9. Hypothesis and Test Procedures A statistical test of hypothesis consist of : 1. The Null hypothesis, H 0 : A null hypothesis is a claim (or statement) about a population parameter that is assumed to be true. 2. The Alternative hypothesis, H 1 : An alternative hypothesis is a claim about a population parameter that will be true if the null hypothesis is false. 3. The test statistic: function of the sample data on which the decision (reject H 0 or do not reject H 0 is to be based. 4. The rejection region and the conclusion. ****************************************************************************** 9.1 Testing Hypothesis on the Population Mean, Null Hypothesis : Alternative hypothesis: H 0 : 0 H 1 : 0 Rejection Region Z < - z 2 or Z > z 2 H1 : 0 H1 : 0 Test Statistic Z > z Z < - z : Z x any population, is known and n is either large or small n x s n x t s n Z any population, large is unknown and n is normal population, n is small is unknown and ****************************************************************************** Note :p-value -we do not reject the null hypothesis, H 0 if The p-value is the smallest significance level at which the null hypothesis is rejected. p-value for one-tailed test Using the p-value approach, -we reject the null hypothesis, H 0 if p-value < for one-tailed test p-value < p-value for two-tailed test 2 for two-tailed test 2 ****************************************************************************** Example 9.1 The average monthly earnings for women in managerial and professional positions is RM 2400. Do men in the same positions have average monthly earnings that are higher than those for women? A random sample of n = 40 men in managerial and professional positions showed x RM 3600 and s = RM 400. Test the appropriate hypothesis using = 0.01 Solution 1.The hypothesis to be tested are, H 0 : = 2400 Z H 1 : > 2400 2 We use normal distribution n>30 3. Rejection Region : Z > z z z 0.01 2.33 x 3600 2400 = = 18.97 s 400 40 n Since 18.97 > 2.33, falls in the rejection region, we reject H 0 and conclude that average monthly earnings for men in managerial and professional positions are significantly higher than those for women. 4. Test Statistic ****************************************************************************** Example 9.2 The daily yield for a local chemical plant has averaged 880 ton for the last several years. The quality control manager would like to know whether this average has changed in recent months. She randomly selects 50 days from the computer database and computes the average and standard deviation of the n = 50 yields as x = 871 tons and s = 21 tons, respectively. Test the appropriate hypothesis using 0.05 . Solution The hypothesis to be tested are, H 0 : = 880 H 1 : 880 The point estimate for test statistic is Z is x . Therefore, x 871 880 = = -3.03 s 21 50 n Rejection Region : Z < - z 2 or Z > z 2 z z0.025 1.96 2 Since – 3.03 < - 1.96, falls in the rejection region, the manager can reject H 0 and conclude that it has changed 9.2 Testing Hypothesis on the Population Proportion, p Null Hypothesis : Alternative hypothesis: H 0 : p p0 H 1 : p p0 Rejection Region Z < - z 2 or Z > z 2 H 1 : p p0 H 1 : p p0 Test Statistic : Z Z > z Z < - z pˆ p0 p0 q0 n Example 9.3 When working properly, a machine that is used to make chips for calculators does not produce more than 4% defective chips. Whenever the machine produces more than 4% defective chips it needs an adjustment. To check if the machine is working properly, the quality control department at the company often takes sample of chips and inspects them to determine if the are good or defective. One such random sample of 200 chips taken recently from the production line contained 14 defective chips. Test at the 5% significance level whether or not the machine needs an adjustment. Solution The hypothesis to be tested are , H 0 : p 0.04 H 1 : p > 0.04 The point estimate for p is p̂ . Therefore, test statistic is Z pˆ p0 0.07 0.04 = = 2.17 p0 q0 0.04(0.96) 200 n Rejection Region : Z > z ; z z 0.05 1.65 Since 2.17 > 1.65, falls in the rejection region, we can reject H 0 and conclude that the machine needs an adjustment. 9.3 Hypothesis Test For the Difference between Two Populations Means, 1 2 H 0 : 1 2 = 0 H 1 : 1 2 0 Null Hypothesis : Alternative hypothesis: H 1 : 1 2 0 H 1 : 1 2 0 Test Statistic x Z 1 x 2 1 2 1 x Z t : 2 n1 1 2 x 2 1 2 2 s1 s 2 n1 n2 1 Z < - z 2 or Z > z 2 Z > z Z < - z For two large and independent samples and 1 and 2 are known. n2 2 x 2 Rejection Region x 2 1 2 2 2 s1 s 2 n1 n2 For two large and independent samples and 1 and 2 are not known. For two small and independent samples taken from two normally distributed populations. Example 9.4: A university conducted an investigation to determine whether car ownership affects academic achievement was based on two random samples of 100 male students, each drawn from the student body. The grade point average for the n1 = 100 non owners of cars had an average and 2 2 variance equal to x1 2.70 and s1 0.36 as opposed to x 2 2.54 and s2 0.40 for the n2 = 100 car owners. Do the data present sufficient evidence to indicate a difference in the mean achievements between car owners and non owners of cars? Test using 0.05 =>The hypothesis to be tested are , H 0 : 1 2 = 0 H1 : Rejection Region : Z < - z 2 or Z > z 2 ; z 0.05 2 = z 0.025 = 1.96 1 2 0 Therefore, test statistic is Z x 1 x 2 1 2 2 2 = s1 s 2 n1 n2 2.70 2.54 = 1.84 Z 0.36 0.40 100 100 Since 1.84 does not exceed 1.96 and not less than - 1.96, we fail to reject H 0 and that is, there is not sufficient evidence to declare that there is a difference in the average academic achievement for the two groups. 9.4 Hypothesis Test For the Difference between Two Population Proportion, p1 p2 H 0 : p1 p2 = 0 H 1 : p1 p 2 0 Null Hypothesis : Alternative hypothesis: Rejection Region Z < - z 2 or Z > z 2 H 1 : p1 p 2 0 H 1 : p1 p 2 0 Test Statistic : Z pˆ1 pˆ 2 p1 p2 pq ˆ ˆ pq ˆˆ n1 n2 where Z > z Z < - z pˆ x1 x2 n1 n2 Example 9.5: Reconsider Example 4.8, At the significance level 1%, can we conclude that the proportion of users of Toothpaste A who will never switch to another toothpaste is higher than the proportion of users of Toothpaste B who will never switch to another toothpaste? Solution The hypothesis to be tested are , H 0 : p1 p2 = 0 H 1 : p1 p 2 0 ( p1 is not greater than p2) ( p1 is greater than p2) Therefore, test statistic is Z pˆ 1 pˆ 2 p1 p 2 1 1 pˆ qˆ n1 n2 = Z 0.20 0.17 0 1 1 (0.187 )(0.813) 500 400 = 1.15 Rejection Region : Z > z ; z 0.01 = 2.33 Since 1.15 < 2.33, we fail to reject H 0 and therefore, we conclude that the proportions of users of Toothpaste A who will never switch to another toothpaste is not greater than the proportion of users of Toothpaste B who will never switch to another toothpaste. Example 9.6: The mean income per month for 30 workers chosen at random from the private sector is RM 650 with a standard deviation of RM 100 per month. The mean income per month for 40 workers from the government sector is RM 500 with a standard deviation of RM 80 per month a) b) Test the hypothesis that the mean income for all workers from government sector is less than RM 520 per month at 5% significance level. Sketch the rejection/critical region. At 0.05 , can we conclude that the mean income per month for private workers is higher than the government workers? Sketch the rejection/critical region. (a) 0.05 (b) 0.05 H o : g 520 H o : p g 0 H1 : g 520 H1 : p g 0 Z test X g g 500 520 sg 80 40 ng 20 12.6491 1.5811 From the table, Z Z 0.05 1.6449 we reject H o if Z test Z . Since Z test 1.5811 Z 0.05 1.6449 so we fail to reject H o . Hence, we accept H o and conclude that there is no sufficient evidence to prove that the mean income for all workers from government sector is less than RM 520 per month. Z test X p X g p g s 2 p np sg2 ng 650 500 0 1002 802 30 40 50 6.7534 22.2111 From the table, Z Z 0.05 1.6449 we reject H o if Z test Z . Since Z test 6.7534 Z 0.05 1.6449 so we reject H o . Hence, we accept H1 and we conclude that the mean income per month for private workers is higher than the government workers. 1.6449 -1.6449 Z Tutorial 7 (Hypothesis Testing) 1) A lecturer claim that the medical students put in more hours studying compared to other students. The mean number of hours spent studying per week for other students is 23 hours with a standard deviation of 3 hours per week. A sample of 25 medical students was selected at random and the mean number of hours spent studying per weeks was found to be 25 hours. Can the lecturer’s claim be accepted at 5% significance level? ( H 0 : 23, H1 : 23; Z 3.333 Reject H 0 ) 2) A sample consists of 36 glasses of orange juice was dispensed from a drink dispenser with mean volume per glass of 21.9 ml and a standard deviation of 1.42 ml. The standard mean volume per glass of orange juice dispensed by the drink dispenser is 22.2 ml. Test the hypothesis that the mean volume per glass is less than the standard mean volume at 0.05 . ( H 0 : 22.2, H1 : 22.2; Z 1.27 Accept H 0 ) 3) A manufacturer of a traditional medicine claimed that his medicine is 90% effective relieving backache for a period of 8 hours. In a sample of 200 people who had backache, the medicine provided relief for 160 people. Determine whether the manufacturer’s claim is legitimate at 1% significance level. ( H 0 : p 0.9, H1 : p 0.9; Z 4.7140 Accept H 0 ) 4) A pharmacy company claimed that not more than 1% that used its drug experience side effect. To prove its claim, 2500 patients were subscripted with the drugs and only 74 experienced side effects. Can the claim by the pharmacy company be accepted at the 5% level of significance? ( H 0 : p 0.01, H1 : p 0.01; Z 9.95 Accept H 0 ) 5) The mean lifetime of 30 bulbs produced by Company A is 500 hours and the mean lifetime of 35 bulbs produced by Company B is 492 hours. If the standard deviation of all bulbs produced by Company A is 10 hours and the standard deviation of all bulbs produced by Company B is 15 hours, test at 1% significance level that the mean lifetime of bulbs produced by Company A is better than that of Company B. ( H 0 : A B 0, H1 : A B 0; Z 2.56 Reject H 0 ) 6) Suppose that we want to investigate whether men’s average earning exceed women’s average earning more than RM 20 per week in a certain workplace. A sample of 60 men earns an average of RM 294 per week with a standard deviation of RM 16, while a sample of 60 women earn an average of RM 265 per week with a standard deviation of RM 18. Test at 1% significance level. ( H 0 : A B 20, H1 : A B 20; Z 2.895 Reject H 0 ) 7) A random sample of 200 screws manufactured by machine A and 100 screws manufactured by machine B showed 19 and 5 defective screws respectively. Test the hypothesis that a) Machine B is performing better than machine A at 5% significance level. b) The two machines are showing different qualities of performance at 5% significance level. ( H 0 : pB pA 0, H1 : pB pA 0; Z 1.3543 Accept H 0 ) ( H 0 : pB pA 0, H1 : pB pA 0; Z 1.3543 Accept H 0 ) 8) Two groups, from urban and from rural area each consists of 100 people who have a disease were taken as samples. A serum is given to the group from urban area but not to the group from rural area (which is called the control group); otherwise, the two groups are treated identically. It was found that in the group from urban area and the group from rural area, 75 and 65 people respectively recover from the disease. Test the hypothesis that the serum helps to cure the disease using 5% significance level. ( H 0 : pu pr 0, H1 : pu pr 0; Z 1.552 Accept H 0 )