1. C HAPTER 2  Hypothesis Testing -Test for one means - Test for two means -Test for one and two proportions

2. WHY WE HAVE TO DO THE HYPOTHESIS? To make decisions about populations based on the sample information. Example :- we wish to know whether a medicine is really effective to cure a disease. So we use a sample of patients and take their data in effect of the medicine and make decisions. To reach the decisions, it is useful to make assumptions about the populations. Such assumptions maybe true or not and called the statistical hypothesis.

3. Type I and Type II error Four outcomes of decision: There are two possibilities correct outcomes and two possibilities incorrect outcomes.

4. Type I error Occurs if you reject null hypothesis Level of significance is the maximum probability of committing a type I error ( ) Type II error occurs if you do not reject the null hypothesis when it is false. The probability of type II error is symbolized by

5. Example Hypothesis Testing and Jury Trial:

6. Definitions Hypothesis Test: It is a process of using sample data and statistical procedures to decide whether to reject or not to reject the hypothesis (statement) about a population parameter value (or about its distribution characteristics). Null HypothesisAlternative Hypothesis states that there is no difference between a parameter and specific value, or that there is no difference between two parameters. states the existence of difference between a parameter and a specific value, or states that there is difference between two parameters.

7. Test Statistic: It is a function of the sample data on which the decision is to be based. Critical/ Rejection region: It is a set of values of the test statistics for which the null hypothesis will be rejected. Critical point: It is the first (or boundary) value in the critical region. P -value: The probability calculated using the test statistic. The smaller the p -value is, the more contradictory is the data to.

8. Procedure for hypothesis testing 1. Define the question to be tested and formulate a hypothesis for a stating the problem. 2. Determine the critical value and critical region. 3. Choose the appropriate test statistic and calculate the sample statistic value. The choice of test statistics is dependent upon the probability distribution of the random variable involved in the hypothesis. 4. Make decision whether to accept or to reject the null hypothesis. 5. Conclusion

9. Hypothesis Testing For One Mean Population 1. Formulate the hypothesis and define claim. 2. Critical value by finding from statistical book. (page 26 or page 28) Two-Tailed Test Right-Tailed Test Left-Tailed Test

10. 3. Test Statistic 4. Rejection Rule

11. 5. Conclusion

12. Example 2.1: A sample of 50 Internet shoppers were asked how much they spent per year on Internet. From this sample, mean expenses per year on Internet is 30460 and sample standard deviation is 10151. It is desired to test whether they spend in mean expenses is RM32500 per year or not. Test at.

13. Solution: 1.The hypothesis tested are: 2. Critical Value : As two tailed (=), so alpha has to divide by two, becomes : 3. Test Statistic: 4. Rejection Region: 5. Conclusion: The Internet Shoppers spend RM32500 per year on the Internet.

14. Example 2.2: A random sample of 10 individuals who listen to radio was selected and the hours per week that each listens to radio was determined. The data are follows: 9 8 7 4 8 6 8 8 9 10 Test a hypothesis if mean hours individuals listen to radio is less than 8 hours at. *Hint : Find

15. Solutions: 1. The hypothesis tested are: 2. Critical Value: 3. Test Statistic: n < 30 4. Rejection Region : 5. Conclusion : Mean hours individuals listen to radio is greater or equal to 8 hours.

16. Exercise 2.1: 1. From the following data, test the null hypothesis that population mean is less than or equal to 100 at 5% significance level. Ans: t = 4.3043, Reject 105 108 112 121 100 105 99 107 112 122 118 105 2. A paint manufacturing company claims that the mean drying time for its paint is at most 45 minutes. A random sample of 35 trials tested. It is found that the sample mean drying time is 49.50 minutes with standard deviation 3 minutes. Assume that the drying times follow a normal distribution. At 1% significance level, is there any sufficient evidence to support the company claim? Ans: z = 8.8741, Reject

17. 3.The quality control manager claim that the daily yield at a local chemical plant was averaged less than 80 tons for the last several years. A random sample of 50 days gives an average yield of 71 tons with a standard deviation of 21 tons. (a) Construct a 95% confidence interval of mean yield at a local chemical plant. Ans : [65.1791, 76.8209] (b) Construct a 99% confidence interval of mean yield at a local chemical plant. Ans : [63.3503, 78.6497] (c) Calculate the width of confidence interval in (a) and (b). Explain the different width. Ans : 11.6418, 15.29 (d) Build a hypothesis testing whether the quality control claim is true or not at 5% significance level. Ans : Z = -3.0305, Reject