(If it's handwritten it should be readable,understandable, and clear)

I. IDENTIFICATION.(15points) Directions: Read and identify what is being described by the following statements.

1. These are data that involve two variables. 2. The level of significance when the confidence level is 90%. 3. This represents the probability of committing a Type II error. 4. A type of error committed when a null hypothesis is rejected when it is true. 5. The test that is generally used when the population started deviation is known or given. 6. It is a graphical representation of the strength and direction of correlation between two variables. 7. A statement (hypothesis) that states that there is a difference between the parameter and a specific value. 8. A type of error committed when a null hypothesis is erroneously rejected. 9. A test of any statistical hypothesis where the alternative hypothesis is written with a not equal sign (?). 10. A test of any statistical hypothesis where the alternative hypothesis is expressed using less than (). 11. TRUE OR FALSE. If the symbol greater than (>) appears in the alternative hypothesis, then the test is right-tailed. 12. TRUE OR FALSE. If a test value or computed value falls in the rejection region, the null hypothesis is rejected, and the alternative hypothesis is accepted. 13. Association that exists between two variables when the points on the scatter plot follow a straight line slanting down to the right. 14. "Giving a Christmas bonus will increase the productivity of employees" is an example of a/an ______ hypothesis. 15. "There is no significant difference between the average weekly allowances of G11 and G12 students from AASMNHS" is an example of a ________ hypothesis.

B. Complete the table below.

Type of Tabular Test Sample Hypotheses Level of Value Decision (one-tailed Size Significance (Reject Ho or or two- Test Value (values n obtained Fail to reject tailed) from the z- Ho) Ho : H = 78 or t-table) H: u 75 H. : P = 0.10 18 0.05 Zo =2.55 H. : p > 0.10 H. : P = 0.60 100 0.05 Zo =-1.02 H . : p * 0.60