Question 28In 2015, the average iceberg weighed 20 million tons. A random sample of 100 icebergs last year hada mean of 19 million tons with a standard deviation of 5.7 million tons. Write the hypotheses youwould use to test if the average weight of icebergs has decreased since 2015.© Ho H = 20, Hiu > 20O Ho: x=20, H₁x > 20O Ho: H20, H₁:μ< 20O Ho: x=20, H₁:x < 20O None of these.2

28) Given,In 2015, the average iceberg weighed, = 20 million tons.Sample size, n = 100Sample mean,…

Answered: Question 28 In 2015, the average…

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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Question 23 A test had 12 items. A new version of a test was made by adding a set of 12 items which resembles the original 12-item test. The new test having 24-items was administered a group of students. The mean and the standard deviation of the new 24-item test scores were 18 and 6, respectively. The difference score between the original 12-item test and the new set of 12 items was calculated. The standard deviation of the difference scores was 3. Estimate reliability.
Question 10 く Suppose that we are studying the amount of time customers wait in line at the checkout at the Gap and Old Navy. The avérage wait time at both stores is five minutes. At the Gap, the standard deviation for the wait time is 2 minutes; at Old Navy the standard deviation for the wait time is 5 minutes. Because Old Navy has a higher standard deviation, we know that there is Select an answer v in the wait times at Old Navy. Overall, wait times at Old Navy are Select an answer v from the average; wait times at the Gap are Select an answer v near the average.
Question 2 Select one answer. Facebook friends: According to Facebook's self-reported statistics, the average Facebook user has 130 Facebook friends. For a statistics project a student at Contra Costa College (CCC) 10 points tests the hypothesis that CCC students will average more than 130 Facebook friends. She randomly selects 3 classes from the schedule of classes and distributes a survey in these classes. Her sample contains 45 students. From her survey data she calculates that the mean number of Facebook friends for her sample is: = 138.7 with a standard deviation of: s = 79.3. She chooses a 5% level of significance. What can she conclude from her data? A. O Nothing. The conditions for use of a t-model are not met. She cannot trust that the p-value is accurate for this reason. B.O We cannot conclude that the average number of Facebook friends for CCC students is greater than 130. The sample mean of 138.7 is not significantly greater than 130. C.O Her data supports her claim. The…
question 16
QUESTION 8 You are interested in seeing whether emotions impact decision making. You have three groups--a happy group, a sad group, and a neutral group. For the happy group there are 5 participants and the mean risky decision making score 5.0 with a standard deviation of 0.7. For the sad group there are 5 participants and the mean risky decision making score 5.4 with a standard deviation of 1.1. For the neutral group there are 5 participants and the mean risky decision making score 5.8 with a standard deviation of 2.8. The sum of squares between samples is equal to 1.6. The sum of squares within samples is equal to 38.0. What is the degrees of freedom total? Please do the calculations on the handout provided.
question 18
Question 7 Use the following information to answer Question 7 and Question 8.The marks of six students in a Mathematics test are represented by the variable x. The following data was produced: n=6, Σx = 279 and Σx2= 13 093.Find the mean and standard deviation of the following data.A. mean =46.5 , standard deviation = 5.82B. mean = 52.5, standard deviation = 4.89C. mean =46.5 , standard deviation = 4.89D. mean =52.5 , standard deviation = 5.82 Question 8 It was later discovered that one of the values was incorrectly recorded as 43. The correct value, in fact, is 34. Find the new values of the mean and standard deviation.A. mean = 45 , standard deviation = 7.07B. mean = 51, standard deviation = 6.84C. mean = 45 , standard deviation = 6.84D. mean = 51 , standard deviation = 7.07
Part 1 of 4 Suppose that you randomly survey death records for people born in 1900 in Virginia and compared the life span of two different ethnicities. Of the 132 individuals sampled from ethnicity 1, the mean life span was 58.5 years with a standard deviation of 16.9 years. Of the 191 individuals sampled from ethnicity 2, the mean life span was 54.3 years with a standard deviation of 10.3 years. Conduct a hypothesis test with a 7% level of significance to see if the mean life spans in Virginia were the same for the two ethnicities. Step 1: State the null and alternative hypotheses. Ho: 12V✓0 Ha: 1₂V0 (So we will be performing a two-tailed Part 2 of 4 Step 2: Assuming the null hypothesis is true, determine the features of the distribution of the differences of sample means. The differences of sample means are t-distributed and distribution standard deviation Question Help: Post to forum C Submit Part test.) 06 ✓✓with distribution mean MacBook Pro Search or type URL ...........
Solve question #9 for part D only.

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