8.1-4 Test of Hypotheses Based on a Single Sample

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1 8.1-4 Test of Hypotheses Based on a Single Sample Example 1 (Example 8.6, p. 312) A manufacturer of sprinkler systems used for fire protection in office buildings claims that the true average system-activation temperature is 130. A sample of n = 9 systems, when tested, yields a sample average activation temperature of F. If the probability distribution of an activation time is normal with standard deviation 1.5 F, does the data contradict the manufacturer s claim that average system-activation temperature is µ = 130? (Is the difference F vs. 130 F statistically significant) 1. We never answer this with 100% certainty, so first we need to decide how sure we want to be that we are right if we conclude that manufacturer s claim is not true. Let us say that if we conclude that manufacturer s claim in not true, we want to limit the probability that our conclusion is incorrect to some small number, for example to = 0.01 (significance level of the test) 2. If µ denotes the true average system-activation temperature (= the mean of the distribution of the activation time), then the hypotheses that we are testing are H 0 : µ = 130 versus H a : µ Test attitude is to assume that the null hypothesis H 0 is true and to claim the alternative hypothesis H a only if the data strongly speaks against H 0. For this reason we say that we reject the null hypothesis H 0 at a significance level (which means that we accept H a ) or that we fail reject the null hypothesis H 0 at a significance level (which means that evidence is not strong enough to claim that H a is true at a given significance level)

2 4. Test will be based on a random sample X 1, X 2,, X n. We need to choose a proper test statistic on which observed value our conclusion is to be made. The test statistic should be such that its distribution under H 0 is known. Because the distribution of activation times is normal, we know that X in normal with mean µ and standard deviation σ, and hence Z = X μ σ n is standard normal N(0,1) Assuming that the null hypothesis H 0 is true i.e. that µ = µ 0 = 130, as a test statistic we choose Z = X If the null hypothesis H 0 : µ = 130 is true then Z is standard normal, that is observed value z of Z should be close to 0. The farer from 0 is z, the stronger evidence that H 0 : µ = 130 is not true. We need to determine the rejection region, that is the region such that if the observed value of the test statistic is in this region then we reject H 0. To meet the requirement that the probability of rejecting H 0 (i.e. concluding that µ 130 ) if H 0 is true (i.e. if in fact µ = 130 ) to be limited to = 0.01, the rejection region must have the property that P(Test statistic Rejection region H 0 is true) =

3 For our test and = 0.01 we have /2 = 0.005, z /2 = z = 2.576, and the rejection region is Rejection region = { z or z 2.576} = { z 2.576} The observed value of the test statistic is z = = 2.16; z is not in the rejection region so we fail to reject H 0. Data do not contradict the manufacturer s claim (at = 0.01.) 6. If we still decide to conclude that data contradict the manufacturer s claim (i.e. that µ 130 ) then the probability that our conclusion is incorrect would be bigger than How big? The smallest level of significance at which H 0 would be rejected is called the P-value or the observed significance level In our case P-value = 2 P(z > 2.16) = 2 normalcdf(2.16,10^99,0,1) = Hence H 0 would be rejected at for example = 0.10, or = 0.05, but H 0 would not be rejected at for example = 0.02 or =0.01. If we conclude that data contradict the manufacturer s claim that the probability that our decision is incorrect is 0.03 (this is the probability of type I error).

4 PARAMETRIC TEST STEP-BY-STEP PROCEDURE 1. Population and sample: Recognize what is known about population and whether the sample size is large or not. 2. Parameter: Recognize parameter tested θ (for example population mean µ, population proportion p) 3. State hypotheses: H a alternative hypotheses (statement that needs strong support from date to be claimed) H 0 null hypotheses (opposite of H a ). For convenience the null hypothesis is always written in the form H 0 : θ = θ 0. Pairs tested in this textbook: H 0 : θ = θ 0 (means H 0 : θ θ 0 ) vs. H a : θ > θ 0 (upper tail test) H 0 : θ = θ 0 (means H 0 : θ θ 0 ) vs. H a : θ < θ 0 (lower tail test) H 0 : θ = θ 0 vs. H a : θ θ 0 (two-tailed test) 4. Test Statistic: Choose a proper test statistics (from Text, lecture notes, Stat Handbook, etc) 5. Distribution of the test statistic under H 0 : Determine the distribution of the test statistic under assumption that H 0 is true 6. Rejection region: Determine the rejection region at a given significance level 7. Observed value of the test statistic: Compute the observed value of the test statistic 8. Decision rule based on rejection region: if the observed value of the test statistic is in rejection region then reject H 0 (= claim H a ) if the observed value of the test statistic is not in rejection region then do not reject H 0 (= uphold H a ). In this case we say that we fail to reject H 0 9. Decision rule based on P-value: Given data, the P-value (or the observed significance level) is the smallest level of significance at which H 0 would be rejected. The smallest means that H 0 is rejected at each P-value and is not if P-value > Computer software will not give you a rejection region but just the P-value.

5 TWO TYPES OF ERRORS Example 2. Court verdict. H 0 : not guilty, H a : guilty. Type I error: say guilty if in fact is not Type II error: say not guilty if in fact is guilty. Type I error usually has much more serious consequences that type II error, and in testing we want to limit probability of type I error to a small number called the significance level of the test Significance level is the upper bound for the probability of type I error. Traditional significance levels are = 0.10, = 0.05, or = 0.01.

6 TESTS ABOUT POPULATION MEAN NORMAL POPULATION WITH KNOWN σ (one sample z test) 1. Population and sample: normal distribution, µ -unknown, σ unknown; sample of any size 2. Parameter: population mean µ 3. Null hypothesis: H 0 : μ = μ 0 4. Test Statistic: Z = X μ 0 σ n 5. Distribution under H 0 : standard normal N(0,1) 6. Rejection region: H a Rejection Region Graph H a : > 0 z > z z H a : < 0 z < - z -z H a : 0 z > z /2 z /2 where z is the critical value for standard normal distribution (Table A.5, bottom row) 7. Computation of P-value: H a P-value H a : > 0 P(Z > z 0 ) H a : < 0 P(Z < z 0 ) H a : 0 P( Z > z 0 ) = 2 P(Z > z 0 ) where z 0 the observed value of the test statistic and Z is N(0,1) TI-83: P( a < Z < b) = normalcdf(a,b) /2 /2

7 TESTS ABOUT POPULATION MEAN NORMAL POPULATION WITH UNKNOWN σ (one sample t test) 1. Population and sample: normal distribution, µ -unknown, σ unknown; sample of any size. 2. Parameter: population mean µ 3. Null hypothesis: H 0 : μ = μ 0 4. Test Statistic: T = X μ 0 s n 5. Distribution under H 0 : t distribution with df = n-1 degrees of freedom 6. Rejection region: below t,n-1 is a critical value for t distribution with df = n-1 (Table A.5) H a Rejection Region Graph H a : > 0 t > t,n-1 t H a : < 0 t < - t,n-1 -t H a : 0 t > t /2,n-1 t /2 where t,n-1 is a critical value for t distribution with df = n-1 (Table A.5) 7. P-value: H a P-value H a : > 0 P(T > t 0 ) H a : < 0 P(T < t 0 ) H a : 0 P( T > t 0 ) = 2 P(T > t 0 ) where t 0 the observed value of the test statistic and T has t distribution with df = n-1 TI-83: P( a < T < b) = tcdf(a,b,df) /2 /2

8 TESTS ABOUT POPULATION MEAN ANY POPULATION, LARGE SAMPLE (large sample z test) 1. Population and sample: any distribution, µ -unknown, σ known or unknown; large sample size (n > 30) 2. Parameter: population mean µ 3. Null hypothesis: H 0 : μ = μ 0 4. Test Statistic: Z = X μ 0 σ n if σ known or Z = X μ 0 s n 5. Distribution under H 0 : approximately standard normal N(0,1) 6. Rejection region: if σ unknown H a Rejection Region Graph H a : > 0 z > z z H a : < 0 z < - z -z H a : 0 z > z /2 z /2 where z is the critical value for standard normal distribution (Table A.5, bottom row) 7. P-value: H a P-value H a : > 0 P(Z > z 0 ) H a : < 0 P(Z < z 0 ) H a : 0 P( Z > z 0 ) = 2 P(Z > z 0 ) where z 0 the observed value of the test statistic and Z is N(0,1) /2 /2

9 TESTS ABOUT POPULATION PROPORTION (one proportion z test) 1. Population and sample: dichotomous made of successes (S) and failures (F), large sample such that np 0 10 and n(1-p 0 ) Parameter: population proportion p of successes, p = P(S) or more precisely parameter p in binomial distribution 3. Null hypothesis: H 0 : p = p 0 4. Test Statistic: Z = p p 0 p 0 (1 p 0 )/n number of succeses where p = n 5. Distribution under H 0 : approximately standard normal N(0,1) 6. Rejection region: H a Rejection Region Graph H a : p > p 0 z > z z H a : p < p 0 z < - z -z H a : p p 0 z > z /2 /2 z /2 where z is the critical value for standard normal distribution (Table A.5, bottom row) 7. P-value: H a P-value H a : p > p 0 P(Z > z 0 ) H a : p < p 0 P(Z < z 0 ) H a : p p 0 P( Z > z 0 ) = 2 P(Z > z 0 ) where z 0 the observed value of the test statistic and Z is N(0,1) /2

10 POWER CURVE. Recall that there are two types of errors Probabilities of type I and type II errors are traditionally denoted by and β, respectively. Because H 0 specifies a unique value of the parameter, there is a single value of. However, there is a different value of β for each value of the parameter consistent with H a. β(θ) = P(not rejecting H 0 if the true value of the parameter tested is θ in the range described by H a ) Experimental data is to consist of drying times from n = 25 test specimens. Suppose that we decided to reject H0 if x Type I error probability for this region is

11 Some values of β are: The graph of β(µ) for µ 75 is below (this is so called power curve)

12 Exercises: