MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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Consider the following questions about CIs.
A researcher tests emotional intelligence (EI) for a random sample of children selected from a population of all students who are enrolled in a school for gifted children. The researcher wants to estimate the
The sample mean M= 130
The sample standard deviation s = 15.
The
The df = N – 1 = 119
- What are the upper and lower limits of the CI and the width of the 95% CI if all the other values remain the same (M = 130, s =15) but you change the value of N to 16?
For N = 16, lower limit = 122.41 and upper limit=137.58 width (upper limit-lower limit = 15,17
- What are the upper and lowers of CI and the width of 95% CI if all the other values remain the same but you change the value of N to 25? . For N = 25, lower limit = and upper limit = width (upper limit-lower limit) =
- What are the upper and lower limits of CI and the width of the 95% CI if all the other values remain the same (M = 130, s = 15) but you change the value of N to 49? For N = 49, lower limit = and upper limit = . width (upper limit – lower limit) =
- Based on the numbers you reported for sample size N of 16, 25, and 49, how does the width of the CI change as N (the number of cases in the sample) increases?
- What are the upper and lower limits and the width of this Ci if you change the confidence level to 80% (and continue to use M = 130, s = 15, and N = 49)? For an 80% CI, lower limit = and upper limit = width (upper limit – lower limit) =
- What are the upper and lower limits and width of the CI if you change the confidence level to 99% (continue to use M= 130, s =15, and N =49)? For a 99% CI, lower limit = and upper limit = width (upper limit – lower limit) =
- How does increasing the level of confidence from 80% to 99% affect the width of the CI?
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- Consider the information above and in question 1. It is conjectured that the mean number of people attending all 2017 Veterans Day celebrations is 145, and of interest is to test this conjecture versus the alternative that the mean number of people attending all 2017 Veterans Day celebrations is different from 145 people To test the hypothesis, a simple random sample of 41 Veterans Day celebrations was selected and the number of people attending each was recorded. The mean number of people attending for this sample of 41 celebrations was 157, with a standard deviation of 23 people. The distribution of the data was bimodal. If appropriate, use this information to test the hypotheses stated in question 3 at the a= .05 level of significance. 4 Are the assumptions met? Yes, because there is a simple random sample Yes, because there is a simple random sample and the population is normally distributed Yes, because there is a simple random sample and the sample size is large enough for the…arrow_forwardExhibit 11-3 The contents of a sample of 26 cans of apple juice showed a standard deviation of 0.06 ounce. We are interested in testing to determine whether the variance of the population is significantly more than .003. Refer to Exhibit 11-3. The p-value for this test is a) .05 b) greater than .10 c) less than .10 d) 1.96arrow_forward2arrow_forward
- Please help me to explain this questions in steps. Thank you.arrow_forwardUse z scores to compare the given values. The tallest living man at one time had a height of 238 cm. The shortest living man at that time had a height of 142.4 cm. Heights of men at that time had a mean of 175.45 cm and a standard deviation of 5.59 cm. Which of these two men had the height that was more extreme? ... Since the z score for the tallest man is z = 0 and the z score for the shortest man is z = the man had the height that was Im- more extreme. (Round to two decimal places.) shortest tallestarrow_forwardConsider the information above and in question 1. It is conjectured that the mean number of people attending all 2017 Veterans Day celebrations is 145, and of interest is to test this conjecture versus the alternative that the mean number of people attending all 2017 Veterans Day celebrations is different from 145 people To test the hypothesis, a simple random sample of 41 Veterans Day celebrations was selected and the number of people attending each was recorded. The mean number of people attending for this sample of 41 celebrations was 157, with a standard deviation of 23 people. The distribution of the data was bimodal. If appropriate, use this information to test the hypotheses stated in question 3 at the a = .05 level of significance. O 7 What is the p-value? 0.05 p-value < 0.10 0.0005 0.03 0.001 < p-value < 0.002 0.0008 4-arrow_forward
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