A sample of size n = 39 has sample mean x = 54 and sample standard deviation s = 6.3.Part: 0/2Part 1 of 2(a) Construct an 80% confidence interval for the population mean μ. Enter the values for the lower and upper limits and the mean to graph. Round theanswers to one decimal place.45.0HHHH45.080% confidence interval54.564.064.0

Given:Sample Size Sample Mean Sample SD

Answered: A sample of size n=39 has sample mean x…

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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Find the 95% confidence interval for estimating the population mean μ If sample mean X = 50, sample size n= 60 and population standard deviation σ is known to be 10.
The weights of a certain type of steer follow a Normal model with a mean weight of 1191 pounds and a standard deviation of 54 pounds. (Use 4 decimals for all answers below.)(a) What weight is required for a steer to be among the heaviest 10% of steers? (b) What weight is required for a steer to be among the lightest 25% of steers? (c) What is the IQR of these weights?
Part 1 of 3 A researcher was investigating the size of math departments in Texas colleges. She believed the population standard deviation was 1.9. A sample of 42 colleges found their mean was 12.5. Find the 3-decimal small value of the 95% confidence interval of the mean. Part 2 of 3: Find the 3-decimal large value of the 95% confidence interval of the mean. Part 3 of 3: Find the 3-decimal margin of error
Find the critical value x²R corresponding to a sample size of 19 and a confidence level of 99 percent. X²R = (Round to 3 decimals)
Develop a 95% confidence interval for the population slope if the followingregression information are given: b1= 41.7, p-value = 0.02, and n = 35.
A sample of n=41n=41 data values randomly selected from a normally distributed population has variance s2=30s2=30. Construct a 95% confidence interval for the population variance. Round your endpoints to one decimal place.
Step 3 The necessary value for Za/2 was determined to be 1.645. Recall the given information. Sample 1 n1 = 400 P1 = 0.56 P2 = 0.41 Substitute these values to first find the lower bound for the confidence interval, rounding the result to four decimal places. P₁(1-P₁) P₂(1-P₂) 1 72 lower bound = P₁-P₂-²a/2 = 0.560.41 Sample 2 n2 = 300 = 0.0559 = 0.56 0.41 = 02441 X Substitute these values to find the upper bound for the confidence interval, rounding the result to four decimal places. P₁(1-P₁) P₂(1-P₂) upper bound = P₁ P₂ + ²a/2√ n1 n2 X + ✔ - 1.645, + 0.56(1 0.56) 0.41(10.41) 300 +1.645. 400 0.56(1 0.56) + 400 + 0.41(1 0.41) 300
find out the confidence interval for population proportion for given data [ 4 decimal places ] population standard deviation =4 xbar = 123.5 sample size= 100 c= 0.95
6.2.13 Use the given confidence interval to find the margin of error and the sample mean. (14.2,24.0) The sample mean is (Type an integer or a decimal.)
Energy drinks: A survey of college students reported that in a sample of 402 male college students, the average number of energy drinks consumed per month was 2.41 with a standard deviation of 3.18, and in a sample of 390 female college students, the average was 1.50 with a standard deviation of 4.77. Part: 0/2 Part 1 of 2 Construct a 80% confidence interval for the difference between men and women in the mean number of energy drinks consumed. Let μ, denote the mean number of energy drinks consumed by men. Use tables to find the critical value and round the answers to at least two decimal places. A 80% confidence interval for the difference between men and women in the mean number of energy drinks consumed is 0<44-₂2 <0. X S Part: 1/2 Part 2 of 2 Based on the confidence interval, is it reasonable to believe that the mean number of energy drinks consumed may be the same for both male and female college students? It (Choose one) reasonable to believe that the mean number of energy drinks…
Paramesan cheese is aged for an average of 24 months to allow it to develop the desire richness in flavor the standard deviation for the length of time Parmesan cheese is aged is 3.7 months How long would Parmesan cheese have to be aged in order to be in the 90th percentile? Assume that the data is normally distributed

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