6b.STATISTICAL INFERENCE A research conducted by SSTA022 Class representative at the University of Limpopo indicatesthat running increases the lifespan among African men. The mean lifespan of 31 African menrunners was 34.0% more than the mean lifespan of 31 inactive African men, with variancesreported to be 110.25% and 104.04%, respectively. Assuming the populations to beapproximately normally distributed with equal variances.i) Construct a 95% confidence interval for μ₁ - ₂.

There are two independent samples which are active and inactive African men. We have to construct…

Answered: A research conducted by SSTA022 Class…

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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In a manufacturing process, a random sample of 9 manufactured bolts has a mean length of 3 inches with a sample variance of .09 and is normally distributed. What is the 90 percent confidence interval for the true mean length of the manufactured bolt?
Taking a random sample of 35 individuals waiting to be serviced by the teller, we find that the mean waiting time was 22.0 minutes and the standard deviation was 8.0 minutes. Construct a 97% confidence interval of the mean waiting time for all individuals waiting in the service line. If an individual waited for 15 minutes, would this be considered unusually fast?
The national fire protection association reported that house fires in the United States result in $6.8billion in losses annually. Suppose a random sample of 50 house fires in San Antonio yields a mean loss of $18503. If the population standard deviation is $3300, what is the margin of error for a 95% confidence interval? What would you recommend if the study required a margin of error of $300 or less
A trucking fleet-owner owns two semi-trucks and wishes to compare the amount of fuel used per week by the two vehicles. In a random sample of 100 weeks, truck #1 used a mean volume of 340 gallons with a standard deviation of 11 gallons. In a second, independent random sample of 100 weeks, truck #2 used a mean volume of 331 gallons with standard deviation 6 gallons.It is of interest to construct a confidence interval for the difference in population means using a confidence level of 90 percent. ?1−?2, where ?1 is the mean fuel volume used by truck #1 and ?2 is the mean volume of fuel used by truck #2. Note: 340−331=9 and 112100+62100‾‾‾‾‾‾‾‾‾‾‾√=1.25299640861417 Your answers below should be entered as numbers. Do not round during intermediate steps. If you round an answer, make sure you do so correctly and keep at least three decimal places. (a) The estimate is: gallons. (b) The standard error is: gallons. (c) The multiplier is: .
Suppose a random sample of 28 customers shows an average transaction amount of $45, with a standard deviation of $12. Find the lower bound of a 95% confidence interval for the population mean of transaction amount
In attempting to improve student services, a college investigates its registration process and finds that for 12 randomly selected students, the mean registration time was 73.2 minutes with a standard deviation of 14.2 minutes. Another college finds that for 15 randomly selected students, the mean registration time was 42.3 minutes with a standard deviation of 9.8 minutes. At the 0.01 significance level, test the claim that the two samples came from populations with the same means. Assume equal population variances.
In attempting to improve student services, a college investigates its registration processand finds that for 12 randomly selected students, the mean registrationi timewas 73.2 minutes with a standard deviation of 14.2 minutes. Another collegei findsthat for 15 randomly selected students, the mean registration time was 42.3 minutes with a standard deviation of 9.8 minutes. At the 0.01 significance level, test the claim that the two samples came from populations with the same means. Assumeequal population variances.
In a random sample of 38 criminals convicted of a certain crime, it was determined that the mean length of sentencing was 67 months, with a standard deviation of 11 months. Construct and interpret a 95% confidence interval for the mean length of sentencing for this crime.
According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this percent is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Suppose we want to calculate a 99% confidence interval for the difference between the proportions of Californians and Oregonians who are sleepdeprived. 1. State the critical value for the interval (3 decimal places): 2. State the difference between the sample proportions, POregon-PCalifornia (3 decimal places): 3. State the standard error of the difference between the proportions (3 decimal places): 4. State the confidence interval being sure to use proper notation (3 decimal places each):
In a current study, a random sample of 10 teachers from Texas and an independent random sample of 10 teachers from Virginia have been asked to report their annual income. The data obtained are as follows. (See attachment for data table and full question). The population standard deviations for the annual incomes of teachers in Texas and in Virginia are estimated as $6100 and $6600, respectively. It is also known that both populations are approximately normally distributed. Construct a 95% confidence interval for the difference −μ1μ2 between the mean annual income of teachers from Texas (μ1) and the mean annual income of teachers from Virginia (μ2). Then find the lower limit and upper limit of the 95% confidence interval. Carry your intermediate computations to at least three decimal places. Round your answers to at least two decimal places. a.) The Lower Limit is: b.) The Upper Limit is:
A paint manufacturer wants to determine the average drying time of a new interior wall paint. If for 12 test areas of equal size he obtained a mean drying time of 66.3 minutes and a standard deviation of 8.4 minutes, conduct a 95% confidence interval for the true mean .
According to a report on sleep deprivation by the Centers for Disease Control and Prevention, the proportion of California residents who reported insufficient rest or sleep during each of the preceding 30 days is 8.0%, while this proportion is 8.8% for Oregon residents. These data are based on simple random samples of 11,545 California and 4,691 Oregon residents. Using Rguroo, and the Large Sample Z method, calculate a 95% confidence interval for the difference between the proportions of Californians and Oregonians who are sleep deprived. Round your answers to 4 decimal places.

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