Quantitative Methods for Economics, Finance and Management (A86050 F86050)

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1 Quantitative Methods for Economics, Finance and Management (A86050 F86050) Matteo Manera Marzio Galeotti 1

2 This material is taken and adapted from Guy Judge s Introduction to Econometrics page at the University of Portsmouth Business School 2

3 Hypothesis Testing There are two branches of statistical inference 1. Estimation 2. Hypothesis testing In estimation we are concerned with obtaining the best possible estimates of population parameters (generally unknown) such as µ, σ 2, σ, etc. There are two types of estimate: A. Point estimate B. Interval estimate

4 Interval Estimation To obtain an interval estimate one determines an interval in which, with a high degree of confidence one predicts the parameter value lies. These intervals are called confidence intervals and the end-points are called confidence limits. In order to obtain interval estimates we need to know more about certain sampling distributions. Recall the normal distribution

5 An interval estimate An interval estimate constitutes a range of values within which the value of the population parameter falls with a specified probability Standard probability intervals (proportions under the curve) are defined by multiples of the standard deviation around the mean In order to be able to produce an interval estimate of µ x, it is necessary to know the statistical properties of the sample mean of X

6 An interval estimate There is a population of values of a continuous random variable, X The population can be described by a normal distribution The first moment of the normal distribution is denoted by µ x, the value of which is unknown The second central moment is denoted by σ x 2

7 An interval estimate A random sample of size n is drawn from the population of values of X The sample values are realisations of the random variables: X 1, X 2,, X n On account of the nature of the population from which the sample has been drawn, X i ~ N(µ x, σ x2 ), i = 1, 2,, n The arithmetic average of the sample values is calculated which provides a point estimate of the unknown value of the population mean

8 An interval estimate An interval estimate - X = (1/n) (X 1 + X X n ) = (1/n)X 1 + (1/n)X (1/n)X n On the basis of the nature of the population from which the sample has been drawn X i ~ N(µ x, σ x2 ), i = 1, 2,, n Result 1: any linear combination of normally-distributed random variables, itself, has a normal distribution Consequently, Z = (X - µ x )/(σ x2 /n) 1/2 ~ N(0, 1)

9 An interval estimate f(z) Prob.(Z < 1.96) Probability statement: P(Z < 1.96) = Prob.(Z > 1.96) Z

10 An interval estimate If P(Z > 1.96) = P(-1.96 < Z < 1.96) then, on the basis of the symmetrical nature of the standardised normal distribution about zero, P(Z < -1.96) = It follows that P(-1.96 < Z < 1.96) = = 0.95

11 A 95% confidence interval for µ x On substitution, then: _ Prob.(-1.96 < (X - µ x )/ (σ x2 /n) < 1.96) = 0.95 It is then possible to obtain the interval estimate: _ Prob {X 1.96 (σ x2 /n) < µ x < X (σ x2 /n)} = 0.95 _

12 A 95% confidence interval for µ x Summarizing: _ X ~ N(µ x, σ x2 /n) _ (X - µ x )/ (σ x2 /n) ~ N(0, 1) _ P(-1.96 < (X - µ x )/ (σ x2 /n) < 1.96) = 0.95 P{X 1.96 (σ x2 /n) < µ x < X (σ x2 /n)} = 0.95

13 An interval estimate We can use the standard error to calculate a range that contains the population mean, at a particular probability, and based on a specific sample: x ± Z α s n (where Z might be 1.96 for.95 probability, for example)

14 Numerical values of the limits In order to be able to obtain numerical values of the lower and upper limits of the confidence interval, a knowledge is required of: - the value of the sample mean; - the size of the sample; -the value of the population variance. However, if the value of the population mean is unknown then it seems unreasonable to assume that the value of the population variance is known.

15 Proposed estimator of the population variance of X If we do not know the value of the population variance of X then we should seek to obtain an estimate using the sample data. The proposed estimator of σ x2 is: _ s x2 = (X i X) n 1 It can be shown that s x2 is an unbiased estimator of σ x 2.

16 Consequences of replacing σ x 2 by s x 2 The foundation of the confidence interval for µ x is the probability density function of the sample mean: _ X ~ N(µ x, σ x2 /n). On the basis of the above: _ X - µ x ~ N(0, 1) (σ x2 /n) However, if σ x2 is replaced by s x2 then the statistic will no longer have a standardised normal distribution, but a t distribution.

17 Characteristics of the Student s t distribution The diagrammatic appearance of a t distribution is very similar to the diagrammatic appearance of a standardised normal distribution The graph of a t distribution is a symmetrical, bell-shaped curve which is centred over the value of zero For a finite sample size, the graph of a t distribution has a flatter appearance than the graph of a standardised normal distribution Thus, there is a greater proportion of the total area underneath the graph falling within the two tails The variance of a variable which has a t distribution is greater than one

18 A 95% confidence interval Let us suppose that we are seeking to construct a 95% confidence interval for the population mean of X From consulting the table of the t distribution, we would find the value on the horizontal axis to the right of which there is 2.5 per cent of the total area underneath the graph On the basis of the symmetrical nature of the t distribution, to the left of minus this value is also 2.5 per cent of the total area underneath the graph

19 Graph of the t distribution f(t) Area = Area = Area = = t t t

20 A 95% confidence interval for µ x Starting point: P(-t < t < t ) = 0.95 _ P(-t < X - µ x < t ) = (s x2 /n) Following a sequence of mathematical manipulations, P{X t (s x2 /n) < µ x < X + t (s x2 /n)} = 0.95

21 Characteristics of the Student s t distribution The form of a t distribution is not fixed The shape of the graph of a t distribution is governed by the number of degrees of freedom The number of degrees of freedom equates with the number of independent sample observations If the sample data have already been used to compute the value of the sample mean then there will be only n 1 independent sample observations As the number of degrees of freedom increases towards infinity, the t distribution more closely resembles the standardised normal distribution

22 Hypothesis Testing The means by which the subject of Economics evolves is by theories being advanced However, it is essential that the validity of a theory be examined The conventional approach towards assessing the validity of a theory is by combining sample data with a suitable statistical procedure Hence, hypothesis testing is central to the development of the subject of Economics Involves testing the relative strength of null vs. alternative hypotheses

23 Null & Alternative hypotheses H 0 - usually highly specific and explicit - often a hypothesis that we suspect is wrong, and wish to disprove e.g.: 1. the means of two populations are the same (H 0 :µ 1 =µ 2 ) 2. two variables are independent 3. two distributions are the same H 1 - what is logically implied when H 0 is false - often quite general or nebulous compared to H 0 the means of two populations are different: H 1 :µ 1 < >µ 2

24 Testing H 0 and H 1 Together, they represent mutually exclusive and exhaustive possibilities You can calculate conditional probabilities associated with sample data, based on the assumption that H 0 is correct P(sample data H 0 is correct) if the data seem highly improbable given H 0, H 0 is rejected, and H 1 is accepted what can go wrong? since we can never know the true state of underlying population, we always run the risk of making the wrong decision

25 Types of error which can be committed When a hypothesis test is being performed, it is possible to make two types of error One type of error is P(rejecting H 0 H 0 is true), the probability of rejecting a true null hypothesis This is known as a type 1 error The probability of a type 1 error is the level of significance P = significance level of the test = alpha (α) This is also referred to as the size of the test The probability of a type 1 error is under the control of the investigator

26 Type I error Smaller alpha values are more conservative from the point of view of Type I errors Compare a alpha-level of.01 and.05: - we accept the null hypothesis unless the sample is so unusual that we would only expect to observe it 1 in 100 and 5 in 100 times (respectively) due to random chance the larger value (.05) means we will accept less unusual sample data as evidence that H 0 is false the probability of falsely rejecting it (i.e., a Type I error) is higher

27 Type II error The more conservative (smaller) alpha is set to, the greater the probability associated with another kind of error - Type II error P(accepting H 0 H 0 is false): failing to reject the null hypothesis when it actually is false The probability of a Type II error (β) is generally unknown 1 P(type 2 error) is referred to as the power of the hypothesis test. The power of the test provides an indication of the ability to reject a false null hypothesis

28 Type I & II errors the relative costs of Type I vs. Type II errors vary according to context in general, Type I errors are more of a problem e.g., claiming a significant pattern where none exists H 0 is correct H 0 is incorrect H 0 is accepted correct decision Type II error (β) H 0 is rejected Type I error (α) correct decision

29 Factors which affect the Probability of a Type 2 Error There is a trade-off between the probability of a type 1 error and the probability of a type 2 error Hence, the lower is the level of significance at which the hypothesis test is being performed then the higher is the probability of a type 2 error (and vice versa) Also, the probability of a type 2 error will be higher, the nearer are the values of the population parameter according to the null and alternative hypotheses

30 Problems with hypothesis testing α χ 2 accept H 0.05 reject H 0 critical: 3.84 observed: 4.84 better to report the actual alpha value associated with the statistic, rather than just whether or not the statistic falls into an arbitrarily defined critical region most computer programs do return a specific alpha level α.016 χ 2 observed: 4.84 you may get a reported alpha of.000 not the same as 0 means α <.0005 (report it like this)

31 Encourages misinterpretation of results it s tempting (but wrong) to reverse the logic of the test having failed to reject the H 0 at an alpha of.05, we are not 95% sure that the H 0 is correct if you do reject the H 0, you can t attach any specific probability to your acceptance of H 1

32 The classical hypothesis testing procedure: steps Step 1: Construct a null hypothesis Step 2: Specify a suitable alternative hypothesis Step 3: Decide upon a test statistic and establish its sampling distribution Step 4: Choose a level of significance at which to perform the test Step 5: In conjunction with the graph of the probability density function of the test statistic, identify the rejection region(s) Step 6: Compute the value of the test statistic Step 7: Observe whether or not the computed value of the test statistic falls within a rejection region

33 Example of application of hypothesis testing procedure There is a population of female students X denotes the height of a female student The population of values of X can be described by a normal distribution The first moment of the normal distribution is denoted by µ x The second central moment of the normal distribution is denoted by σ x 2 Neither of the values of the parameters is known

34 Example (continued) A random sample of size n=81 is drawn from the population of values of X The arithmetic average of the sample values is calculated to be 5 2 (or 62 inches) The value of the sample variance is calculated to be 49 inches squared An expert on the subject of the height of a female student claims that the population average height is at least 5 3 Implement the classical hypothesis testing procedure for the purpose of assessing the validity of the expert s claim

35 Step 1 Step 2 S1. Construct a null hypothesis. H 0 : µ x = 63 inches S2. Specify a suitable alternative hypothesis. H 1 : µ x < 63 inches

36 Step 3 S3. Decide upon a test statistic and establish its sampling distribution. The sample values are realisations of the random variables: X 1, X 2,, X n, where X i ~ N(µ x, σ x2 ), i = 1, n _ X - µ x ~ N(0, 1) (σ x2 /n) However, this is not a practical recommendation for a test statistic. Practical form of test statistic: _ X - µ x ~ t df = n-1 (s x2 /n)

37 Step 4 Step 5 S4. Choose a level of significance at which to perform the hypothesis test. α= 0.05 (The probability of rejecting the null hypothesis when this is true) S5. In conjunction with the graph of the probability density function of the test statistic, identify the rejection region(s). The alternative hypothesis contains the < symbol. Hence, the rejection region is the left-hand tail of the graph of the t distribution. Because α = 0.05 then this tail accommodates five per cent of the total area underneath the graph.

38 Graph of the t distribution f(t) Reject Ho Area = 0.05 Do not reject Ho t

39 Step 6 Step 7 S6. Compute the value of the test statistic: t = = -9/7 = (49/81) S7. Observe whether or not the computed value of the test statistic falls within the rejection region: > Therefore, at the 5% level of significance, it is not possible to reject H 0 in favour of H 1. There is insufficient evidence to be able to refute the claim which is made by the expert.

40

41 Note: it turns out that the variance has a χ 2 distribution being the sum of squares of normal distributions (residuals are normally distributed). The ratio of normal to χ 2 distributions is distributed t

42

43

44 More generally: 1. Two-tailed test H 0 : β 1 = b 1 H 1 : β 1 b 1 Choose α and construct k α σ and check whether it includes the hypothesized value b 1 under H 0 - k α σ b 1 + k α σ Reject if: > b 1 + k α σ or: > < b 1 + k α σ The H 0 : β 1 = 0 is a special case of the above. Just replace b 1 with 0. k α depends on the distribution of the test. Here is the Student s t as the variance is estimated

45 More generally: 2. One-tailed test H 0 : β 1 = b 1 H 1 : β 1 > b 1 or β 1 < b 1 Set up confidence interval: P( < b 1 + k α σ )=1-α or P( > b 1 + k α σ )=1-α If: > b 1 + k α σ reject H 0 or < b 1 + k α σ reject H 0

46 Forecasting using the simple regression model Once a model has been estimated (and carefully validated using economic and statistical tests) it can be used for prediction or forecasting. For example our estimated relationship between sales and ads is (approximately) sales = ads + residual We can use this to predict sales for some particular level of advertising, say ads = 70 The disturbance term is assumed to take its expected value so we put the residual = 0.

47 Forecasting using the simple regression model sales(ads=70) = * 70 = This is just a point forecast. We can create a forecast confidence interval by taking 95% forecast interval = point forecast ± s F t n-2, Here that would give ± * = ± 5.75 i.e. [54.29, 65.79] This interval is quite large because it is based on a rather small sample. Hence both s F and t n-2, will be fairly large. Forecasts based on larger samples will be more precise.

48 More on the standard error of the forecast The formula for S F (for the simple bivariate model) is S F = $σ u [ 1 + n -1 + (X F - X ) 2 / Σ (X i - X ) 2 ] Notice that S F is smaller the smaller is $σ u the closer is X F to the sample mean X the greater is the sample size n the greater is the in sample variation in X

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