Hypothesis testing Contents 1 Principle of hypothesis testing One sample tests 3.1 Tests on Mean of a Normal distribution..................... 3. Tests on Variance of a Normal distribution.................... 3.3 Large Sample Test on Mean........................... 4.4 Test on a Population Proportion......................... 5 3 Two Samples Tests 5 3.1 Testing Equality of Variances........................... 5 3. Testing Equality of Means (σ 1 = σ )....................... 6 3.3 Testing Equality of Means (σ 1 σ )....................... 6 3.4 Large Samples Tests on Means.......................... 7 3.5 Testing Equality of Means Paired Samples................... 8 4 Hypothesis Testing of Distribution 9 4.1 Chi-Square Goodness of Fit Test......................... 9 4. Tests of Skewness and Kurtosis.......................... 9 5 Excercises 11
1 Principle of hypothesis testing A statistical hypothesis is a claim (or statement) about a population parameters (µ σ π λ... ) a distribution (normal Poisson... ). A null hypothesis H is a claim (or statement) about a population parameter that is assumed to be true until it is declared false... for example H : µ = µ 0. An alternative hypothesis A is a claim about a population parameter that will be true if the null hypothesis is false A : µ µ 0 both-sided test A : µ > µ 0 one-sided test A : µ < µ 0 one-sided test. reality H is true H is false decision about H prob. prob. not reject correct decision 1 type II error β reject type I error correct decision 1 β If we reject the null hypothesis which is true we call this type I error. The probability of this error is a significance level. A number 1 is probability that we do not reject the true hypothesis H. If we accept the null hypothesis although it is false we call this type II error. The probability of this error is β. A number 1 β a power of the test is the probability that we reject the null hypothesis H if it is false. To test the null hypothesis we use a function of a random sample T = T (x 1 x... x n ) so called test statistic which has under the null hypothesis H known distribution (usually t u χ F ). We divide the all possible values of the test statistic into W 1 - non of H the set of values connected with the hypothesis H W - of H the set of values connected with the hypothesis A. Steps to Perform a Test of Hypothesis 1. State the null and alternative hypothesis H and A.. Select a significance level (usually 0.05 a 0.01). 3. Choose the test statistic. 4. Determine the W. 5. Calculate the value of the test statistic. 6. Make a decision: If the value of the test statistic falls in the we reject the null hypothesis H and say that we accept the alternative hypothesis A with the probability 1. If the value of the test statistic falls in the non we do not reject the null hypothesis H.
alternative hypothesis A : µ > µ 0 A : µ < µ 0 A : µ µ 0 W = {t t t 1 (ν)} W = {t t t 1 (ν)} W = { t t t 1 (ν)} Steps to Perform a Test of Hypothesis via p-value 1. State the null and alternative hypothesis H and A.. Select a significance level (usually 0.05 a 0.01). 3. Calculate the p-value statistical software e.g. STAT1 R etc. 4. Make a decision: If > p-value we reject the null hypothesis H with the probability 1 i.e. significance level. If < p-value we do not reject the null hypothesis H. at One sample tests.1 Tests on Mean of a Normal distribution Let X 1 X... X n be a random sample from a normal distribution N(µ σ ). A statistic T = X µ n S has a Student distribution with ν = n 1 degrees of freedom. We use this statistic for testing of the parameter µ. Let x 1 x... x n be values of a random sample (measured data) x denotes an arithmetic mean and s a sample standard deviation. We test the hypothesis that the parameter µ is equal to a constant µ 0 : H : µ = µ 0 the test statistic t = x µ 0 n s has under the null hypothesis H a Student t-distribution with ν = n 1 degrees of freedom. According to the alternative hypothesis we construct following regions of rejection: where t 1 t 1 are quantiles of the Student distribution.. Tests on Variance of a Normal distribution Let X 1 X... X n be a random sample from a normal distribution N(µ σ ). A statistic χ = (n 1)S σ 3
alternative hypothesis A : σ > σ0 W = { χ χ χ 1 (ν) } A : σ < σ0 W = {χ χ χ { (ν)} } A : σ σ0 W = χ χ χ (ν) or χ χ 1 (ν) alternative hypothesis A : µ > µ 0 W = {u u u 1 } A : µ < µ 0 W = {u u u 1 } A : µ µ 0 W = { } u u u 1 has Pearson distribution with ν = n 1 degrees of freedom. We use this statistic for testing of the parameter σ. Let x 1 x... x n be values of a random sample (measured data) s denotes sample variance. We test the hypothesis that the parameter σ is equal to a constant σ 0: the test statistic H : σ = σ 0 χ = (n 1)s σ 0 has under the null hypothesis H a Pearson χ -distribution with ν = n 1 degrees of freedom. According to an alternative hypothesis we construct following regions of rejection: where χ 1 χ 1 χ χ are quantiles of the Pearson χ -distribution..3 Large Sample Test on Mean Let X 1 X... X n be a random sample from any distribution with the mean µ. A statistic U = X µ n S has for large n approximately a normal distribution N(0 1) see the central limit theorems. We use this statistic for testing of the parameter µ. Let x 1 x... x n be values of a random sample (measured data) x denotes an arithmetic mean and s a sample standard deviation. We test the hypothesis that the parameter µ is equal to a constant µ 0 : H : µ = µ 0 the test statistic u = x µ 0 n s has under the null hypothesis H asymptotically a normal distribution N(0 1). According to an alternative hypothesis we construct following regions of rejection: where u 1 u 1 are quantiles of N(0 1). 4
alternative hypothesis.4 Test on a Population Proportion A : π > π 0 W = {u u u 1 } A : π < π 0 W = {u u u 1 } A : π π 0 W = { } u u u 1 Suppose that a random sample of size n has been taken from a large (possibly infinite) population and that m observations in this sample belong to a class of interest. Then p = m is a point n estimator of the proportion of the population π that belongs to this class. A random variable U = p π π(1 π)/n has for n approximately a normal distribution N(0 1) see central limit theorems. We use this statistic for testing of the population proportion. Let ˆπ = m be a point estimator of population proportion. We test the hypothesis that the n parameter π is equal to a constant π 0 : H : π = π 0 a test statistic u = ˆπ π 0 π0 (1 π 0 )/n has under the null hypothesis H asymptotically a normal distribution N(0 1). According to an alternative hypothesis we construct following regions of rejection: where u 1 u 1 are quantiles of N(0 1). 3 Two Samples Tests 3.1 Testing Equality of Variances Let X 1 X... X n1 be a random sample from N(µ 1 σ1) and Y 1 Y... Y n be a random sample from N(µ σ). SX and S Y are corresponding sample variances. The statistic F = S X S Y σ σ 1 has a Fisher-Snedecor distribution with ν 1 = n 1 and ν = n 1 degrees of freedom. Let x 1 x... x n1 be values of a random sample from N(µ 1 σ 1) y 1 y... y n be values of a random sample from N(µ σ ) s x and s y corresponding values of sample variances. We test a hypothesis that the parametr σ 1 is equal to the parameter σ : H : σ 1 = σ the test statistic is F = s x s y 5
alternative hypothesis A : σ 1 > σ W = {F F F 1 (ν 1 ν )} A : σ1 < σ W = {F F F (ν 1 ν )} A : σ1 σ W = { F F F (ν 1 ν ) F F 1 (ν 1 ν ) } which has under the null hypothesis H a Fisher-Snedecor distribution with ν 1 = n 1 1 and ν = n 1 degrees of freedom. According to the alternative hypothesis we construct following regions of rejection: where F F 1 F F 1 are quantiles of the Fisher-Snedecor distribution ν 1 = n 1 1 ν = n 1. 3. Testing Equality of Means (σ 1 = σ ) Let X 1 X... X n1 be a random sample from N(µ 1 σ 1) and Y 1 Y... Y n be a random sample from N(µ σ ). We assume that these random samples are independent. X Y S X a S Y are corresponding sample means and variances. If σ 1 = σ then a statistic where T = X Y (µ 1 µ ) n1 n S n 1 + n S = [ (n1 1)S X + (n 1)S Y n 1 + n (it is co called pooled estimator of the common σ) has a Student distribution with ν = n 1 +n degrees of freedom. Let x 1 x... x n1 be values of a random sample from N(µ 1 σ 1) y 1 y... y n be values of a random sample from N(µ σ ) x y s x a s y are corresponding values of sample means and variances. We test a hypothesis that the parameter µ 1 is equal to the parameter µ (σ 1 = σ ): a test statistic is where H : µ 1 = µ t = x y n1 n S n 1 + n [ (n1 1)s x + (n 1)s y S = n 1 + n has under the null hypothesis H the Student distribution with ν = n 1 + n degrees of freedom. According to the alternative hypothesis we construct following regions of rejection: where t 1 t 1 are quantiles of the Student distribution ν = n 1 + n. 3.3 Testing Equality of Means (σ 1 σ ) Let X 1 X... X n1 be a random sample from N(µ 1 σ 1) and Y 1 Y... Y n be a random sample from N(µ σ ). We assume that these random samples are independent. 6 ] 1/ ] 1/
alternative hypothesis A : µ 1 > µ A : µ 1 < µ A : µ 1 µ alternative hypothesis A : µ 1 > µ A : µ 1 < µ A : µ 1 µ W = {t t t 1 (ν)} W = {t t t 1 (ν)} W = { t t t 1 (ν)} W = {t t t 1 (ν)} W = {t t t 1 (ν)} W = { t t t 1 (ν)} X Y S X a S Y are corresponding sample means and variances. If σ 1 = σ then a statistic T = X Y (µ 1 µ ) SX n1 + S Y n has approximately a Student distribution with ν degrees of freedom. Let x 1 x... x n1 be values of a random sample from N(µ 1 σ 1) y 1 y... y n be values of a random sample from N(µ σ ) x y s x a s y are corresponding values of sample means and variances. We test a hypothesis that the parameter µ 1 is equal to the parameter µ (σ 1 σ ): a test statistic H : µ 1 = µ t = x y s x n 1 + s y n has under the null hypothesis H approximately a Student distribution with ν degrees of freedom. Degrees of freedom are given by a formula ν ( ) s x n 1 + s y n ( ) 1 s ( ) x n 1 1 n 1 + 1 s y n 1 n rounded down to the nearest integer number. According to the alternative hypothesis we construct following regions of rejection: where t 1 t 1 are quantiles of the Student distribution with ν degrees of freedom (see the previous page). 3.4 Large Samples Tests on Means Let X 1 X... X n1 be a random sample from a distribution with the mean µ 1 and Y 1 Y... Y n be a random sample from a distribution with the mean µ. We assume that these random samples are independent and samples are large enough. 7
X Y S X a S Y alternative hypothesis A : µ 1 > µ W = {u u u 1 } A : µ 1 < µ W = {u u u 1 } A : µ 1 µ W = { } u u u 1 are corresponding sample means and variances. A statistic U = X Y (µ 1 µ ) SX n1 + S Y n has approximately a normal distribution N(0 1). Let x 1 x... x n1 are values of a random sample from the first distribution y 1 y... y n are values of a random sample from the second distribution x y s x and s y are corresponding values of sample means and variances. We test a hypothesis that the parameter µ 1 is equal to the parameter µ : H : µ 1 = µ a test statistic u = x y s x n 1 + s y n has under the null hypothesis H approximately a normal distribution N(0 1). According to the alternative hypothesis we construct following regions of rejection: where are quantiles of N(0 1). u 1 u 1 3.5 Testing Equality of Means Paired Samples Let us have two dependent samples two data values one for each sample are collected from a source (or an element). These are also called paired or matched samples. We assume two dependent random variables X and Y with means µ 1 and µ the difference D = X Y is a random variable too. Let D 1 D... D n be a random sample where differences D i = X i Y i have a normal distribution N(µ σ ) where µ = µ 1 µ (σ is not needed). A statistic T = D µ S D n where D is a sample mean of differences and S D is a sample standard deviation of differences has a Student distribution with ν = n 1 degrees of freedom. Let d 1 = x 1 y 1 d = x y... d n = x n y n be measure valued of differences d is its sample mean and s d is its sample standard deviation. We test a hypothesis that the parameter µ 1 is equal to the parameter µ : a test statistic H : µ 1 = µ t = d s d n has under the null hypothesis H a Student distribution with ν = n 1 degrees of freedom. According to the alternative hypothesis we construct following regions of rejection: where t 1 t 1 are quantiles of the Student distribution ν = n 1. 8
alternative hypothesis A : µ 1 > µ A : µ 1 < µ A : µ 1 µ W = {t t t 1 (ν)} W = {t t t 1 (ν)} W = { t t t 1 (ν)} 4 Hypothesis Testing of Distribution 4.1 Chi-Square Goodness of Fit Test We divide values of a random sample x 1 x... x n into k disjoint classes where n j j = 1... k is frequency of the class j and π j is a probability that the random variable X has value from the class j calculated under the condition that X has an assumed distribution. The main idea of the test is to compare relative frequencies n j /n with theoretical probabilities π j. We state the null and alternative hypothesis: H : the random variable X follows an assumed distribution A : the random variable X does not follow an assumed distribution. The test statistic is k χ (n j nπ j ) = nπ j j=1 which has under the null hypothesis H for large n (asymptotically) a Pearson χ -distribution with ν = k c 1 degrees of freedom where c is a number of estimated parameters of the assumed distribution. A is W = { χ χ χ 1 (ν) } where χ 1 (ν) is a quantile of the Pearson χ -distribution. Recommendation: nπ j > 5 j = 1... k. If this condition is not satisfied it is necessary to join the classes. 4. Tests of Skewness and Kurtosis The normal distribution has 3 = 0 a 4 = 0. We can use these properties to test normality. We calculate a sample skewness and kurtosis (they are estimates of 3 and 4 ) We state hypothesis: ˆ 3 = a 3 = 1 ns 3 n n i=1 H 1 : 3 = 0 A 1 : 3 0 Test statistic is u 3 = (x i x) 3 ˆ 3 = a 4 = 1 ns 4 n a 3 D(a3 ) where D(a 3) = n (x i x) 4 3. i=1 6(n ) (n + 1)(n + 3) 9
which has under the null hypothesis H 1 asymptotically normal distribution N(0 1). A is W = { } u 3 u 3 u 1 where u 1 is a quantile of N(0 1). H : 4 = 0 A : 4 0 Test statistic is u 4 = a 4 + 6 n+1 D(a4 ) where D(a 4) = 4n(n )(n 3) (n + 1) (n + 3)(n + 5) which has under the null hypothesis H asymptotically normal distribution N(0 1). A is W = { } u 4 u 4 u 1 where u 1 is a quantile of N(0 1). Compound Tests of Skewness and Kurtosis We state hypothesis: H : a random variable X has a normal distribution A : a random variable X has not a normal distribution. Test statistic is C = u 3 + u 4 which has under the null hypothesis H approximately χ distribution with two degrees of freedom. u 3 and u 4 are test statistics defined above. A is W = { C C χ 1 () } where χ 1 () is a quantile of the Pearson χ -distribution. 10
5 Excercises 1. Repeat all three basic procedures 1 from section 1 for testing of statistical hypotheses (use the application STAT1). Use your own data sets.. The desired average moisture content of roasted coffee is 4.% and a standard deviation of 0.4%. In 0 samples were measured following actual values of moisture in percents: 4.5 4.3 4.1 4.9 4.6 3. 4.4 5.1 4.8 4.0 3.7 4.4 3.9 4.1 4. 4.1 4.7 4.3 4. 4.4 Assume that a random sample is from a normal distribution (check). Use the significance level of 0.05 to determine whether the underlying file from which the samples come exhibits the desired (a) average moisture (b) variability of moisture. 3. Representatives of the environmental movement actively oppose the construction of new factories in the area which environment is already quite marked by industrial activities. They assume that one of the consequences of an unhealthy environment could be also the low birth weight of newborns in the given area. Does it make sense to use a lower birth weight as an argument against the construction of a new factory knowing that the birth weight of the healthy population has a normal distribution with a mean of 3500 g? Their claim would be supported via the sample of 50 randomly selected newborns born in this area measured an average weight is 3310 g and a standard deviation of 500 g. Perform the test at a significance level of 0.01. 4. The operation was reduced in a factory which largely employs commuters from the area. The transport company is afraid of the fact that it would decreas the average number of passengers in one bus on certain lines. For this reason an investigation was carried out in 40 randomly selected buses and the relevant lines in rush hours with the following results: Number of passengers 5 8 9 34 35 38 40 4 45 Number of cases 4 7 10 6 5 3 1 For the past years it is known that the average number of passengers per one bus (under comparable conditions) was 36 persons. In the case that the survey will prove that the number of passengers decreased the transport company will have to reduce traffic. What is the decision ( = 0.05)? 5. According to an unnamed company that is engaged in the survey of consumer habits 33% of households prefer shopping in hypermarkets. The company anticipates further growth in their popularity. To verify that idea they randomly selected 50 people and found that hypermarkets prefer 93 of all respondents. Is the result of a survey in accordance with that assumption ( = 005)? 6. For two technicians we are supposed to verify the accuracy of their measurement of chemical concentration of given substance. The first technician performed 15 measurements with a standard deviation of 03 and the second technician performed 0 measurements with a standard deviation of 0.55. With 95% and 99% confidence level make sure 1 1. via the confidence interval. via the 3. via the p-value 11
that the first technician performs measurements more accurately than the second technician. Measurement errors have a normal distribution. 7. Two years ago in June a chemical analysis of 85 samples of water from different parts of the town lake was conducted. The main reason was to investigate the amount of chlorine in the water. The use of salt for road maintenance in winters was significantly reduced over the last two years. This year also in June 110 water samples from the lake was a re-analysed. Following results were obtained: years ago This year Average 18.3 17.8 Sample standard deviation 1. 1.8 Verify the claim that lower salt use in winter reduces the amount of chlorine in the lake ( = 0.05 0.01 respectively). 8. As a part of the pre-election survey there was the key question whether voters in the upcoming elections give priority to the Left or Right. The survey was conducted in two regions. It is possible on the basis of the data insist that the Left has in the region A significantly more supporters than in region B? Perform the test at a significance level of 0.05. The survey results are listed in the table below: Respondents Left Right Does not know Region A 150 50 70 30 Region B 00 50 90 60 9. The thickness of two tree types were compared (the thickness of the tree trunk at a height of 1.3 m above the heel strain). Test at the 0.05 level of significance that both tree types are well advanced in thickness. Assume that samples have a normal distribution. Stand A 33.5 8.6 36. 41.4 41.0 43.7 4.1 33.8 40.5 9.6 31.5 6.5 Stand B 9.5 30.4 1.6 4.4 8.5 6.8 5.5.6 6.1 19. 4.6 10. During the test of the reliability of the altimeter 15 control measurements were done when compared with other standard altimeter. Perform the paired test that both altimeters measure well. Assume that these samples are from normal distributions. Altimeter A 6.8.1 19.5 18.5 9.4 33. 16.3 18.5 35.4 9.5 19. Altimeter B 6..4 19.5 19.0 9.3 33.5 16. 18.9 36.0 9.8 18.9 Altimeter A 18.4 8.6 17.5 6. Altimeter B 18.5 9.0 18.0 6.0 11. In the group of 110 students pocket money was monitored. From the identified data were calculated sample skewness a 3 = 0.446 and sample kurtosis a 4 = 0.31. Using tests of zero skewness zero kurtosis and C-test verify at the significance level of 0.05 that the sample comes from a normal distribution. Do the same using the modified variants of the tests. 1
1. The shooter shoots at a target and it is recorded the number of hits in 3 shots. The shooter performed 50 triples shots with the following results: Number of hits 0 1 3 Number of cases 1 4 3 Estimate the probability ˆπ of hit in one shot. Is it possible to describe the distribution of hits in 3 shots by the binomial distribution? Verify the hypothesis using χ -goodness of fit test at a significance level of 0.05. Use the fact that the sample mean is an unbiased estimate of the mean which in the case of the binomial distribution is equal to nπ where n is the number of shots and π is the probability of hitting in one shot. 13. 10 measurements of the milk fat were performed (in g/100 g milk) with the following results: 1.59 1.57 1.47 1.65 1.59 1.39 1.49 1.48 1.69 1.3. Verify the normality at the significance level of 0.05 use the Kolmogorov-Smirnov s test. 14. For the sample of 50 civil servants was find out the amount of the monthly salary. From the collected data were determined the sample skewness a 3 = 0.73 and the sample kurtosis a 4 = 1.311. Use the C-test modified C-test and verify whether a random variable indicating the monthly salary of civil servants could be described by a normal distribution. 13