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…
The test statistic of z = - 1.06 is obtained when testing the claim that p = 4/7. a. Using a significance level of a = 0.01, find the critical value(s). b. Should we reject Ho or should we fail to reject Ho? Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard normal distribution table. a. The critical value(s) is/are z = (Round to two decimal places as needed. Use a comma to separate answers as needed.) b. Choose the correct conclusion below. A Fail to reject Ho. There is sufficient evidence to warrant rejection of the claim that p = 4/7. O B. Reject Họ. There is not sufficient evidence to warrant rejection of the claim that p = 4/7. C. Reject Ho. There is sufficient evidence to warrant rejection of the claim that p = 4/7. D. Fail to reject Ho. There is not sufficient evidence to warrant rejection of the claim that p = 4/7.
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
It is claimed that the proportion of adults with at least one chronic disease in a certain country might be more than the reported number, which is only 25%. A random sample of 85 adults are surveyed, of which 28 are found to suffer from at least one chronic disease. Use a 0.01 level of significance to test the above claim. Click here to view page 1 of the standard normal distribution table. Click here to view page 2 of the standard nomal distribution table. (a) Identify the null and alternative hypotheses. O A. H,: p= 0.25 against H,: p 0.25 OD. Hp: p>0.25 against H,: p=0.25 O E. H,:p= 0.25 against H,: p 0.25 OF. H:p OC. z (c) The z value for test can be calculated as (Round to two decimal places including any zeros.) (d) Make an appropriate conclusion. O A. Reject the null hypothesis because there is not significant evidence that more than 25% of the adults suffer from at least one chronic disease. O B. Do not reject the null hypothesis because there is not significant evidence that…
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…

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