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 30 investors found that 18 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,.2 0.6< 0.6Ho: IH,: I(Type integers or decimals. Do not round.)Compute the test statistic.(Round to two decimal places as needed.)

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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For the given scenario, determine the type of error that was made, if any. (Hint: Begin by determining the null and alternative hypotheses.) Researchers state 2.5 hours as the mean amount of television watched by children per night. One parent claims that the mean amount of television watched by children per night is less than 2.5 hours. The parent conducts a hypothesis test and fails to reject the null hypothesis. Assume that in reality, the mean amount of television watched by children per night is 2.3 hours. Was an error made? If so, what type?
QUESTION 3 You may need to use the appropriate appendix table or technology to answer this question. Given that z is a standard normal random variable, compute the following probabilities. (Round your answers to four decimal places.) (a) P(z ≤ −3.0) (b) P(z ≥ −3) (c) P(z ≥ −1.5) (d) P(−2.6 ≤ z) (e) P(−2 < z ≤ 0)
Suppose that in a recent poll, 550 of the 915 randomly selected U.S. adults agreed with the statement “social media sites like Facebook and Twitter do more to divide the country” (as opposed to unite the country, or neither). a. Use this information to construct a 90% confidence interval.
1. In a study of perception, 110 men are tested and 24 are found to have red/green color blindness (based on data from USA Today). Assume that this was done in a simple random manner. A 95% confidence interval estimate of the proportion of all men with red/green color blindness was found to be 0.14 to 0.30. If we learn that the actual population proportion of men with red/green colorblindness is equal to 0.12, then what can we conclude? A. The confidence interval was wrong. B. We would need to see the sample data to make any conclusions Pick either answer A or B as your answer and show your work to go with it
5.A random sample of 362 married couples found that 286 had two or more personality preferences in common. In another random sample of 550 married couples, it was found that only 30 had no preferences in common. Let p1 be the population proportion of all married couples who have two or more personality preferences in common. Let p2 be the population proportion of all married couples who have no personality preferences in common. Find a 90% confidence interval for p1 – p2. (Use 3 decimal places.) lower limit upper limit
8. A sample of 200 light bulbs was tested and it was found that 11 were defective. Calculate a 95% confidence interval around this sample proportion. a. 0.055 + 1.96 b. 0.055 +0.016 c. 0.055 + 0.032 d. 11 1.96 e. 200 + 11
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?
A 95% confidence interval for how much time on average GRC students spend doing homework every day is 75 to 105 minutes.Choose a correct interpretation of the statement above. A. If random sampling is repeated, 95% of the confidence intervals created will capture the true mean. B. If random sampling is repeated, 95% of the samples will have a mean that is between 75 and 105 minutes. C. 95% of GRC students study between 75 and 105 minutes on average every day. D. There is 95% chance that the true mean is between 75 and 105 minutes.
1. Suppose a random sample of 32 National Football League players had an average weight of 252.5 pounds. Assume that the standard deviation for weight of all players in the league is 31 pounds. Determine the margin of error for a 99% confidence interval for this sample.

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