A sample of size n = 31 has sample mean x = 53 and sample standard deviation 5 = 8.1.Part: 0/2Part 1 of 2(a) Construct a 95% 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.45.045.095% confidence interval54.564.064.0›B199 Fake Twitter followers: Many celebrities and public figures have Twitter accounts with large numbers of followers. However, some of these followers are fake,resulting from accounts generated by spamming computers. In a sample of 48 twitter audits, the mean percentage of fake followers was 13.6 with a standarddeviation of 9.2.Part: 0 / 2Part 1 of 2(a) Construct a 98% confidence interval for the mean percentage of fake Twitter followers. Round the answers to at least one decimal place.A 98% confidence interval for the mean percentage of fake Twitter followers is

Answered: Image /qna-images/answer/b3a18348-89b8-4e24-94c7-bbfcb5a2c762.jpg

Answered: ple of size 12=31 has sample mean = 53…

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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One year, the mean age of an inmate on death row was 40.5 years. A sociologist wondered whether the mean age of a death-row inmate has changed since then. She randomly selects 32 death-row inmates and finds that their mean age is 39.4, with a standard deviation of 9.5. Construct a 95% confidence interval about the mean age. What does the interval imply? Question Help v E Click the icon to view the table of critical t-values. Choose the correct hypotheses. Ho: O Table of Critical t-Values %3D H: (Type integers or decimals. Do not round.) Area in vight tail -Distribution Area in Right Tail Degrees of Freedom 0.25 0.20 0.15 0.10 0.05 0.025 0.02 001 0.005 0.0025 0.001 0.0005 1.000 1.376 1.061 0.978 0.941 0.920 1.963 1.386 1.250 1190 1.156 3.078 1886 1638 1.533 1476 6.314 12.706 15.894 4849 3.482 2.999 2.757 2.612 2.517 2.449 2.398 2.359 2.328 2.303 2282 2.264 2.249 31821 63657 6.965 4.541 3.747 3.365 127321 318.309 636.619 22.327 10.215 Z173 5.893 5.208 4.785 4.501 4.297 4.144 2 0816 0.765…
Babies: A sample of 25 one-year-old girls had a mean weight of 24.1 pounds with a standard deviation of 4.3 pounds. Assume that the population of weights is normally distributed. A pediatrician claims that the standard deviation of the weights of one-year-old girls is greater than 7 pounds. Do the data provide convincing evidence that the pediatrician's claim is true? Use the =α0.05 level of significance.
Shoulder height is known to be normally distributed with a mean of 56.65 inches for males and 52.5 inches for females. The males vary with a standard deviation of 2.46 inches and females at 2.49 inches. find the z-score representing a male whose shoulder height is 49 inches a standardized curve N(0,1) is shown below use this information to find an equivalent shoulder height for females and explain the relevance of this value and shading in the context of the problem
Random samples of size 46 are drawn from a population with mean 161.9 and standard deviation 16.1. Find the mean and the standard deviation of the sample mean. Round to four decimal places.Mean = Standard Deviation = Enter your answer as a number accurate to 4 decimal places.
Adult men have heights with a mean of 69.0 inches and a standard deviation of 2.8 inches. Find the z-score of a man 63.4 inches tall. (to 2 decimal places)
Please help me asap. Thank you
population mean is 533.7 and standard deviation is 129.4. suppose you do not know population mean in a random sample of 75 the average spent is 550. find a 98.1% CI for population mean
Newspaper Reading Times A survey taken several years ago found that the average time a person spent reading the local daily newspaper was 10.8 minutes. The standard deviation of the population was 3 minutes. To see whether the average time had changed since the newspaper's format was revised, the newspaper editor surveyed 40 individuals. The average time that the 40 people spent reading the paper was 12.4 minutes. At α=0.10 , is there a change in the average time an individual spends reading the newspaper? Find the 90% confidence interval of the mean. (a) State the hypothesis and identify the claim with the correct hypothesis. :H0 ▼(Choose one) :H1 ▼(Choose one) This hypothesis test is a ▼(Choose one) test. (b)Find the critical value(s). Round the answer to 2 decimal places, if necessary. Critical values: (c)Compute the test value. Round the answer to at least 2 decimal places, if necessary. z= (d)Make the decision.…
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From generation to generation, the mean age when smokers first start to smoke varies. However, the standard deviation of that age remains constant around 2.1 years. A survey of 40 smokers of this generation was done to see if the mean starting age is at least 19. The sample mean was 18.1 with a sample standard deviation of 1.3. Does the data support the claim at the 5% level? (Give the p-value, your decision, the reason for your decision and your conclusion)
Newspaper Reading Times A survey taken several years ago found that the average time a person spent reading the local daily newspaper was 10.8 minutes. The standard deviation of the population was 3 minutes. To see whether the average time had changed since the newspaper's format was revised, the newspaper editor surveyed 40 individuals. The average time that the 40 people spent reading the paper was 12.4 minutes. At α=0.10 , is there a change in the average time an individual spends reading the newspaper? Find the 90% confidence interval of the mean. **NOTE** I already have answer for a,b, and c (d)Make the decision. ▼(Choose one) the null hypothesis. (e)Summarize the results. There ▼(Choose one) enough evidence to support the claim that there is a change in the average time an individual spends reading the newspaper. Find the 90% confidence interval of the mean. Round the answers to nearest whole number. <u< (f)Do the results agree?…
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