A personal trainer determines that an individual will get the most benefit from a workout if they keep their heart rateat an average of 150 beats per minute during workouts. To determine if the individual is doing so successfully, arandom sample of 30 workouts is selected from their fitness watch. A 95% confidence interval for these workoutsreveals that the true mean heart rate while working out is between 158 and 167 beats per minute. Based upon thisinterval, what conclusion should be made about the hypotheses: Ho = 150 versus H 150 where u = thisindividual's true mean heart rate during working out at a = 0.05?Reject Ho. There is convincing evidence that the mean heart rate from these 30 workouts differs from 150.Reject Ho. There is convincing evidence that this individual's true mean heart rate while working out differs from150.Fail to reject Ho. There is not convincing evidence that the mean heart rate from these 30 workouts differs from150.Fail to reject Ho. There is not convincing evidence that this individual's true mean heart rate while working outdiffers from 150.SubmitNextSave and ExitMark this and return

We have given that Sample n = 30 mu= 50 Confidence interval between ( 158 , 167 ) level of…

Answered: A personal trainer determines that an…

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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You measure 27dogs' weights, and find they have a mean weight of 63 ounces. Assume the population standard deviation is 6.1 ounces. Based on this, what is the maximal margin of error associated with a 95% confidence interval for the true population mean dog weight.
A researcher hypothesizes that performance on a cognitive test will be better when people have had a full night's sleep compared to when these same people are sleep deprived. A sample of 12 people is deprived of sleep for one night before taking the test. Under these circumstances, the mean number of questions they answered correctly was 6.08 with an SD of 1.38. A week later, these same people complete the test afler a full night's sleep. The mean number of questions answered correctly under these circumstances was 14.25 with an SD of 3.7. The raw scores and Z-scores were as follows: 3. No Sleep 0 66 139 Sleep 16 0.47 19 1 29 0.20 0 20 -0.07 0 20 -0.34 0.66 15 1.39 15 -0.79 14 15 0.06 0.06 -0.79 13 12 -0.61 -2.77 0 20 3. -2.24 4. -0.06 15 6. -0.06 16 047 -0.06 17 0.74 Test the researcher's hypothesis at a -05.
A study of 40 English professors showed that they spend, on average, 12.6 minutes correcting a student's term paper. The standard deviation was 2.5 minutes. Construct the 99% confidence interval of the mean for all term papers.
An apartment hunter is shopping for either a one-bedroom or two-bedroom apartment in varying apartment buildings. She wants to estimate the additional amount she would have to pay, on average, for a two-bedroom apartment versus a one-bedroom apartment. The table shows the prices, in dollars, for one- and two-bedroom apartments from the nine buildings she visited. A 95% confidence interval for the true mean difference between two-bedroom and one-bedroom apartments for apartment buildings like the ones in the sample was (255.93, 508.51). A friend of the apartment hunter tells her to expect to pay about $500 more per month for a two-bedroom apartment than for a one-bedroom apartment. Does the confidence interval provide evidence for the friend’s claim? A. No, the confidence interval does not contain 0, so the friend’s claim is not plausible. B. No, the confidence interval contains many more numbers below 500 than above 500, so it is unlikely that the friend’s assertion is correct. C.…
You measure 38 textbooks' weights, and find they have a mean weight of 58 ounces. Assume the population standard deviation is 3.9 ounces. Based on this, construct a 90% confidence interval for the true population mean textbook weight .
In a poll, 1100 adults in a region were asked about their online vs. in-store clothes shopping. One finding was that 44% of respondents never clothes-shop online. Find and interpret a 95% confidence interval for the proportion of all adults in the region who never clothes-shop online. Click here to view page 1 of the table of areas under the standard normal curve. Click here to view page 2 of the table of areas under the standard normal curve. The 95% confidence interval is from to (Round to three decimal places as needed.)
Sam is the manager of a used clothing store. He wants to estimate the average amount a customer spends per visit. A random sample of 70 customers’ receipts gave a mean of = $7.10 with a standard deviation s = $2.30. Find a 99% confidence interval for the average amount spent by all customers. Interpret your findings.
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