Sampling, Confidence Interval and Hypothesis Testing

Transcription

1 Sampling, Confidence Interval and Hypothesis Testing Christopher Grigoriou Executive MBA HEC Lausanne

2 Sampling : Careful with convenience samples! World War II: A statistical study to decide where to put more armor on bombers It seems bombers came back with holes in them, sometimes lots of holes. Somebody in the brass suggested that they put more armor in the places that had the most holes, figuring those places must be getting hit more often. 2

3 Distribution of the sample mean A machine is set up so that the average content of juice per bottle equals µ. The actual amounts per bottle are distributed around this average with a standard deviation σ = 5cl. Consider a sample of 50 bottles. x i : content of a randomly selected bottle _ x: average content per bottle for the sample Distribution of x: E( x ) = µ SD( x ) = σ/ n Shape: Normal Assumptions: Population standard deviation σ known Large sample Large population 3

4 Example: Fraction of heads when tossing a fair coin repeatedly n = n = n = n =

5 Confidence intervals: Motivation The latest poll (1,100 respondents) reveals that 54% of the population supports the government's budgetary decisions. The margin of error is ± 3%. ==> Point estimate: 54% Margin of error: ± 3% ==> Confidence interval: [51%, 57%] Interpretation: If many such polls were taken the true % of respondents supporting the government would lie in the calculated interval in most instances 5

6 Confidence intervals: Motivation (2) 100 measurements of the time required to check a customer in Average check-in time = 3.5 minutes (a) How reliable is this estimate? (b) How certain can we be that on a given day the actual average check-in time is between 3 and 4 minutes? (c) Can we define a duration such that we are 99% certain that the actual average check-in time does not exceed this duration? 6

7 Confidence intervals: The basic idea Population µ, σ, p Sampling Sample Inference: Confidence interval Estimates _ x, s, p^ _ SE = SD( x ) Summarising data First focus on sample mean Then do a similar analysis for proportions 7

8 Example A machine is set up such that the average content of juice per bottle equals µ. A sample of 100 bottles yields an average content of 48cl. Calculate a 90% and a 95% confidence interval for the average content. Assume that the population standard deviation σ = 5cl. 8

9 Sample size One day there was a fire in a wastebasket in the dean s office and in rushed a physicist, a chemist and a statistician. The physicists immediately started to work on how much energy would have to be removed from the fire to stop combustion. The chemist worked on which reagent would have to be added to the fire to prevent oxidation. While they were doing this, the statistician was setting fires to all the other wastebaskets in the office. What are you doing? they asked? The statistician replied, well to solve the problem, obviously, you need a large sample size 9

10 Sample size What sample size is required to estimate the average contents to within 0.5cl at the 95% confidence level? 10

11 Hypothesis testing Population µ, σ, p Sampling Sample Hypotheses: H 0, H a Inference Estimates _ x, s, p^ _ SE = SD( x ) Summarising data Hypothesis test: Reject H 0? Observed data 11

12 Carrying out a hypothesis test: The classical approach Step 1. Null hypothesis: H 0 : µ = m, Alternative hypothesis: H a : µ m ==> If hypothesis true, then observed sample mean x ~ N(m, σ/ n ) Step 2. Sample size ==> Calculate SD( x ) = σ/ n Step 3. Significance level: How unlikely does the observed value have to be to decide to reject H 0 Step 4. Acceptance region: Determine the range of values for which H 0 will not be rejected Step 5. Take the sample and see if the observed value justifies rejection of H 0 12

13 Example A machine is set up such that the average content of juice per bottle equals µ. A sample of 36 bottles yields an average content of 48.5cl. Test the hypothesis that the average content per bottle is 50cl at the 5% significance level. Assume that the population standard deviation σ = 5cl. 13

14 The impact of sample size A machine is set up such that the average content of juice per bottle equals µ. A sample of 100 bottles yields an average content of 48.8cl. Test the hypothesis that the average content per bottle is 50cl at the 5% significance level. Compare the conclusion to that based on the 36 bottle sample. Assume that the population standard deviation σ = 5cl. 14

15 One-tailed tests A machine is set up such that the average content of juice per bottle equals µ. A sample of 36 bottles yields an average content of 48.5cl. Can you reject the hypothesis that the average content per bottle is less than or equal to 45cl in favour of the alternative that it exceeds 45cl (5% significance level)? Assume that the population standard deviation σ = 5cl. 15

16 Exercise: Formulating H 0 The manager claims that the average content of juice per bottle is less than 50cl. The machine operator disagrees. A sample of 100 bottles yields an average content of 49cl per bottle. Does this sample allow the manager to claim he is right (5% significance level)? Assume that the population standard deviation σ = 5 cl. 16

17 How do you choose a significance level? D. Bowers, Statistics from Scratch 17

18 Computer based approach: How likely am I to observe such an extreme value? Compute significance level of sample result Reject H 0 if significance level lower than the required level Result stated as: The observed difference is significant at the x% level ==> P-value P/2 P/2 P _ m x 2-tailed test: H 0 : µ = m, H a : µ m _ m x 1-tailed test: H 0 : µ m, H a : µ > m 18

19 Example A machine is set up such that the average content of juice per bottle equals µ. A sample of 100 bottles yields an average content of 48.8cl. Calculate the P-value for the hypothesis that the average content per bottle equals 50cl. Assume that the population standard deviation σ = 5cl. 19

20 Unknown population standard deviation σ If large sample: estimate σ s = n ( xi x) i= 1 n 1 2 Replace σ by s in the analysis 20

21 Small samples Degrees of freedom t-score Population must follow a Normal distribution 2a. σ known: Sample mean follows Normal distribution 2b. σ unknown: Replace σ by estimate s Sample mean follows t-distribution with n-1degrees of freedom N(0,1) Tail Degrees of freedom probability % % % % % t(2) t(10) 21

22 Confidence intervals for Proportions Example: Sample of 80 customers 60 reply they are satisfied with the service they received Calculate a 95% confidence interval for the proportion of satisfied customers Observed value ^p: Standard deviation of p: ^ Distribution of p: ^ 95% confidence interval for the true proportion p: 22

23 Newspaper reports The latest poll (1,100 respondents) reveals that 54% of the population supports the government's budgetary decisions. The margin of error is ± 3%. ==> Point estimate: 54% Margin of error: ± 3% ==> Confidence interval: [51%, 57%] Observed value p: ^ Standard deviation of p: ^ Confidence level of interval: 23