KIn a random sample of 36 criminals convicted of a certain crime, it was determined that the mean length of sentencing was 63 months, with a standard deviation of 12 months. Construct and interpret a 95% confidenceinterval for the mean length of sentencing for this crime.Click here to view the standard normal distribution table (page 1).Click here to view the standard normal distribution table (page 2).Click here to view the table of critical t-values.Select the correct choice below and fill in the answer boxes to complete your choice.(Use ascending order. Round to one decimal place as needed.)OA. 95% of the sentences for the crime are betweenand monthsB. There is a 95% probability that the mean length of sentencing for the crime is betweenOC. We can be 95% confident that the mean length of sentencing for the crime is betweenandmonths.andmonths.View an example Get more help -Q Search14480GIBNMDI7Kno12insertprt sc741PA4LClear allCheck answerdeletehomebackspacenumlockenterand73:12 AM4/5/2024PRE

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Answered: K In a random sample of 36 criminals…

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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Areas under the Normal Curve .00 .01 .02 .03 .04 .05 .06 .07 .08 .09 0.5040 0.5438 0.5832 0.5199 0.5596 0.5239 0,5636 0.6026 0.5279 0,5675 0.6064 0.0 0.5000 0.5080 0.5120 0.5319 0.5714 0.6103 0.5160 0.5359 0.0 0.5753 0.1 0.2 0.5398 0.5478 0.5517 0.5557 0.1 0.5793 0.5871 0.5910 0.5948 0.5987 0.6141 0.2 0.6331 0.6700 0.6368 0.6480 0.6517 0.6879 0.6179 0.6217 0.6255 0.6628 0.6293 0.6664 0.6406 0.6443 0.6808 0.3 0.4 0.3 0.4 0.6554 0.6591 0.6736 0.6772 0.6844 0.6985 0.7324 0.7642 0.7224 0.7549 0.7852 0.5 0.6915 0.6950 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.5 0.6 0.7257 0.7291 0.7357 0.7389 0.7422 0.7734 0.7454 0.7486 0.7794 0.8078 0.7517 0.7823 0.6 0.7 0.7580 0.7611 0.7673 0.7704 0.7764 0.7 0.7995 0.8264 0.7939 0.8 0.9 0.8051 0.8315 0.8133 0.8389 0.7881 0.7910 0.7967 0.8023 0.8106 0.8 0.8159 0.8186 0.8212 0.8238 0.8289 0.8340 0.8365 0.9 1.0 0.8413 0.8485 0.8554 0.8770 0.8962 0.8599 0.8810 0.8438 0.8461 0.8508 0.8531 0.8577 0.8621 1.0 1.1 0.8643 0.8665 0.8686 0.8708 0.8830 0.8729 0.8925…
In a random sample of 81 audited estate tax returns, it was determined that the mean amount of additional tax owed was $3454 with a standard deviation of $2528. Construct and interpret a 90% confidence interval for the mean additional amount of tax owed for estate tax returns. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Click here to view the table of critical t-values. Find and interpret a 90% confidence interval for the mean additional amount of tax owed for estate tax returns. Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to the nearest dollar as needed.) OA. One can be 90% confident that the mean additional tax owed is between $ OB. There is a 90% probability that the mean additional tax owed is between $ OC. 90% of taxes owed for estate tax returns are between $ and $ and $ and $
Salaries of 39 college graduates who took a statistics course in college have a mean, x, of $64,800. Assuming a standard deviation, ơ, of $13,783, construct a 95% confidence interval for estimating the population mean µ. Click here to view a t distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. ..... $>1> <µ< $ (Round to the nearest integer as needed.) %24
Salaries of 43 college graduates who took a statistics course in college have a mean, x, of $66,900. Assuming a standard deviation, o, of $12,687, construct a 90% confidence interval for estimating the population mean u. Click here to viewat distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. S
Salaries of 32 college graduates who took a statistics course in college have a mean, x, of $61,600. Assuming a standard deviation, o, of $11,936, construct a 99% confidence interval for estimating the population mean p. Click here to view a t distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. ..... $ <µ < $ (Round to the nearest integer as needed.)
Choose the appropriate statistical test. When computing, be sure to round each answer as indicated. A dentist wonders if depression affects ratings of tooth pain. In the general population, using a scale of 1-10 with higher values indicating more pain, the average pain rating for patients with toothaches is 6.8. A sample of 30 patients that show high levels of depression have an average pain rating of 7.1 (variance 0.8). What should the dentist determine? 1. Calculate the estimated standard error. (round to 3 decimals). [st.error] 2. What is thet-obtained? (round to 3 decimals). 3. What is the t-cv? (exact value) 4. What is your conclusion? Only type "Reject" or Retain"
Salaries of 38 college graduates who took a statistics course in college have a mean, x, of $65,900. Assuming a standard deviation, o, of $14,369, construct a 95% confidence interval for estimating the population mean μ. Click here to view a t distribution table. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. <μ<$ (Round to the nearest integer as needed.)
In a random sample of 36 criminals convicted of a certain crime, it was determined that the mean length of sentencing was 54 months, with a standard deviation of 13 months. Construct and interpret a 90% confidence interval for the mean length of sentencing for this crime. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Click here to view the table of critical t-values. Select the correct choice below and fill in the answer boxes to complete your choice. (Use ascending order. Round to one decimal place as needed.) O A. There is a 90% probability that the mean length of sentencing for the crime is between and months. O B. We can be 90% confident that the mean length of sentencing for the crime is between and months. O C. 90% of the sentences for the crime are between and months.
A simple random sample of size n = 40 is drawn from a population. The sample mean is found to be x = 120.5 and the sample standard deviation is found to bes = 13.2. Construct a 99% confidence interval for the population mean. The lower bound is (Round to two decimal places as needed.)
In a survey, 34% of the respondents stated that they talk to their pets on the telephone. A veterinarian believed this result to be too high, so he randomly selected 210 pet owners and discovered that 70 of them spoke to their pet on the telephone. Does the veterinarian have a right to be skeptical? Use the a = 0.05 level of significance. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). Because npo (1- Po): V 10, the sample size is V 5% of the population size, and the sample the requirements for testing the hypothesis satisfied. (Round to one decimal place as needed.) What are the null and alternative hypotheses? versus H: Ho: (Type integers or decimals. Do not round.) Determine the test statistic, Zo Z0 = (Round to two decimal places as needed.) Determine the critical value(s). Select the correct choice below and fill in the answer box to complete your choice. (Round to two decimal places as…
Salaries of 49 college graduates who took a statistics course in college have a mean, x, of $63,500. Assuming a standard deviation, o, of $14,619, construct a 90% confidence interval for estimating the population mean u. Click here to view at distribution table. Click here to view page 1 of the standard normal distribution table Click here to view page 2 of the standard normal distribution table. $

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