The formula used to calculate a confidence interval for the mean of a normal population when n is small is the following.S√nWhat is the appropriate t critical value for each of the following confidence levels and sample sizes? (Round your answers to two decimal places.)x + (t critical value)USE SALT(a) 98% confidence, n = 17(b) 90% confidence, n = 11(c) 99% confidence, n = 24(d) 90% confidence, n = 25(e) 80% confidence, n = 13(f) 95% confidence, n = 10

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A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Use technology to construct the confidence intervals for the population variance o2 and the population standard deviation o. Assume the sample is taken from a normally distributed population. c = 0.95, s2 = 15.21, n= 30 The confidence interval for the population variance is (Round to two decimal places as needed.) The confidence interval for the population standard deviation is (Round to two decimal places as needed.)
The Z-score tells you the number of standard deviations away from the mean (and X-μ can be used to find the in what direction) a data value is. The formula: Z Z-score for a single member of the population. Notice X - μ tells you the signed distance the data value (X) is from the mean. When you divide that by the size of each chunk (the standard deviation) you are measuring the distance in units of the size of the standard deviations (how many standard deviations fit in the distance between the data value and the mean) - which gives you the Z-score. σ The formula X = μ+Z. can be used to find the value in the population when given the Z-score (signed number of standard deviations). Notice that Z. tells you how far from the mean the data value is (since Z is the number of standard deviations and is the size of each standard deviation the product tells the total distance). So adding the distance from the mean to the mean gives the location along the number line for the data value. A…
the distribution of scores on a standardized test is approximately Normal with standard deviation of σ = 21. If a sample of 135 students had a mean score of x¯ = 73.6, use this to find a 99% confidence interval for the mean score of all students on this test. We are 99% confident that the the mean score of students on this test was in the interval ______ to _____. What is the margin of error for your confidence interval? Margin of error =
Find the 95% confidence interval for the population mean if the sample is × = 70.4 and the sample size is n = 24. Assume that the population is normally distributed and the sample standard deviation is 0 = 5. Left Left: Right:
Beetles' weight has a normal distribution with mean of 85g and standard deviation of 6.4g. Suppose we repeatedly took random samples of 12 of these beetles. The percentage of samples that have a sample mean weight less than 83.2g is closest to: (You should use the empirical rules to estimate this percentage) A: 2.5% B: 85% C: 100% D: 15%
A large automobile insurance company selected samples of single and married male policyholders and recorded the number who made an insurance claim over the preceding three-year period. Single Policyholders Married Policyholders n1=550 n2=1,000 Number making claims = 110 Number making claims = 150 a. Use a=0.05 Test to determine whether the claim rates differ between single and married male policyholders. z-value (to 2 decimals) p-value (to 4 decimals) b. Provide a 95% confidence interval (to 4 decimals) for the difference between the proportions for the two populations. Enter negative answer as negative number. ( , )
A distribution of scores is approximately normal with a mean of 78 and a standard deviation of 8.6. Which of the following equations can be used to find the score x above which 33 percent of the scores fall? (A) 0.44 = x – 78 (8.6)? (B) 0.67 x – 78 (8.6)? x – 78 8.6 (C) 0.33 x – 78 (D) 0.44 = 8.6 x – 78 8.6 (E) 0.67 %3D
What sample size is needed to give a margin of error within +3% in estimating a population proportion with 90% confidence? An initial small sample has p = 0.77. %3D Round your answer up to the nearest integer. Sample size =
The formula used to calculate a confidence interval for the mean of a normal population when n is small is the following. xt (t critical value) √n What is the appropriate t critical value for each of the following confidence levels and sample sizes? (Round your answers to two decimal places.) USE SALT (a) 98% confidence, n = 17 (b) 90% confidence, n = 12 (c) 98% confidence, n = 24 (d) 90% confidence, n = 25 (e) 80% confidence, n = 13 (f) 95% confidence, n = 10
Use the t-table or any statistical tool and report the t-score that you multiple by the standard error to form the margin of error for a 95% confidence interval for a mean with 5 observations. a. 2.132 b. 2.776 c. 2.015 d. 2.571
The formula used to calculate a confidence interval for the mean of a normal population when n is small is the following. x ± (t critical value) s n What is the appropriate t critical value for each of the following confidence levels and sample sizes? (Round your answers to two decimal places.) d. 90% confidence, n = 23 e. 95% confidence, n = 13 f. 95% confidence, n = 10
You want to obtain a sample to estimate a population mean. Based on previous evidence, you believe the population standard deviation is approximately σ=49.8σ=49.8. You would like to be 90% confident that your estimate is within 5 of the true population mean. How large of a sample size is required?n = Round up to the next whole number.

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