1. A random sample of size 15 is selected from a normal population. The population standard deviation is unknown. Assume the null hypothesis indicates a two-tailed test and the researcher decided to use the 0.10 significance level. For what values of t will the null hypothesis not be rejected?
For a chi-square test involving a contingency table, suppose the null hypothesis is rejected. We conclude that the two variables are __________. a) related b) curvilinear c) linear d) not related 4. A sales manager for an advertising agency believes that there is a relationship between the number of contacts that a salesperson makes and the amount of sales dollars earned. What is the dependent variable?
5. A test statistic is a value determined from sample information used to reject or not reject the null hypothesis. 6. The region or area of rejection defines the location of all those values that are so large or so small that the probability of their occurrence under a true null hypothesis is rather remote. 7.
D. Do not reject the null hypothesis if the z test statistics is 1.96 or greater. 3) The area of rejection, on a bell shaped curve, defines the location of all those values that are A. so small or so large that the probability of their occurrence under a true null hypothesis is rather slim B. so small or so large that the probability of their occurrence under a true null hypothesis is to be expected C. so small or so large that the probability of their occurrence under a false null hypothesis is rather remote D. within the selected confidence interval for the test 4) The Roman Senate has become concerned about the loyalty of the army in Gaul commanded by Julius Caesar. They claim that, of the 80,000 men in the army, at least 28,000 are foreign
Alternatively, we reject the null hypothesis, if the p value is less than the significance level Substituting the value we get t = 43.74-5014.6396/50 = -3.02 The p value corresponding to t = -3.02 and 49 d.f. is 0.002 which is smaller than the significance level. The value of the test statistic is in the critical region and hence it is significant. Therefore, we reject the null hypothesis at 5% level of significance. The p-value is 0.002 which is smaller than the significance level.
So, 2000 = 30000/Square root of sample size. Solving for the Square root of sample size, we get Square root of sample size = 30000/2000 = 15. Taking its square, the sample size is found as 225. Chapter 9 Exercise 1 No it is not a good defense. If you choose 40 random employees from the corporation, the standard error would equal 6/Square root of 40 = .95 days.
16 – 56 Ethics and Standard Costs Farmer Frank’s produces items from local farm products and distributes them to supermarkets. Over the years, price competition has become increasingly important, so Susan Kramer, the company’s controller, is planning to implement a standard cost system for Farmer Frank’s. She asked her cost accountant, Margaret Chang, to gather cost information on the production of blueberry preserves (Farmer Frank’s most popular product). Margaret reported that blueberries cost $.75 per quart, the price she intends to pay to her good friend who has been operating a blueberry farm that has been unprofitable for the last few years. Because of an oversupply in the market, the price for blueberries has dropped to $.60 per quart.
A very large majority of banana varieties are not really able to grow for international trade, according to the article, their skin is too thin, or their pulp is very bland. Big banana companies like Dole and Chiquita are working on developing a replacement variety of the Cavendish, which has become the mass-market banana of choice for farmers and distributors because it provides a lot of fruit. Although Cavendishes need a lot of caring to grow, they are the only variety that provides farmer with a large amount of palatable fruit that can endure overseas trips without ripening or bruising too easily. The Cavendish is rich in Vitamins B6 and C, it contains a lot of potassium, magnesium, and fiber in it. It is also not very expensive, that is almost 60 cents per pound.
TOPIC 8 Chi-Square goodness-of-fit test Problem 12.1 Use a chi-square goodness-of-fit to determine whether the observed frequencies are distributed the same as the expected frequencies (α = .05) Category | fo | fe | 1 | 53 | 68 | 2 | 37 | 42 | 3 | 32 | 33 | 4 | 28 | 22 | 5 | 18 | 10 | 6 | 15 | 8 | Step 1 Ho: The observed frequencies are distributed the same as the expected frequencies Ha: The observed frequencies are not distributed the same as the expected frequencies Step 2 df = k – m – 1 Step 3 α = 0.05 x 2 0.05, 5df = 11.0705 Step 4 Reject Ho if x 2 > 11.0705 Category | fo | fe | | 1 | 53 | 68 | | 2 | 37 | 42 | | 3 | 32 | 33 | | 4 | 28 | 22 | | 5 | 18 | 10 | | 6 | 15 | 8
Answer | | | | | Selected Answer: | False | Correct Answer: | False | | | | | Question 6 2 out of 2 points | | | If we are solving a 0-1 integer programming problem, the constraint x1 ≤ x2 is a conditional constraint. Answer | | | | | Selected Answer: | True | Correct Answer: | True | | | | | Question 7 0 out of 2 points | | | If we are solving a 0-1 integer programming problem, the constraint x1 ≤ x2 is a __________ constraint.Answer | | | | |