3. Mr. James Lynch conducts a study to find that the average timepeople wait in line before they can talk to a clerk at the DMV officeThe study involved 25 customers. The average was 45 minutes with astandard deviation of 12 minutes.Find a 99% confidence interval for the true mean time thatcustomers have to wait in line

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Answered: r. James Lynch conducts a study to find…

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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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.
The researcher took a random sample of 50 residents and found that the mean height of men is 175cm. Suppose that the team is made aware of the fact that the distribution of heights is distributed normally. 1, The sample median is an unbiased estimate of the mean height of men. Is the following statement about the estimator true? 2, Determine whether or not the median or mean is an efficient estimator of the population median.
A researcher wants to test whether a certain sound will make rats do worse onlearning tasks. It is known that an ordinary rat can learn to run a particular maze correctly in 18 trials, with a standard deviation of 6. (The number of trials to learn this maze is normally distributed.) The researcher now tries an ordinary rat in the maze, but with the sound. The rat takes 38 trials to learn the maze.Using the .05 level, what should the researcher conclude? (b) Then explain your answer to someone who has never had a course in statistics (but who is familiar with mean, standard deviation, and Z scores).
A researcher is interested in studying the use of hypnosis to relieve pain. She took a random sample of 16 subjects and measured their sensory ratings. The mean of the sensory ratings was found to be 7.5 and standard deviation to be 2.3. It is known that the sensory ratings are normally distributed. Construct a 99% confidence interval for the mean sensory ratings of the population. a) (5.81, 9.19) b) (6.27, 8.73) c) (6.36, 8.64) d) (6.02, 8.98)
A study of the career paths of hotel general managers sent questionnaires to an simple random sample of 260 hotels belonging to major U.S. hotel chains. There were 155 responses. The average time these 155 general managers had spent with their current company was 11.85 years. (Take it as known that the standard deviation of time with the company for all general managers is 3.8 years.) Find the margin of error for an 80% confidence interval to estimate the mean time a general manager had spent with their current company. Round your answer to 2 decimal places.
Suppose that Jerry (a baker) wants to find a 95% Confidence Interval for the mean baking time for all of his cider donuts. He takes a random sample of 40 cider donuts over the course of several weeks. He found that the average baking time of the sample was 16.4 minutes with a standard deviation of 2.2 minutes. d. Suppose Jose, another donut maker, states that his cider donuts have a mean baking time of 15 minutes. Is there evidence at the 95% confidence level, to support the claim that the true mean baking times for Jerry’s and Jose’s cider donuts are different? Justify. e. Suppose Lee, another donut maker, states that her cider donuts have a mean baking time of 17 minutes. Is there evidence at the 95% confidence level, to support the claim that the true mean baking times for Jerry’s and Lee’s cider donuts are different? Justify.
A new small business wants to know if its current radio advertising is effective. The owners decide to look at the mean number of customers who make a purchase in the store on days immediately following days when the radio ads are played as compared to the mean for those days following days when no radio advertisements are played. They found that for 1010 days following no advertisements, the mean was 18.618.6 purchasing customers with a standard deviation of 1.51.5 customers. On 77 days following advertising, the mean was 20.320.3 purchasing customers with a standard deviation of 0.90.9 customers. Test the claim, at the 0.020.02 level, that the mean number of customers who make a purchase in the store is lower for days following no advertising compared to days following advertising. Assume that both populations are approximately normal and that the population variances are equal. Let days following no advertisements be Population 1 and let days following advertising be Population 2.…
The lecturer are concerned that students who enrolled for the course this semester are spending more time playing online games than revising for their upcoming test. In order to pass the test, students need to spend at least 6 hours per week on the course materials. They conducted a survey of 120 students from Section 10G and 15P for which the result shows that the mean time they spent studying for the course in a week is 5.7667 hours with a standard deviation of 1.6487 hours. Using the critical value z0.025 =1.96 , perform hypothesis testing for mean to test whether or not students spent at least 6 hours weekly to study for the course.
In the natural gas consumption case, the estimated consumption is 12.003 in a week when the average hourly temperature is 30 degrees. If the distance value is 0.2642 when xo 30, compute a 95% confidence interval for the mean consumption for all weeks having average hourly temperature of 30 degrees. Recall that n= 8 and s=0.6542 O 135, 12.656) Reason: Use tozss ) distance = 12.003 : (2.447),6542)5140) = 12.003 823 = (1.180, 12.826] O m 580, 12.426) Reason: Use 9 = tozss) distance = 12.003 (2.447K.6542 5140) = 12.003 1823 = [11180, 12.826) O 18012 826) (11.739, 12.267) Reason: Use tozss) distance #12.003 1(2.4476542 5140) = 12.003823 [11180. 12.826) Correct Answer Read
5.37 Dental anxiety study. To gauge their fear of going to a den- tist, a random sample of adults completed the Modified Dental Anxiety Scale questionnaire (BMC Oral Health, Vol. 9, 2009). Scores on the scale range from zero (no anx- iety) to 25 (extreme anxiety). The mean score was 11 and the standard deviation was 4. Assume that the distribution of all scores on the Modified Dental Anxiety Scale is ap- proximately normal with µ = 11 and o = 4. a. Suppose you score a 10 on the Modified Dental Anxiety Scale. Find the z-value for your score. b. Find the probability that someone scores between 10 and 15 on the Modified Dental Anxiety Scale. c. Find the probability that someone scores above 20 on the Modified Dental Anxiety Scale.
A report states that the mean yearly salary offer for students graduating with a degree in accounting is $48,722. Suppose that a random sample of 50 accounting graduates at a large university who received job offers resulted in a mean offer of $49,870 and a standard deviation of $3900. Do the sample data provide strong support for the claim that the mean salary offer for accounting graduates of this university is higher than the national average of $48,7227 Ho: = Ha: Select an answer V (Put in the correct symbol and value) a = P-value =
A researcher is interested in finding a 98% confidence interval for the mean number of times per day that college students text. The study included 110 students who averaged 33.1 texts per day. The standard deviation was 16.8 texts. Round answers to 3 decimal places where possible

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