The statistic was calculated for the variables above and was reported as .93. Using a = .05 (for atwo-tailed test), should the null hypothesis be rejected?YesNoCannot be determined from the information given. The following data applies to the next 3 problems. A researcher was interested in examining if arelationship existed between two variables, X and Y. Five people were selected, and their scores onX and Y are shown below.X11.0010.008.007.0014.00Y21.0017.0017.0012.0023.00

The statistic was calculated for the variables was reported as 0.93.

Answered: The statistic was calculated for the…

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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In general, men have greater income than women. So, a researcher hypothesizes that men’s occupational prestige scores on average should be significantly higher than women’s. The following table shows the OPS by gender. The researcher wants to test the hypothesis at the .05 level. The estimated standard error of the difference between means is 1.0. Is it going to be a one-tailed test, or a two-tailed test? A. One-tailed test B. Two-tailed test
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The data in the accompanying table give the weights (in g) of randomly selected quarters that were minted after 1964. The quarters are supposed to have a median weight of 5.670 g. Use the sign test and a 0.01 significance level to test the claim that the median is equal to 5.670 g. Do quarters appear to be minted according to specifications? Click the icon to view the data. What are the null and alternative hypotheses? O A. Ho: Median weight = 5.670 g H₁: Median weight 5.670 g More Info OB. Ho: Median weight ≤ 5.670 g H₁: Median weight = 5.670 g O D. Ho: Median weight = 5.670 g H₁: Median weight # 5.670 g 0 Post-1964 Quarters 5.5863 5.5764 5.6582 5.5649 5.5844 5.7402 5.7367 5.6057 5.5995 5.6165 5.6448 5.5974 5.6609 5.6363 5.6352 5.5986 5.5797 5.5501 5.6238 5.6462 5.6423 5.7021 5.5481 5.7078 5.6055 5.5324 5.7417 5.6776 5.5924 5.5691 5.5666 5.7287 5.6312 5.6017 5.6626 5.6436 5.6634 5.6132 5.5441 5.6363 -
In creating the null hypothesis for a hypothesis test for the standard deviation, it is wrong to write s = 30. Why is this so? ○ It just doesn't look and feel right. O Because my professor said so. O It should be the inequality sign, and not the equality sign. O Because hypothesis statements are about the popula- tion and not about the sample.
A Shapiro-Wilk result had a p-value of 0.05. Which of the following statements is best? O O O There is a 5% chance that the data is truly different from the normal curve There is a 5% chance that the data presented is only due to coincidence There is a 95% chance that the data is truly different from the normal curve There is a 95% chance that the data presented is only due to coincidence 7
Q5. With 0.05 significance level, H1: p > 0.5, and the value of test statistic is z = 1.93. Then theP-value is equal to 0.9732.a. Trueb. False
The mean price of mid-sized cars in a region is $57,600. A test is conducted to see if the claim is true. Identify the null hypothesis. ? v? v Identify the alternative hypothesis. ? v:?v| ? v Identify the type I error. O You decide that the mean price of mid-sized cars is not $57,600, when the mean price of mid- sized cars actually is $57,600. O You decide that the mean price of mid-sized cars is $57,600, when the mean price of mid- sized cars actually is not $57,600. O You decide that the mean price of mid-sized cars is not $57,600, when the mean price of mid- sized cars actually is nor $57,600. O You decide that the mean price of mid-sized cars is $57,600, when the mean price of mid- sized cars actually is $57,600. Identify the type II error. O You decide that the mean price of mid-sized cars is not $57,600, when the mean price of mid- sized cars actually is $57,600. O You decide that the mean price of mid-sized cars is $57,600, when the mean price of mid- sized cars actually is not…
A golf analyst claims that the standard deviation of the 18-hole scores for a golfer is at most 3.1 strokes. State Ho and H, in words and in symbols. Then determine whether the hypothesis test for this claim is left-tailed, right-tailed, or two-tailed. Explain your reasoning. ..... State the null hypothesis in words and in symbols. Choose the correct answer below. A. The null hypothesis expressed in words is, "the standard deviation of the 18-hole scores for a golfer is not 3.1 strokes." The null hypothesis is expressed symbolically as, "Ho: σ7 3.1." B. The null hypothesis expressed in words is, "the standard deviation of the 18-hole scores for a golfer is at least 3.1 strokes." The null hypothesis is expressed symbolically as, "Ho: o2 3.1." C. The null hypothesis expressed in words is, "the standard deviation of the 18-hole scores for a golfer is at most 3.1 strokes." The null hypothesis is expressed symbolically as, "Ho: os3.1." D. The null hypothesis expressed in words is, "the…
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A golf analyst claims that the standard deviation of the 18-hole scores for a golfer is less than 1.5 strokes. State H, and H, in words and in symbols. Then determine whether the hypothesis test for this claim is left-tailed, right-tailed, or two-tailed. Explain your reasoning. State the null hypothesis in words and in symbols. Choose the correct answer below. O A. The null hypothesis expressed in words is, "the standard deviation of the 18-hole scores for a golfer is at least 1.5 strokes." The null hypothesis is expressed symbolically as, "Ho: 02 1.5." O B. The null hypothesis expressed in words is, "the standard deviation of the 18-hole scores for a golfer is at most 1.5 strokes." The null hypothesis is expressed symbolically as, "Ho:os1.5." O C. The null hypothesis expressed in words is, "the standard deviation of the 18-hole scores for a golfer is less than 1.5 strokes." The null hypothesis is expressed symbolically as, "Ho: o1.5." The hypothesis test is because the V contains

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The statistic was calculated for the variables above and was reported as .93. Using a = two-tailed test), should the null hypothesis be rejected? Yes No Cannot be determined from the information given. .05 (for a