Lesson 9: Test of hypothesis PDF PDF PDF pdf, Lecture notes of Mathematics

Download Lesson 9: Test of hypothesis PDF PDF PDF pdf and more Lecture notes Mathematics in PDF only on Docsity! LESSON 9: TEST OF HYPOTHESIS UNIT 1 – BASIC CONCEPTS OF STATISTICAL HYPOTHESIS TESTING Introduction: Researchers are interested in answering many types of questions. For example, an educator might wish to see whether a new teaching technique is better than a traditional one. A retail merchant might want to know whether the public prefers a certain color in a new line of fashion. Automobile manufacturers are interested in determining whether seat belts will reduce the severity of injuries caused by accidents. These types of questions can be addressed through statistical hypothesis testing, which is a decision-making process for evaluating claims about a population. In hypothesis testing, the researcher must define the population under study, state the particular hypotheses that will be investigated, give the significance level, select a sample from the population, collect the data, perform the calculations required for the statistical test, and reach a conclusion. Learning Objectives: After successful completion of this lesson, you should be able to: 1. understand the concept of hypothesis testing 2. state the null and alternative hypotheses. Course Materials: STATISTICAL HYPOTHESIS It is a statement about the numerical value of a population parameter. It is a statement or tentative assertion which aims to explain facts about a certain phenomenon. It needs to be resolved whether it is true or not. It must be subjected to statistical testing procedure known as test of hypothesis or hypothesis testing. TWO KINDS OF HYPOTHESIS 1. NULL HYPOTHESIS It denoted by Ho, is a statement that there is no difference between a parameter and a specific value. 2. ALTERNATIVE HYPOTHESIS It denoted by Ha, is the opposite or negation of the null hypothesis. It is a statement that there exists a difference between a parameter and a specific value. EXAMPLE. CLAIM: The average monthly income of Filipino families who belong to low income bracket is Php 8,000 NULL HYPOTHESIS  Ho:The average monthly income of Filipino families who belong to low income bracket is Php 8, 000.  𝜇 = 8000 ALTERNATIVE HYPOTHESIS  Ha: The average monthly income of Filipino families who belong to low income bracket is not Php 8, 000.  𝜇 ≠ 8000 TYPES OF TESTS 1. DIRECTIONAL TEST  A test of any statistical hypothesis is expressed, using less than < or greater than > is called directional test or one-tailed test since the critical or rejection region lies entirely in one tail of the sampling distribution Example. Males will score significantly higher than Females on IQ scores. 2. NONDIRECTIONAL TEST  A test of any statistical hypothesis where the alternative hypothesis is written with a not equal sign ≠ is called a nondirectional test or two tailed test since there is no assertion made on the direction of the difference. The rejection region is split into two equal parts, one in each tail of the sampling distribution. Example. Males will score significantly different than Females on IQ scores. TYPES OF ERROR 1. TYPE I ERROR  It occurs when we rejected the null hypothesis when it is true. It also called alpha error (𝛼 𝑒𝑟𝑟𝑜𝑟). LESSON 9: TEST OF HYPOTHESIS UNIT 2 – TESTING A HYPOTHESIS ON THE POPULATION MEAN Introduction: In Inference for Means, our focus is on inference when the variable is quantitative, so the parameters and statistics are means. In “Estimating a Population Mean,” we learned how to use a sample mean to calculate a confidence interval. The confidence interval estimates a population mean.(lumencandela,2020) Learning Objectives: After successful completion of this lesson, you should be able to: 1. test hypotheses about population means, 2. formulate inferences about population parameters. Course Materials: When a researcher wishes to test whether an acceptable or established mean value for a certain product is still true then a test of a single population mean or z-test can be used. The formula can be used when the sample is equal or greater than 30 (n >≥30) provided the sample data are normally distributed and the population standard deviation is known. 𝑧 = (?̅? − 𝜇) 𝜎 √𝑛 𝜇 = population mean 𝜎 = population standard deviation 𝑛 = sample size Example. A principal at a school claims that the students in his school are above average intelligence. A random sample of 50 students’ IQ scores have a mean score of 110.5. Is there sufficient evidence to support the principal’s claim? The mean population IQ is 100 with a standard deviation of 15 and alpha level of 0.05. 1. Formulate the null and alternative hypothesis Ho: 𝜇 = 100 Hα: 𝜇 > 100 2. Set the level of significance, test statistic and critical values. The level of significance is 0.05. Z-test will be use. And the critical value is 1.96. 3. Compute the test value. 𝑧 = (?̅? − 𝜇) 𝜎 √𝑛 𝑧 = (110.5 − 100) 15 √50 = 4.95 4. Make a decision whether to accept or reject the null hypothesis. Since z of 4.95 is greater than 1.96, we reject the null hypothesis 5. Formulate a conclusion. The claim of the principal is incorrect. The sample standard deviation can also be used to estimate the population standard deviation, f I is known. In cases where a small size is used and the population standard deviation is unknown, then the t-test as given in formula will be used 𝑡 = (?̅? − 𝜇) 𝑠 √𝑛 𝜇 = population mean 𝑠 =sample standard deviation 𝑛 = sample size Usually, t-tests are most appropriate when dealing with problems with a limited sample size (n < 30). Example. A sample of five measurement randomly selected from an approximately normally distributed population, resulted in the summary statistics: ?̅? = 4.6, 𝑠 = 1.5. Test the null hypothesis that the mean of the population is 6 against the alternative hypothesis 𝜇 ≠ 6. Use 𝛼 = 0.05. 1. Formulate the null and alternative hypothesis Ho: 𝜇 = 6 Hα: 𝜇 ≠ 6 2. Set the level of significance, test statistic and critical values. The level of significance is 0.05. T-test will be used. And from the t-table with df = 5-1 = 4 then the critical value is ± 2.776. 3. Compute the test value. 𝑡 = (4.6 − 6) 1.5 √5 𝑡 = −2.087 4. Make a decision whether to accept or reject the null hypothesis. Since t of -2.087 is greater than -2.776, we failed to reject the null hypothesis 5. Formulate a conclusion. There is no significant difference between the means. Watch:  Hypothesis tests on one mean: t test or z test https://www.youtube.com/watch?v=vw2IPZ2aD-c  Statistics 101: Two Populations, z-test with Hypothesis https://www.youtube.com/watch?v=5NcMFlrnYp8  Statistics 101: Two Populations, t-test with Hypothesis https://www.youtube.com/watch?v=oJjkjkY6mmA Read:  z Test for a Mean Note: qo = 1- po = 1-.10=0.90 4. Make a decision whether to accept or reject the null hypothesis. Since z of -0.91 is greater than -1.96, we failed to reject the null hypothesis. 5. Formulate a conclusion. There is no significant difference between the sample proportion and population proprtion. Watch:  Hypothesis Testing With Two Proportions https://www.youtube.com/watch?v=pCbNUnZ98oE  Hypothesis Test for Proportions https://www.youtube.com/watch?v=t09Vyd7H52A Read:  z Test for a Proportion Bluman, A. G. (2012). Descriptive and Inferential Statistics. In Bluman, A. G., ELEMENTARY STATISTICS: A STEP BY STEP APPROACH, EIGHT EDITION (pp. 437-441). New York: McGraw-Hill Education LESSON 9: TEST OF HYPOTHESIS UNIT 4 – TESTING OF INDEPENDENCE (CHI-SQUARE TEST) Introduction: The Chi-square test of independence determines whether there is a statistically significant relationship between categorical variables. It is a hypothesis test that answers the question—do the values of one categorical variable depend on the value of other categorical variables? Learning Objectives: After successful completion of this lesson, you should be able to: 1. test a distribution for goodness of fit, using chi-square 2. formulate inferences about population parameters. Course Materials: In the test for independence, the claim is that the row and column variables are independent of each other. This is the null hypothesis. The test statistic used is the same as the chi-square goodness-of-fit test. The principle behind the test for independence is the same as the principle behind the goodness- of-fit test. The test for independence is always a right tail test. In fact, you can think of the test for independence as a goodness-of-fit test where the data is arranged into table form. This table is called a contingency table. The number of degrees of freedom for the test of independence is: df = (number of columns – 1) (number of rows – 1) The following formula calculates the expected number (E): E = (row total)(column total) total number surveyed FORMULA FOR TEST FOR INDEPENDENCE 𝒙𝟐 = ∑ (𝑶 − 𝑬)𝟐 𝑬 (𝒊∙𝒋)  O = observed values  E = expected values  i = the number of rows in the table  j = the number of columns in the table Examples. Find out if there is a significant relationship between the extent of cultural practices and religious involvement of the Bagos in the hinterland municipalities of Ilocos Sur. Cultural practices are categorized into three scales (Always Practiced, Moderately Practiced and Not Practiced) while religious involvement include Catholic, Non- Catholic, and not at all. RELIGIOUS INVOLVEMENT AND CULTURAL PRACTICES RELIGIOUS INVOLVEMENT CULTURAL PRCTICES TOTAL ALWAYS (AP) SOMETIMES (SP) NOT (SP) CATHOLIC 112 120 29 261 NON- CATHOLIC 131 101 90 322 NOE AT ALL 16 25 13 54 TOTAL 259 246 132 637 Solution.