Hypothesis Testing

HYPOTHESIS TESTING. Hypotheses are assumptions, guesses, or statements or assertion which may or may not be true concerning one or more population parameters. Hypothesis can be subdivided into two; null hypothesis and alternative hypothesis. The null hypothesis is denoted by H0 and the alternative is denoted by H1. The null hypothesis is a statistical hypothesis which can be in some way be tested or it is the hypothesis that we are suspicious about and wish to disprove. It is easily testable hypothesis. It is also a statement that claims the existence of no disagreement with the true conditions in the population of interest. H0 and H1 are complementary meaning the acceptance of one leads to the rejection of the other. We state the null hypothesis in such a way as to specify the exact value hypothesized for the population parameter of interest whereas the alternative hypothesis allows for the possibility of one or several values. Alternative hypothesis is just the hypothesis that is alternatively accepted if the null hypothesis is rejected. There exist two possible types of alternative hypothesis; one tailed /one sided hypothesis and two tailed /two sided hypothesis. A two- tailed is the alternative hypothesis that considers any changes in the parameter whether it be an increase or decrease. That is, a test with not equal to sign ( ) is called two tailed or two sided test because the rejection region is divided into two equal parts, one to each tail or side of the distribution of the test statistic. For one tailed/ one sided, the rejection region for the alternative hypothesis containing a greater than symbol > lies entirely to the right tail or side of the distribution while the rejection region for the alternative hypothesis containing a less than < symbol lies to the left tail or side of the distribution.

A test of a null hypothesis is simply a rule based on the results of a random sample whereby we decide whether to accept or reject the particular null hypothesis (H0) under question. Test statistic depends on the nature of data and it helps in decision making since the decision to reject or accept H0 depends on the magnitude of the test statistic. Critical or Rejection region : is the set of all possible values of the test statistic that lie to one side of the critical value. The values that lie to the other sides of the critical value constitute the acceptance region. The critical value of a test statistic is that value or values that lies on the boundary or boundaries of the critical region. Significance level of a test: This is the probability that the test statistic lies in the critical region under H0 and the most commonly used levels are 5% and 1%. Decision rule for a statistical test is a model given values of a sample statistics ( not necessarily the test statistics) that will lead to either the acceptance or rejection of the stated null hypothesis. The decision rule tells us to reject H0 if the value of the test statistic lies in the critical region. On the other hand, we accept H0 if the computed value of the test

statistic lies in the acceptance region. Simple and Composite Hypothesis In describing hypothesis, a simple hypothesis specify one value for the population parameter for example µ = 0 or µ = 30 while a composite hypothesis specifies a range of values for the population parameter, for example µ < 30, µ > 0 etc. Errors in Hypothesis Testing. We always run the risk of committing an error when we either reject or accept H0, the error we commit when we reject H0 that should have been accepted is called a TYPE I error and the error we commit when we accept H0 that should have been rejected is called the TYPE II error. We denote type I error by α that is probability(reject true H0) = α type II error by β that is probability(accept false H0) = β The probability of committing a type I error α is called the level of significance of the test or the significance level. It should be noted that α is the more serious error and deliberate effort is usually made to keep it as small as possible. The knowledge of α helps us to determine the critical value and hence the rejection and the acceptance regions for our test. For a one-sided test, the total value of α is used to determine the critical value since in this case, the entire rejection lies to one side of the distribution of the test statistic. For a two-sided, we use α/2 (half of α) to determine the required critical value because in this case, the rejection region is divided into two equal parts, one to each side of the distribution of the test statistic.

Accept H0

TABLE FOR ERRORS H0 is true Correct Decision (1 – β)

Reject H0

Incorrect decision Type I error α (I – β) gives the power of the test.

Standard Format of Hypothesis Testing. Step 1: State the null and alternative hypothesis. Step 2: Determine the test statistic.

H0 is false Incorrect decision Type II error β Correct Decision (1 – α)

Step 3: Determine the critical region using the cumulative distribution table from the test statistic. This step involves specifying the values of the test statistic that leads to the acceptance or rejection of H0. Step 4: Compute the value of the test statistic based on the sample information. Step 5: You make the statistical decision and interpretation. Example 33. It has been found from experience that the mean lifetime of a sample of 50 bulbs produced by a company is computed to be 1570 hours with a standard deviation of 100 hours. If µ is the mean lifetime of all bulbs produced by the company. Test the hypothesis: µ= 1000 hours against the alternative µ 1000 hours. Solution. H0 : µ = 1000 hours H1 : µ 1000 hours It is a two tailed test.

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We REJECT H0 since it lies on the critical part. Example 34. A certain manufacturing company claimed that the breaking strength of cables they produced have mean of 1000 and standard deviation of 10. The company also claimed that the breaking strength can also be increased by a new technique, a random sample of 30 cables is tested and the mean was found to be 1050. To test the claim, does the data support the company’s claim or not at 0.01 level of significance. Solution. H0 : µ = 1000 H1 : µ > 1000 It is a one tailed test.

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We reject the null hypothesis and conclude that the data does not support the company’s claim. Example 35. In the past, the tensile strength of the polythene bag produced by a businessman was 4500kg. To determine whether the machine producing these polythene bags is in proper working condition, a sample of 17 polythene bags are measured. The sample mean was 4482kg and the standard deviation was 115kg. Is the machine in proper working order at 0.05 level of significance? Solution. H0 : µ = 4500 (machine is in proper working order) H1 : µ 4500 (machine is not in proper working order) It is a two tailed test.

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We accept the null hypothesis; the machine is in proper working order. Example 36. Specifications given by Standard Organization of Nigeria (SON) That industrial catalysts must contain at least 40.5% Tungsten. A sample of 12 brands of catalysts from different companies showed a mean Tungsten content of 41.2% and standard deviation of 0.95%. Can we conclude at 0.05 significance that the catalysts meet the required specifications? Solution. H0 : µ = 40.5% H1 : µ 40.5% It is a two tailed test.

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We reject the null hypothesis and conclude that the catalysts do not meet the stipulated specifications. Example 37. A company manufactures fluorescent tubes which have life span that is normally distributed with a mean of 700 hours and a standard deviation of 50 hours. A random sample of 100 tubes showed an average life span of 720 hours. Does this suggest at the 0.01 level of significance that the average life span of the fluorescent tubes manufactured by the company has increased? Solution. H0 : µ = 700 H1 : µ > 700 It is a one tailed test.

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We reject the null hypothesis and conclude that the mean life span of the fluorescent tube manufactured by the company has increased. Example 38. It has been found from experience that the mean breaking strength of a particular brand of thread is 9.72kg with a standard deviation of 1.4kg. Recently a sample of 36 pieces of the thread showed a mean breaking strength of 8.93kg. Can one conclude at a significant level of 0.05 that the thread has become inferior? Solution. H0 : µ = 9.72 kg H1 : µ < 9.72 kg It is a one tailed test.

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We accept the null hypothesis and conclude that the brand has not become inferior.