MAT X Hypothesis Testing - Part I

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1 MAT X Hypothesis Testing - Part I Definition : A hypothesis is a conjecture concerning a value of a population parameter (or the shape of the population). The hypothesis will be tested by evaluating evidence that comes from an experiment or a study. In these notes, we consider hypotheses concerning a mean µ or a proportion p. Formulating the null hypothesis and the alternative hypothesis : The null hypothesis, that we denote H 0, is a hypothesis that we often would like to reject. The rejection of H 0 will then implies the acceptance of an alternative hypothesis H 1. The alternative hypothesis is often our research hypothesis. Remarks : We often state the null hypothesis as a simple hypothesis (that is a simple value) and the alternative hypothesis will be stated an composite hypothesis (i.e. a set of many values). The null hypothesis is usually a statement of no effect. Suppose that θ is the unknown population parameter and that θ 0 is a real number. For example, θ could be the population mean. We will consider three types of alternative. 1. Suppose that a priori, our research hypothesis is that θ > θ 0. Then, we will consider a right sided alternative. That is, we will test H 0 : θ = θ 0 against θ > θ Suppose that a priori, our research hypothesis is that θ < θ 0. Then, we will consider a left sided alternative. That is, we will test H 0 : θ = θ 0 against θ < θ 0. 1

2 3. A two sided alternative would give the following test : H 0 : θ = θ 0 against θ θ 0. The research hypothesis is often (but not always) the alternative hypothesis. Example 1 : (a) Often the investigator would like to verify a change concerning a population parameter under new experimental conditions. For example, an investigator states that the use of a new treatment will produce taller plants on average compared to some old treatment. The mean height under the old treatment is 20 cm. (i) Formulate the hypotheses that we will test to verify the statement of the investigator. (ii) We suspect the opposite that is that the new mean will be smaller. Formulate the hypotheses that we will test to verify our hypothesis. (b) Let us now consider an example concerning a proportion. Consider an experiment where we cross green pees with yellow pees. Formulate a hypothesis to verify that the proportion of yellow offsprings is p = 25%. 2

3 Implementation of a hypothesis test : Formulate a null hypothesis to test against an alternative hypothesis. We start by assuming that the null hypothesis H 0 is true. Often this is a statement of no effect. For example, if we are testing H 0 : µ = 20 against H 1 : µ > 20, so we will assume that µ = 20 is true. The next step is to evaluate the evidence from our experiment or study against H 0 in favour of H 1. For our example, suppose that x = So we have evidence against H 0 in favour of H 1 since x > 20. (Recall : x is an estimate of µ). A value of the sample mean larger than 22.5, would be considered as even stronger evidence against H 0 in favour of H 1. We must then evaluate the significance of our evidence. To do so, we will evaluate of p-value. Definition : The p-value of the test is the probability that a value is as extreme (i.e. in favour of H 1 ) as the current observed value assuming that H 0 is true. What does a small p-value mean? It means that assuming that H 0 is true, our chances of observing such evidence is highly unlikely. However, most events that occur are likely by definition. So most likely, our calculation of the probability is wrong. Thus, our original assumption that H 0 is true is most likely wrong. Thus, we should reject H 0 to accept H 1. We will interpret small p-values as evidence against H 0 in favour of H 1. How small should a p-value be to be considered as small enough? To determine if the evidence is significant, we will need to fix a level of significance α such that a p-value smaller than this level will be considered as significant. (Often we use α = 5%). Decision Rule : If p < α, then we have significant evidence against H 0. We reject H 0 in favour of H 1. If p α, so then we do not have evidence against H 0. So we fail to reject H 0. 3

4 Test statistic : To compute the p-value, we will need a test statistic. A test statistic is a statistic that is used to test our hypotheses. It is often based on an estimator of the parameter concerned in the hypotheses. For example, if we are testing a hypothesis on µ, then can use a test statistic based on the sample mean X. Let µ be the unknown population mean. Say, that we want to test a null hypothesis H 0 : µ = µ 0, where µ 0 is some real number. The test statistic that we will use will be Z = X µ 0 σ/ n, if the population is normal (or n 30) and σ is known. if n 40 and σ is known. Z = X µ 0 S/ n, T = X µ 0 S/ n, if the population is normal and σ is unknown. Remark : If the corresponding underlying assumptions are true and that the null hypothesis is true (i.e. µ = µ 0 ), then the z-test statistic has a standard normal distribution N(0, 1) and the t-test statistic has a t distribution with ν = n 1 degrees of freedom. 4

5 Observed value of the test statistic and the p-value : Consider H 0 : µ = 20. We will use either a z-test statistic or a t-test statistic to test H 0. We interpret the z (or t) value as a measure of distance between the mean in the null hypothesis and the observed mean x (in standard deviation). Furthermore, the sign of z (or t) also indicates a direction (+ means x > µ 0 and means x < µ 0 ). Suppose that the population is normal and that we observe 22.5 and s = 5.1 for a sample of size n = 26. The observed value of the t-test statistic is t 0 = x 20 s/ n = / 26 = (a) [Right-sided alternative] : Suppose that we want to test H 0 : µ = 20 against H 1 : µ > 20. Since the observed value of the t-test statistic is positive, then we have evidence against H 0 in favour of H 1. Even stronger evidence would be even larger t-values. So we compute the p-value is follows : p-value = P (T > 2.50), where T has a T (n 1) = T (25) distribution. Using Table 17.4, we get < p-value < Suppose that the significance level of α = 5%. This means that the p-value is smaller than the significance level, so we can reject H 0 to accept H 1. So we conclude that µ > 20. (b) [Left-sided alternative] : Suppose that we want to test H 0 : µ = 20 against H 1 : µ < 20. Since the observed value of the t-test statistic is positive, then we do not have evidence against H 0 in favour of H 1. Yet, we can still compute the corresponding p-value. We interpret t-values even more to the left of the observed value t 0 = 2.50, as stronger evidence against H 0 in favour of H 1. So we compute the p-value is follows : p-value = P (T < 2.50), 5

6 where T has a T (n 1) = T (25) distribution. Using Table 17.4, we get 0.99 < p-value < Suppose that the significance level of α = 5%. This means that the p-value is not smaller than the significance level, so we should not reject H 0. We do not have sufficient evidence to conclude that µ > 20. (c) [Two-sided alternative] : Suppose that we want to test H 0 : µ = 20 against H 1 : µ 20. In this case, both positive and negative t-values are considered as evidence against H 0 in favour of H 1. We will consider values which are even further away from zero as the observed value t 0 = 2.50, as even stronger evidence in favour of H 1. So we compute the p-value is follows : p-value = P ({T < 2.50} or {T > 2.50}) = 2 P (T > 2.50), where T has a T (n 1) = T (25) distribution. Using Table 17.4, we get Therefore, < P (T > 2.50) < < p-value < Suppose that the significance level of α = 5%. This means that the p-value is smaller than the significance level, so we should reject H 0 and accept H 1. We have significant evidence that µ 20. Computing the p-value : 1. Suppose that we are using a t-test statistic with ν degrees of freedom and let t 0 be the observed value of the test statistic. We will compute the p-value as follows : If it is a right-sided alternative : p-value = P (T > t 0 ), where T has a T (ν) degrees of freedom. 6

7 If it is a left-sided alternative : p-value = P (T < t 0 ), where T has a T (ν) degrees of freedom. If it is a two-sided alternative : p-value = P ({T > t 0 } or T < t 0 ) = 2 P (T > t 0 ), where T has a T (ν) degrees of freedom. 2. Suppose that we are using a z-test statistic and let z 0 be the observed value of the test statistic. We will compute the p-value as follows : If it is a right-sided alternative : If it is a left-sided alternative : If it is a two-sided alternative : p-value = P (Z > z 0 ) = 1 Φ(z 0 ). p-value = P (T < z 0 ) = Φ(z 0 ). p-value = P ({Z > z 0 } or Z < z 0 ) = 2 P (Z > z 0 ) = 2 [1 Φ( z 0 )]. 7

8 Example 2 : Consider a particular type of tree that usually grows on average 3 meters per year. Investigators suspect that this mean has decreased in the last few years because of deforestation and climate change. (a) Formulate the null hypothesis and altervative hypothesis to verify the investigators conjecture. (b) A random sample of n = 15 observations gave a mean growth rate of 2.55 meters per year and a standard deviation of 0.63 meters per year. Assume that the growth rate is normally distributed, do we have significant evidence conclude that the mean growth rate has decreased. Use a level of significance of α = 5%. 8

9 Hypothesis Testing concerning a proportion Assuming that we have a binomial experiment with a probability of success p, then we can use a z-test statistic Z = ˆP p 0 p0 (1 p 0 )/n to test H 0 : p = p 0, where p 0 is some real number between 0 and 1. If the number of trials is large and H 0 is true, then Z has a N(0, 1) distribution approximately. Example 3 : Consider an experiment where we cross green pees with yellow pees. (a) Formulate a hypothesis to verify that the proportion of yellow offsprings is p = 25%. (b) Among 580 offspring, 152 are yellow. What should we conclude? Use a level of significance of 5%. 9