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Advanced Math Topics 10.5 Tests Concerning Means for Small Samples If you understood yesterday’s lesson about accepting/rejecting a company’s claim from a large sample, then today’s lesson should come quickly, as it is very similar. If you did not understand yesterday’s lesson, today’s lesson may help because the only difference is which chart to use in the back of the book. Refresher: A large sample is… greater than 30. A small sample is… less than or equal to 30. A manufacturer claims that each bag of mixed nuts sold contains an average of 10 cashew nuts. A sample of 15 cans has an average of 8 cashew nuts with a standard deviation of 3. Does this indicate that we should reject the manufacturer’s claim? Use a 5% level of significance. Null Hypothesis, H0: μ = 10 Alternative Hypothesis, H1: μ ≠ 10 It does not state that we are looking for strictly MORE THAN or LESS THAN 10 cashews, and we are only looking to reject the claim that there is an average of 10 cashews, therefore this is a two-tail test and we use ≠. (x ) y t s 0.025 0.025 t = -2.145 μ = 10 z = 2.145 (8 10 ) 15 Since the sample is small, we use the t-table in the t back. The degrees of freedom is 15 - 1 = 14. You can 3 either draw a picture and label each rejection region or divide the significance level by 2 (because t = -2.58 it is a two-tail test) to find the 0.05/2 = t0.025. Since the t-score is in the rejection region, our decision is that must reject the claim that each bag has an average of 10 cashews. A new health pill is being used at a hospital to help fight cholesterol. The manufacturer claims that anyone with high cholesterol who takes this pill will lose 15 mg of cholesterol within a month. A doctor believes this is inflated, & gives this pill to six people with high cholesterol and finds that they lose an average of 12 mg with a standard deviation of 4 mg. Should we reject the manufacturer’s claim? Use a 5% level of significance. Null Hypothesis, H0: μ = 15 Alternative Hypothesis, H1: μ < 15 Since the doctor feels the manufacturer’s average is inflated, if we reject the claim, we will state that the average is less than 15 mg. Therefore this is a one-tail test. (x ) y t s 0.05 t = -2.015 μ = 15 (12 15 ) 6 t Since the sample is small, we use the t-table in the 4 back. The degrees of freedom is 6 - 1 = 5. You can either draw a picture and label the rejection region or divide the significance level by 1 (because t = -1.84 it is a one-tail test) to find the 0.05/1 = t0.050. Since the t-score is in the acceptance region, we cannot reject the manufacturer’s claim that the average amount of cholesterol lost is 15 mg. From the HW P. 497 1) Welfare officials in a certain city claim that the average number of cases of child abuse handled daily is 7.62. A newspaper tests this claim by selecting 10 days and they finds the average for those days is 5.32 claims with a standard deviation 2.06. Using a 5% level of significance, should we reject the welfare officials’ claim? t = -3.53 Which is outside the acceptance region of t = -2.262 and t = 2.262. We must reject the officials’ claim. HW P. 497 #1-8 You should get 4 two-tail tests and 4 one-tail tests.
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