**Question 2 of 20 Step 1 of 3**A children's clothing company sells hand-smocked dresses for girls. The length of one particular size of dress is designed to be 29 inches. The company regularly tests the lengths of the garments to ensure quality control, and if the mean length is found to be significantly longer or shorter than 29 inches, the machines must be adjusted. The most recent simple random sample of 18 dresses had a mean length of 33.46 inches with a standard deviation of 5.89 inches. Assume that the population distribution is approximately normal. Perform a hypothesis test on the accuracy of the machines at the 0.025 level of significance.**Step 1 of 3:** State the null and alternative hypotheses for the test. Fill in the blank below.\( H_0: \mu = 29 \)\( H_a: \mu \) _____ 29**Options:**- \( > \)- \( < \)- \( \neq \)*Answer: 2 Points*- Tables- Keypad© 2021 Hawkes Learning---In this step, you need to define the alternative hypothesis (\( H_a \)) for the statistical test, indicating whether the mean dress length is different from the target value of 29 inches. The options include a greater than, less than, or not equal to sign.

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Answered: Question 2 - of 20 Step 1 of 3 01:34:36…

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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SCENARIO 10-15 The table below presents the summary statistics for the starting annual salaries (in thousands of dollars) for individuals entering the public accounting and financial planning professions. Sample I (public accounting): 1= 60.35, S1 = 3.25, n1 = 12 Sample II (financial planning): x2= 58.20, S2 = 2.48, n2 = 14 Test whether the mean starting annual salaries for individuals entering the public accounting professions is higher than that of financial planning assuming that the two population variances are the same. Referring to Scenario 10-15, suppose a = 0.10. Which of the following represents the correct conclusion?
Chapter 5, Section 2, Exercise 043 Hearing Loss in TeenagersA recent study1 found that, of the 1771 participants aged 12 to 19 in the National Health and Nutrition Examination Survey, 19.5% had some hearing loss (defined as a loss of 15 decibels in at least one ear). This is a dramatic increase from a decade ago. The sample size is large enough to use the normal distribution, and a bootstrap distribution shows that the standard error for the proportion is SE=0.009.Find a 95% confidence interval for the proportion of teenagers with some hearing loss.Round your answers to two decimal places.The 95% confidence interval is to .
Question 3 A fruit farmer has a large number of almond trees, all of the same variety and the same age. One year he wishes to estimate the mean yield of his trees. He collects all the almonds from 8 trees and records the following weights (kg); 36 53 78 67 92 77 59 66 a) Use these data to estimate the mean and standard deviation of the yields of all the farmer's trees. b) Construct a 95% confidence interval for the mean yield. c) What statistical assumption is required for your procedure to be valid? d) How might you select a sample of eight trees from those growing in a large field. Criteria [2.1, 2.2, 2.3]
QUESTION 1 A health researcher read that a 150-pound college-aged, male can burn an average of 530 calories per hour playing tennis with a standard deviation of 25 calories. 40 college-aged males were randomly selected at a commuter school and the mean number of calories burned per hour playing tennis was 540 calories. Do college-aged male tennis players at the commuter school burn more calories per hour than the college-aged, male tennis player population? Test with a 0.05 significance level. What are the hypotheses? Null and Alternate
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Question 3 A. The management of White Industries is considering a new method of assembling its golf cart. The present method requires a mean time of 42.3 minutes to assemble a cart. The mean assembly time for a random sample of 24 carts, using the new method, was 40.6 minutes, and the standard deviation of the sample was 2.7 minutes. Using the .10 level of significance, can we conclude that the assembly time using the new method is faster? B. Do married and unmarried women spend the same amount of time per week using Facebook? A random sample of 45 married women who use Facebook spent an average of 3.0 hours per week on this social media website. A random sample of 39 unmarried women who regularly use Facebook spent an average of 3.4 hours per week. Assume that the weekly Facebook time for married women has a population standard deviation of 1.2 hours, and the population standard deviation for unmarried Facebook users is 1.1 hours per week. Using the .05…

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