Suppose that the proportion of investors who are risk-averse (that is, try to avoid risk in their investment decisions) is though to be at least 0.6. A financial advisorbelieves that the proportion is actually less than 0.6. A survey of 31 investors found that 19 of them were risk-averse. Formulate a one-sample hypothesis test for aproportion to test this belief.Determine the null hypothesis, H,, and the alternative hypothesis, H,.H,:(Type integers or decimals. Do not round.)

Answered: Image /qna-images/answer/9a058d6d-448b-4bb9-bf68-cbb5d31030ec.jpg

Answered: Suppose that the proportion of…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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4. A poll was conducted in a very small community in country X to see if residents are for or against a new policy that requires them to wear a mask. A total of 500 persons were surveyed. Of the 500 surveyed 200 are against wearing a mask. Compute a 90% confidence interval for the true percent of residents who are against the new policy, and interpret the confidence interval.
4. Ten students on a low-fat diet designed to lower their cholesterol and recorded their before and after diet data in the following table: Before diet cholesterol level: After diet cholesterol level: 140 140 220 230 110 120 240 220 200 190 180 150 190 200 360 300 280 300 260 240 Based on a 95% confidence level, after hypothesis testing, what would you tell the students about the diet? Does the null hypothesis reject H0 or fail to reject H0?
In a poll of 7 randomly selected students showed that 31.3 % of the respondents believe in global warming. Based on the sample results, the margin of error E that corresponds to a 95% confidence level is equal to: O a. 0.45 O b. 0.34 O c. 0.29 O d. 0.22
In September 2011, Gallup surveyed 1,004 American adults and asked them whether they blamed Barack Obama a great deal, a moderate amount, not much, or not at all for U.S. economic problems. The results showed that 53% of respondents blamed Barack Obama a great deal or a moderate amount. Calculate the 99% confidence interval for the proportion of all American adults who blame Barack Obama a great deal or a moderate amount for U.S. economic problems. O (0.504, 0.556) O (0.499, 0.561) O (0.489, 0.571) O The confidence interval cannot be determined from the information given.
3. Stainless steels can be susceptible to stress corrosion cracking under certain conditions. A materials engineer is interested in determining the proportion of steel alloy failures that are due to stress corrosion cracking. a. In the absence of preliminary data, how large a sample must be taken so as to be sure that a 98% confidence interval will specify the proportion to within ±0.05? Round up the answer to the nearest integer. b. In a sample of 200 failures, 30 of them were caused by stress corrosion cracking. Find a 98% confidence interval for the proportion of failures caused by stress corrosion cracking. Round the answers to three decimal places. c. Based on the data in part (b), estimate the sample size needed so that the 95% confidence interval will specify the proportion to within ±0.03. Round up the answer to the nearest integer.
7. A company that has many department stores in the southern states wanted to find at two such stores the percentage of sales for which at least one of the items was returned. A sample of 800 sales randomly selected from Store A showed that for 280 of them at least one item was returned. Another sample of 900 sales randomly selected from Store B showed that for 279 of them at least one item was returned. a. Construct a 98% confidence interval for the difference between the proportions of all sales at the two stores for which at least one item is returned. b. Using the 10% significance level, can you conclude that the proportions of all sales for which at least one item is returned is higher for Store A than for Store B?
2. You are going to construct a 90% confidence interval for the proportion ofadults that have a post-secondary education. You want the margin of errorto be .03. Your initial guess is that 35% of the population has a postsecondary education. What should the sample size be to achieve thismargin of error?
Suppose that the proportion of investors who are risk-averse (that is, try to avoid risk in their investment decisions) is though to be at least 0.6. A financial advisor believes that the proportion is actually less than 0.6. A survey of 30 investors found that 18 of them were risk-averse. Formulate a one-sample hypothesis test for a proportion to test this belief. ..... Determine the null hypothesis, H,, and the alternative hypothesis, H,. 2 0.6 < 0.6 Ho: I H,: I (Type integers or decimals. Do not round.) Compute the test statistic. (Round to two decimal places as needed.)
3. Consider a sample set of (96; 83; 79; 48; 53; 63; 102; 96; 87; 114; 103; 54; 81; 55; 69]. Compute the 95% confidence interval (see Example 8.8) for the mean of the population useing the Student's T-test
1.A poll questioned 800 students about their views on pizza for lunch at school. The results indicated that 65% of respondents felt that pizza was a must for lunch at school and would quit school if there was no pizza at lunch. a) Determine the 90% confidence interval. b) What is the margin of error for this response at the 90% confidence level?
A study regarding the average cost of online streaming services in two states (A and B) is conducted. The results are shown in the following table. X and Y are independent.
2. One of the restaurants participating in Love Week wants to know proportion of Allen residents who plan on coming to their restaurant on their Love Week Spirit Night. If they plan on creating a 96% confidence interval and they want the margin of error to be no more than 3.8%, how large of a sample do they need to get? the lA lo nolh

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