6c 1 2 3 6. A sample size of n 11, X= 40.1 with sample standard deviation o 1.6 at 99% confidence level7. A sample size of n 121, X 825 with sample standard deviation o=116 at 98% confidence leved8. A sample size of n 20, X 101 with sample standard deviation o=26 at 98% confidence leval%3Dob of nC. Solve each given problem below. Show your solution and state the interpretation for your answer.babnrRSere obtained from a random sample of 151. A mean of 115.4 and a standard deviation 10.8observations, which is believed to be normally distributed. Find the confidence interval forthe population mean p using the given confidence level.A. 90%d sanmbiteoВ. 95%C. 99%2. The following data are selected randomly from the population of normally distributed values.23 2832 36293119 2233384129223731 39Construct a 95% confidence interval to estimate the population mean.3. A popular restaurant takes a random sample of 25 customers and records how long eachcustomer occupied a table. They found a sample mean of 1.2 hours and a sample standarddeviation of 0.3 hour. Construct a 95% confidence interval estimate of the mean of thepopulation.4. The mean prize of 25 computer desktop for gaming online is 55,450deviation of 3,576 pesos. Find the 99% confidence interval of the true mean price of computerdesktop.pesoswith a standard5. The average profit earned by an online seller per day with 100 followers is 3,865witha standard deviation of 890 pesos. Find the 95% confidence interval of the true mean prontpesosearn per day.

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Answered: nins 8S 1, A mean of 115.4 and a…

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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Assume we wanted to show that the true proportion of virginica iris seeds that fail to germinate is less than 20%. Let's say we tested at an alpha of 0.05, and found a p-vale of 0.082. Which of the following would be an appropriate conclusion and interpretation? A) With an alpha of 0.05, and a p-value of 0.082, we fail to reject the null and state we have insufficient evidence to support that the true proportion of virginica iris seeds that fail to germinate is less than 20%. B) With an alpha of 0.05, and a p-value of 0.0082, we reject the null and state we have sufficient evidence to support that the true proportion of virginica iris seeds that fail to germinate is less than 20%. C) With an alpha of 0.05, and a p-value of 0.082, we reject the null and state we have sufficient evidence to support that the true proportion of virginica iris seeds that fail to germinate is less than 20%. D) With an alpha of 0.05, and a p-value of 0.082, we fail to reject the null and state…
18. Solve the following problems.a. After 6 points have been added to every score ina sample, the mean is found to be M 5 70 and the standard deviation is s 5 13. What were the values for the mean and standard deviation for the original sample?b. After every score in a sample is multiplied by 3, the mean is found to be M 5 48 and the standard devia- tion is s 5 18. What were the values for the mean and standard deviation for the original sample?
Based on a random sample of size 50, you observe a sample mean of 143.6 and a sample standard deviation of 38.1. You then calculate a 95% confidence interval for the population mean, and it has length L1. Now suppose that the sample size had instead been 150, but with the same sample mean and the same sample standard deviation. Let L2 be the length of a 95% confidence interval for the population mean based on this larger sample. Which of the following is closest to the ratio L2/L1? a. 0.78 b. 0.57 c. 0.86 d. 0.65
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The desired percentage of SiO2 in a certain type of aluminous cement is 5.5. To test whether the true average percentage is 5.5 for a particular production facility, 16 independently obtained samples are analyzed. Suppose that the percentage of SiO2 in a sample is normally distributed with ? = 0.32 and that x = 5.21. (Use ? = 0.05.) (a) Does this indicate conclusively that the true average percentage differs from 5.5?State the appropriate null and alternative hypotheses. H0: ? = 5.5Ha: ? ≠ 5.5H0: ? = 5.5Ha: ? ≥ 5.5 H0: ? = 5.5Ha: ? < 5.5H0: ? = 5.5Ha: ? > 5.5 Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimal places.) z = P-value = State the conclusion in the problem context. Do not reject the null hypothesis. There is sufficient evidence to conclude that the true average percentage differs from the desired percentage.Reject the null hypothesis. There is sufficient evidence…
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nins 8S 1, A mean of 115.4 and a standard deviation 10.8 were obtained from a random sample of 15 observations, which is believed to be normally distributed. Find the confidence interval for the population mean p using the given confidence level. bcuq А. 90% В. 95% С. 99% el sambiteo