Question 23In a survey sample of 1,000 respondents, the mean annual salary is $32,245, with a standard deviation of 4021.54.Questions (number as follows):1. Calculate and interpret 95% confidence interval (both upper and lower limit) for these data.2. Calculate and interpret 99% confidence interval (both upper and lower limit) for these data.3. Which confidence interval would be best to report and why?10 pts

The given values are:Sample size (n) = 1,000Sample mean (x̄) = $32,245Standard deviation (σ) =…

Answered: Question 23 In a survey sample of 1,000…

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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If n = 480 and X = 384, construct a 90% confidence interval for the population proportion, p.Give your answers to three decimals < p <
Next question In a random sample of eleven people, the mean driving distance to work was 25.7 miles and the standard deviation was 4.1 miles. Assume the population is normally distributed and use the t-distribution to find the margin of error and construct a 99% confidence interval for the population mean u. Interpret the results. + Identify the margin of error. (Round to one decimal place as needed.) Construct a 99% confidence interval for the population mean. (Round to one decimal place as needed.) Interpret the results. Select the correct choice below and fill in the answer box to complete your choice. (Type an integer or a decimal. Do not round.) O A. With % confidence, it can be said that the population mean driving distance to work (in miles) is between the interval's endpoints. O B. % of all random samples of eleven people from the population will have a mean driving distance to work (in miles) that is between the interval's endpoints. O C. It can be said that % of the…
8. Construct the confidence interval for the population variance for the given values. Round your answers to one decimal place. Answer: Lower endpoint: n = 24, s² = 3.9, and c = 0.99 Upper endpoint:
Question 1, 8.3.36
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suppose you obtained a mean of $167 spent on transportation and a standard deviation of $40. Calculate a 95% confidence interval for the mean and select the values that come closest to those that would fill the spaces in the following interpretation: we can be 95% confident that they mean amount of money spent on transportation lies between _____ and _____ .
A random sample of 82 women showed that the mean number of children reported was 1.87 with a standard deviation of 1.7 . (Interestingly, a sample of 82 men showed a mean of 1.3 children.) Complete parts (a) through (c) below. LOADING... Click the icon to view the t-table. Question content area bottom Part 1 a. Find a 95% confidence interval for the population mean number of children for women. Because the sample size is so large, you can use 1.96 for the critical value of t (which is the same as the critical value of z) if you do the calculations manually. Choose the correct answer below and, if necessary, fill in the answer boxes to complete your choice. (Type integers or decimals rounded to the nearest hundredth as needed. Use ascending order.)
Answer 4 , 5 and 6
If we want to provide a 95% confidence interval for the mean of a population, the Z value is, .95 1.645 .485 1.96
3. Given the following GPA for 4 students: 2, 3, 3, 4. Age of students are: 10, 11,12, 15. There are 10,000 total students. Average GPA: Unknown, StandardDeviation of GPA: 0.02 and it’s Normal.a. Find the mean and standard deviation for GPA.b. Calculate a 95% confidence interval for the populations mean GPA.c. Use Correlation formula to find r between GPA and Age.d. Find Regression Line Equation.e. If a student’s age is 13, what is the corresponding GPA?

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