ACTIVITY 4 Find the p-value given the following information. 1. Left -tailed test; computed z statistic = -1.62 2. Right-tailed test; computed z statistic = 2.06 3. Two-tailed test; computed z statistic = 1.22 4. Left-tailed test; computed z statistic = -1.41 5. Two-tailed test; computed z statistic = -1.62 ACTIVITY 5 Find the approximate p-value given the following information. 1. A two-tailed test; computed t-value=1.845; df =10 2. A left-tailed test; computed t-value =2.295; df=21 3. A right-tailed test; computed t-value=1.565; df=27 4. A two-tailed test; computed t-value=2.452; df=12 5. A left-tailed test; computed t-value=2.125; df =2715. What does the symbol n stand for in the formula z = =,"? A. sample mean B. sample size C. sample variance D. sample standard deviation ACTIVITY 10 In the problem below, (a) state the null and the alternative hypotheses; (b) compute the test statistic; (c) determine the critical value and the rejection region; and (d) draw a conclusion. 1. It is claimed that that the average weight of babies at birth is 3.4kg. The average weight of a random sample of 30 newly born babies was determined. It was found out that the average weight was 3.1 kg. Is there a reason to believe that the average weight of babies at birth is not 3.4 kg? Assume that the population standard deviation is 1.1 kg. Use 0.05 level of significance. 11...CONTINUATION OF ACTIVITY 7 variance is assumed to be unknown and n 390 4. What is the t-value? A. 2.3 B. 2.77 C. 2.5 D. -2.326 5. What is the critical value/s? A. -2.493 B. 2.499 C. 2.656 D. -2.492 6. What is the p-value? A. 0.025 > p > 0.01 B. 0.025 > p > 0.1 C. 0.025 > p > 0.03 D. 0.025 > p > 0.001 7. What can be concluded about the data? A. Do not reject the null hypothesis. B. Reject the null hypothesis. C. Accept the alternative hypothesis. D. Not enough information For Items 8-13, refer from the statement below. The Head of the Math Department announced that the mean score of Grade 9 students in the first periodic examination in Mathematics was 89 and the standard deviation was 12. One student who believed that the mean score was less than this, randomly selected 34 students and computed their mean score. She obtained a mean score of 85. At 0.01 level of significance, test the student's belief. COACTIVITY 6 Find the critical value of the ff. 1. A right-tailed test; a = 0.05; df = 24 2. A left-tailed test; a = 0.01; df = 14 3. A two-tailed test; ( = 0.1; df = 18 4. A left-tailed test; ( = 0.05; df = 28 5. A right-tailed test; a = 0.01; df = 25 ACTIVITY 7 Let us see what you have learned in this module by completing the following statements. 1. is a procedure used by statisticians to determine whether or not to reject a statement about a population. 2-7. The following are steps for a hypothesis test for a population mean when the variance is known and the population is assumed to follow normal distribution: Step 1 State the (2)_ and (3) hypothesis Step 2. Choose the (4) Step 3. Compute the (5). Step 4. Determine the (6) Step 5. Draw a (7) 8-10. Assuming the population follows a normal distribution, the appropriate test statistic for testing a claim about a population mean when the population8. What is the population variance? A. 12 B. 34 C. 144 D. 140 9. What is the null hypothesis? A. Hat ! = 89 B. Hot ! = 89 C. Ho: 1 = 85 D. Ho: # > 85 10. What is the z-value? A. 1.94 B. -1.90 C. -1.94 D. 1.90 1 1. What is the critical value/s? A. -2.33 B. 2.33 C. 1.33 D. -1,33 12. What is the p-value? A. 0.0262 B. 0.0056 C. 0.088 D. 0.0234 13. What can be concluded about the data? A. Do not reject the null hypothesis. B. Reject the null hypothesis. C. Accept the alternative hypothesis. D. Not enough information 14. In the t-test formula, the symbol $ stands for? A. sample standard deviation B. population standard deviation C. population variance D. sample variance 10