3. (Chihara and Hestelberg Exercise 7.6.4) In R, the qt command computes quantiles of the t (a) F...
Please show all work!!! 11.* A random sample of size n 64 is drawn from a population with mean μ and standard deviation σ. The mean and standard deviation of the sample are X = 308.9 and s 31.9 a. Find a 90%confidence interval for the mean μ. Interpret this interval. b. Find a 95%confidence interval for the mean μ. Interpret this interval. c. Find a 99%confidence interval for the mean μ. Interpret this interval. d. Compare the widths of...
Hey the correct answer is given for both, but can you please do the steps on how to get to the answer, Im confused on how to solve? Thank you! 3. At least what sample size is necessary in order to obtain a 90% confidence interval of length 6, when SD = 10? A. 8 B. 16 C. 60 D. 15 E. 61 Answer: E 4. With which combination of confidence level and sample size will the corresponding confidence interval...
When is unknown and the sample is of size n 230, there are two methods for computing confidence intervals for u. (Notice that, When is unknown and the sample is of size n<30, there is only one method for constructing a confidence interval for the mean by using the student's t distribution with d.f. = n - 1.) Method 1: Use the Student's t distribution with d.f. = n - 1. This is the method used in the text. It...
For the following sample sizes and confidence levels, find the t-values suitable for building confidence intervals: a) n = 15; 90%. b) n = 6; 95%. c) n = 19; 99%. d) n = 25; 98%. e) n = 10; 99%. f) n = 41; 90%.
6. + -16.66 points MIntroStat97E004.Tut What critical value t* from Table D should be used to construct the following? (Round your answers to three decimal places. (a) a 95% confidence interval when n-29 (b) a 99% confidence interval when n 17 (c) a 90% confidence interval when n-61
A 90 % confidence interval (a t interval) for the mean lives (in minutes) of Kodak AA batteries is ( 440, 480 ). Assume that this result is based on a sample of size 15 . 1) What is the value of the sample standard deviation? a) 43.9784 b) 57.7796 c) 57.5895 d) 44.1856 2) Construct the 99% confidence interval. a) (426.1974,493.8026) b) (430.1984,489.8016) c) (443.2129,476.7871) d) (444.7772,475.2228) 3) If the confidence interval (442.7050 ,477.2950) is obtained from the same...
a. Find a 95% confidence interval for μ. b. What do you mean when you say that a confidence coefficient is .90? c. Find a 99% confidence interval for μ. d. What happens to the width of a confidence interval as the value of the confidence coefficient is increased while the sample size is held fixed? e. Would your confidence intervals of parts a and c be valid if the distribution of the original population were not normal? Explain. A...
When σ is unknown and the sample is of size n ≥ 30, there are two methods for computing confidence intervals for μ. Method 1: Use the Student's t distribution with d.f. = n − 1. This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method. Method 2: When n ≥ 30, use the sample standard deviation s as an estimate for σ, and then use the...
*** round 3 decimal places as needed **** IGNORE FIRST PICTURE. need anwsers for n=144 n=576 The mean and standard deviation of a random sample of n measurements are equal to 34.4 and 3.8, respectively. a. Find a 99% confidence interval for win81. b. Find a 99% confidence interval for if n = 324 c. Find the widths of the confidence intervals found in parts a and b. What is the effect on the width of a confidence interval of...
Please generalize ci95.t to ci.t so that we can calculate confidence interval at any confidence level given a sample using the KORGER.csv data The function prototype looks like, ci.t <- function(n, mu, sd_mu, conf) { ##your code goes here } The sample outputs need to be replicated are, setwd("…") kg <- read.csv("KORGER.csv") kd <- split(kg$Distance, kg$Team)$KOREA n <- length(kd); m <- mean(kd); s <- sd(kd)/sqrt(n); >ci.t(n, m, s, 0.95) 95%CILower 95%CIUpper 8448.693 11307.852 >ci.t(n,m,s,0.90) 90%CILower 90%CIUpper 8715.393 11041.152 >ci.t(n,m,s,0.99)...