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Statistical inference: confidence intervals and hypothesis testingPowerPoint Presentation

Statistical inference: confidence intervals and hypothesis testing

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Statistical inference: confidence intervals and hypothesis testing

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Statistical inference: confidence intervals and hypothesis testing

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Statistical inference: confidence intervals and hypothesis testing

The objective of this session is

- Inference statistic
- Sampling theory
- Estimate and confidence intervals
- Hypothesis testing

- Descriptive
calculate various type of descriptive statistics in order to summarize certain quality of the data

- Inferential
use information gained fromthe descriptive statistics of sample datato generalize to the characteristics of the whole population

2 broad areas

- Estimation
- create confidence intervals to estimate the true population parameter

- Hypothesis testing
- test the hypotheses that the population parameter has a specified range

mean:

standard deviation:

- When working withthe samples of datawe have to rely onsampling theoryto give us theprobability distributionpertaining to the particular sample statistics
- This probability distribution is known as
“the sampling distribution”

- Assume there is a population …
- Population size N=4
- Random variable, X,is age of individuals
- Values of X: 18, 20,22, 24 measured inyears

C

B

D

A

Summary measures for thePopulation Distribution

P(X)

.3

.2

.1

0

A B C D

(18) (20) (22) (24)

Population mean Distribution

Summary measures of sampling distribution

Sampling distribution of thesample arithmetic mean

Sampling distribution of thestandard deviation of the

sample means

- Estimation of thepopulation parameters:
- point estimates
- confidence intervalsorinterval estimators

- Confidence intervals for:
- Means
- Variance

Large or Small samples ???

large samples (n >= 30)

- apply Z-distribution

Probability distribution

confidence interval

large samples (n >= 30)

- From the normally distributed variable, 95% of the observations will be plus or minus 1.96 standard deviations of the mean

large samples (n >= 30)

- The confident interval is given as

Probability distribution

2.5% in tail

95% confidence interval

2.5% in tail

-1.96 SE

+1.96 SE

large samples (n >= 30)

Probability distribution

2.5% in tail

95% confidence interval

2.5% in tail

-1.96 SE

+1.96 SE

large samples (n >= 30)

- Thus, we can state that:
“the sample mean will lie within an interval plus or minus 1.95 standard errors of the population mean 95% of the time”

large samples (n >= 30)

Example

we have data on 60 monthly observations of the returns to the SET 100 index. The sample mean monthly return is 1.125% with a standard deviation of 2.5%. What is the 95% confidence interval mean ???

large samples (n >= 30)

Example (cont’d)

- Standard error is calculated as
- the confidence interval would be
- The probability statement would be

large samples (n >= 30)

Example (cont’d)

- The probability statement would be
- How does the analyst use this information ???

What about small samples (n < 30)

- apply t-distribution

Probability distribution

What about small sample ??? (n < 30)

- Apply t-distribution
- The confidence intervalbecomes
- Theprobability statementpertaining to this confidence interval is

Example

- From 20 observations, the sample mean is calculated as 4.5%. The sample standard deviation is 5%. At the 95% level of confidence:
the confidence intervalis …

the probability statementis …

- Apply a distribution
- Theconfidence intervalis given as
- The probability statementpertaining to this confidence interval is

Example

- From a sample of 30 monthly observations the variance of the FTSE 100 index is 0.0225. With n-1 = 29 degrees of freedom (leaving 2.5% level of significant in each tail)
the confidence interval is …

the probability statement is …

2 Broad approaches

- Classical approach
- P-valueapproach
- is an assumption about the value of a population parameter of the probability distribution under consideration

- When testing, 2 hypotheses are established
- the null hypothesis
- the alternative hypothesis

- The exact formulation of the hypothesis depends upon what we are trying to establish
- e.g. we wish to know whether or not a population parameter, , has a value of

- How aboutwe wish to know whether or not a population parameter, , is greater than a given figure , the hypothesis would then be …
- Andif we wish to know whether or not a population parameter is greater than a given figure , the hypothesis would then be …

- In hypothesis testing we have to standardizing the test statistic so that the meaningful comparison can be made with the
- Standard normal (z-distribution)
- t-distribution
- distribution

- The hypothesis test may be
- One-tailed test
- Two-tailed test

MEAN

VARIANCE

Two-tailed test of the mean

- Set up the hypotheses as
- Decide on the level of significance for the test (10, 5, 1% level etc.) and establish 5, 2.5, 0.5% ineach tail
- Set the value of in thenull hypothesis
- Identify the appropriatecritical valueof z (or t) from the tables (reflect the percentages in the tails according to the level of significance chosen)

Two-tailed test of the mean

- Applying the followingdecision rule:
Accept H0if

Reject H0 if otherwise

Example

- Consider a test ofwhether or notthe mean of a portfolio manager’s monthly returns of 2.3% is statistically significantly different from the industry average of 2.4%. (from 36 observations with a standard deviation of 1.7%)

Example

- An analyst claims that the average annual rate of return generated by a technical stock selection service is 15% and recommends that his firm use the services as an input for its research product. The analyst’s supervisor is skeptical of this claim and decides to test its accuracy by randomly selecting 16 stocks covered by the service and computing the rate of return that would have been earned by following the service’s recommendations with regards to them over the previous 10-year period. The result of this sample are as follows:
- The average annual rate of return produced by following the service’s advice on the 16 sample stocks over the past 10 years was 11%
- The standard deviation in these sample results was 9%
Determine whether or not the analyst’s claim should be accepted or

rejected at the 5% level of significant ???

One-tailed test of the mean (Right-tailed tests)

- Set up the hypotheses as
- Applying the followingdecision rule:
Accept H0if

Reject H0 if

Example

- If we wish to test that the mean monthly return on the FTSE 100 index for a given period ismore than1.2. From 60 observations we calculate the mean as 1.25% and the standard deviation as 2.5%.

Example

- We wish to test that the mean monthly return on the S&P500 index isless than1.30%. Assume also that the mean return from 75 observations is 1.18%, with a standard deviation of 2.2%.

Two-tailed test Applying the followingdecision rule:

Accept H0if

Reject H0 if otherwise

One-tailed test Applying the followingdecision rule:

Accept H0if

Reject H0 if

Left or right tailed test ???

How ‘bout the other ???

Two-tailed test

- The standardized test statistic for the population variance is
- This standardized test statistic has a distribution

Example

- If we wish to test the variance of share B isbelow25. The sample variance is 23 and the number of observation is 40

- The p-value is the lowest level of significance at which
the null hypothesis is rejected

- If the p-value ≥ the level of significance (α)
accept null hypothesis

- If the p-value < the level of significance (α)
reject null hypothesis

- If the p-value ≥ the level of significance (α)

- If we wish to find an investment give at least 13.2%. Assume that the mean annualized monthly return of a given bond index is 14.4% and the sample standard deviation of those return is 2.915%, there were 30 observations an the returns are normally distributed.

- The test statistic is:
- With degree of freedom = 29
a t-value of 2.045 leaves 2.5% in the tail

a t-value of 2.462 leaves 1% in the tail

- Calculate p-value from interpolation
- P-value = 0.025 – (0.50 x (0.025 – 0.01) = 0.0175 = 1.75%
- P-value (1.75%) < α (5%), thus reject null hypothesis

- Meaning of statistical inference
- Sampling theory
- Application of statistical inference
- Confidence intervals
- Estimation
- Hypothesis testing
- Two-tailed
- One-tailed

Z-distribution t-distribution X2-distribution

means variance