardless of shape of population x Ch. 6-25

26 If the Population is not Normal (continued) Sampling distribution properties: Central Tendency Variation µ x = µ σ x = σ n Population Distribution Sampling Distribution (becomes normal as n increases) Smaller sample size µ Larger sample size x µ x x Ch. 6-26

27 Example n Suppose a large population has mean µ = 8 and standard deviation σ = 3. Suppose a random sample of size n = 36 is selected. n What is the probability that the sample mean is between 7.8 and 8.2? Ch. 6-27

28 Example Solution: (continued) n Even though the population is not normally distributed, we use the central limit theorem to get an approximated solution n the sampling distribution of approximately normal x is µ x n with mean = 8 n and standard deviation σ 3 σ x = = = n Ch. 6-28

29 Example Solution (continued): (continued) P(7.8 < X < 8.2) X - µ = P < < 3 3 σ 36 n 36 = P(-0.4 < Z < 0.4) = Population Distribution???????????? Sampling Distribution Sample Standard Normal Distribution Standardize µ = 8 X µ X = 8 µ z = 0 x Z Ch. 6-29

30 Acceptance Intervals n Goal: determine a range within which sample means are likely to occur, given a population mean and variance n n n By the Central Limit Theorem, we know that the distribution of X is approximately normal if n is large enough, with mean µ and standard deviation Let z α/2 be the z-value that leaves area α/2 in the upper tail of the normal distribution (i.e., the interval - z α/2 to z α/2 encloses probability 1 α) Then σ X µ ± zα /2σ X is the interval that includes X with probability 1 α Ch. 6-30

31 6.3 Sampling Distributions of Sample Proportions Sampling Distributions Sampling Distribution of Sample Mean Sampling Distribution of Sample Proportion Sampling Distribution of Sample Variance Ch. 6-31

32 Sampling Distributions of Sample Proportions p = the proportion of the population having some characteristic n Sample proportion ( pˆ ) provides an estimate of p: ˆp = number of items in the sample having the characteristic of interest sample size n 0 pˆ 1 n pˆ has a binomial distribution, but can be approximated by a normal distribution when n is large Ch. 6-32

33 Sampling Distribution of p ^ n Normal approximation: P(ˆp) Sampling Distribution ˆp Properties: E(ˆp) = p and σ 2 ˆp = Var ( ˆp ) = p(1 p) n (where p = population proportion) Ch. 6-33

34 Z-Value for Proportions Standardize pˆ to a Z value with the formula: Z = ˆp p σ ˆp = ˆp p p(1 p) n Ch. 6-34

35 Example n If the true proportion of voters who support Proposition A is p =.4, what is the probability that a sample of size 200 yields a sample proportion between.40 and.45? i.e.: if p =.4 and n = 200, what is P(.40 pˆ.45)? Ch. 6-35

36 n Example if p =.4 and n = 200, what is pˆ P(.40.45)? (continued) σp ˆ Find : σ ˆp = p(1 p) n =.4(1.4) 200 = Convert to standard normal: P(.40 pˆ.45) = = P Z P(0 Z 1.44) Ch. 6-36

37 n Example if P =.4 and n = 200, what is pˆ P(.40.45)? (continued) Use standard normal table: P(0 Z 1.44) =.4251 Sampling Distribution Standardized Normal Distribution.4251 Standardize pˆ Z Ch. 6-37

38 6.4 Sampling Distributions of Sample Variance Sampling Distributions Sampling Distribution of Sample Mean Sampling Distribution of Sample Proportion Sampling Distribution of Sample Variance Ch. 6-38

39 Sample Variance n Let x 1, x 2,..., x n be a random sample from a population. The sample variance is s 2 = 1 n 1 n i= 1 (x i x) 2 n the square root of the sample variance is called the sample standard deviation n the sample variance is different for different random samples from the same population Ch. 6-39

40 Sampling Distribution of Sample Variances n The sampling distribution of s 2 has mean σ 2 n 2 E(s ) = If the population distribution is normal, then Var(s 2 σ 2 4 2σ ) = n 1 n If the population distribution is normal then 2 (n-1)s 2 σ has a χ 2 distribution with n 1 degrees of freedom Ch. 6-40

41 The Chi-square Distribution n The chi-square distribution is a family of distributions, depending on degrees of freedom: n d.f. = n 1 χ 2 χ 2 χ d.f. = 1 d.f. = 5 d.f. = 15 n Text Table 7 contains chi-square probabilities Ch. 6-41

42 Degrees of Freedom (df) Idea: Number of observations that are free to vary after sample mean has been calculated Example: Suppose the mean of 3 numbers is 8.0 Let X 1 = 7 Let X 2 = 8 What is X 3? If the mean of these three values is 8.0, then X 3 must be 9 (i.e., X 3 is not free to vary) Here, n = 3, so degrees of freedom = n 1 = 3 1 = 2 (2 values can be any numbers, but the third is not free to vary for a given mean) Ch. 6-42

43 Chi-square Example n A commercial freezer must hold a selected temperature with little variation. Specifications call for a standard deviation of no more than 4 degrees (a variance of 16 degrees 2 ). A sample of 14 freezers is to be tested What is the upper limit (K) for the sample variance such that the probability of exceeding this limit, given that the population standard deviation is 4, is less than 0.05? Ch. 6-43

44 Finding the Chi-square Value χ 2 = (n 1)s 2 σ 2 Is chi-square distributed with (n 1) = 13 degrees of freedom n Use the the chi-square distribution with area 0.05 in the upper tail: χ 2 13 = (α =.05 and 14 1 = 13 d.f.) probability α =.05 χ 2 χ 2 13 = Ch. 6-44

45 Chi-square Example (continued) χ 2 13 = (α =.05 and 14 1 = 13 d.f.) So: P(s 2 2 (n 1)s > K) = P > χ = 0.05 or (n 1)K 16 = (where n = 14) (22.36)(16) so K = = (14 1) If s 2 from the sample of size n = 14 is greater than 27.52, there is strong evidence to suggest the population variance exceeds 16. Ch. 6-45