Mathematical Statistics with Applications
7th Edition
ISBN: 9780495110811
Author: Dennis Wackerly, William Mendenhall, Richard L. Scheaffer
Publisher: Cengage Learning
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 13, Problem 79SE
a
To determine
Find the number of replication required to estimate any treatment mean correct to within
b
To determine
Calculate the degrees of freedom will be available for estimating
c
To determine
Calculate the approximate half-width of a 95% confidence interval for the difference in the mean response for the two treatments.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Suppose we took a random sample of 200 students and ask if they preferred morning, evening, or night classes. a.If 76 students preferred morning classes what is the 95% CI for the proportion of students who prefer morning classes? b. Based on this confidence interval what would be the decision to the hypothesis: H0 = .3 vs Ha ≠ .3 using alpha = .05? c.How large a sample should be taken in order to estimate the proportion to within 2%, with 99% confidence?
A 30 month study is conducted to determine the difference in rates of accidents per month between two departments in an assembly plant. If a sample of 12 from the first department averaged 12.3 accidents per month with a standard deviation of 3.5, and a sample of 9 from the second department averaged 7.6 accidents per month with standard deviation of 3.4, find: The t-value for a 95% confidence interval is:
A. 2.09302
B. 1.724718
C. 2.08596
D. 1.729133
E. none of the preceding
In a survey of
460
drivers from the South,
397
wear a seat belt. In a survey of
340
drivers from the Northeast,
281
wear a seat belt. At
alpha equals 0.06
,
can you support the claim that the proportion of drivers who wear seat belts is greater in the South than in the Northeast? Assume the random samples are independent. Complete parts (a) through (e).
Chapter 13 Solutions
Mathematical Statistics with Applications
Ch. 13.2 - The reaction times for two different stimuli in a...Ch. 13.2 - Prob. 2ECh. 13.4 - State the assumptions underlying the ANOVA of a...Ch. 13.4 - Prob. 4ECh. 13.4 - Prob. 5ECh. 13.4 - Suppose that independent samples of sizes n1, n2,,...Ch. 13.4 - Four chemical plants, producing the same products...Ch. 13.4 - Prob. 8ECh. 13.4 - Prob. 9ECh. 13.4 - A clinical psychologist wished to compare three...
Ch. 13.4 - It is believed that women in the postmenopausal...Ch. 13.4 - If vegetables intended for human consumption...Ch. 13.4 - One portion of the research described in a paper...Ch. 13.4 - The Florida Game and Fish Commission desires to...Ch. 13.4 - Prob. 15ECh. 13.4 - An experiment was conducted to examine the effect...Ch. 13.5 - Prob. 17ECh. 13.5 - Refer to Exercise 13.17 and consider YiYi for i ...Ch. 13.5 - Refer to the statistical model for the one-way...Ch. 13.7 - Refer to Examples 13.2 and 13.3. a Use the portion...Ch. 13.7 - Refer to Examples 13.2 and 13.4. a Use the portion...Ch. 13.7 - a Based on your answers to Exercises 13.20 and...Ch. 13.7 - Refer to Exercise 13.7. a Construct a 95%...Ch. 13.7 - Prob. 24ECh. 13.7 - Prob. 25ECh. 13.7 - Prob. 26ECh. 13.7 - Prob. 27ECh. 13.7 - Prob. 28ECh. 13.7 - Prob. 29ECh. 13.7 - Prob. 30ECh. 13.7 - Prob. 31ECh. 13.7 - Prob. 32ECh. 13.7 - Prob. 33ECh. 13.7 - Prob. 34ECh. 13.7 - Prob. 35ECh. 13.8 - Prob. 36ECh. 13.8 - Prob. 37ECh. 13.8 - Prob. 38ECh. 13.8 - Prob. 39ECh. 13.8 - Prob. 40ECh. 13.9 - Prob. 41ECh. 13.9 - The accompanying table presents data on yields...Ch. 13.9 - Refer to Exercise 13.42. Why was a randomized...Ch. 13.9 - Prob. 44ECh. 13.9 - Prob. 45ECh. 13.9 - Prob. 46ECh. 13.9 - Prob. 47ECh. 13.9 - Prob. 48ECh. 13.9 - Prob. 49ECh. 13.9 - Prob. 50ECh. 13.9 - Prob. 51ECh. 13.10 - Prob. 52ECh. 13.10 - Prob. 53ECh. 13.10 - Prob. 54ECh. 13.10 - Refer to Exercise 13.46. Construct a 95%...Ch. 13.10 - Prob. 56ECh. 13.10 - Prob. 57ECh. 13.11 - Prob. 58ECh. 13.11 - Prob. 59ECh. 13.11 - Prob. 60ECh. 13.11 - Prob. 61ECh. 13.11 - Prob. 62ECh. 13.12 - Prob. 63ECh. 13.12 - Prob. 64ECh. 13.12 - Prob. 65ECh. 13.12 - Prob. 66ECh. 13.12 - Prob. 67ECh. 13.12 - Prob. 68ECh. 13.13 - Prob. 69ECh. 13.13 - Prob. 70ECh. 13.13 - Refer to Exercise 13.42. Answer part (a) by...Ch. 13.13 - Refer to Exercise 13.45. Answer part (b) by...Ch. 13 - Prob. 73SECh. 13 - Prob. 74SECh. 13 - Prob. 75SECh. 13 - Prob. 77SECh. 13 - A study was initiated to investigate the effect of...Ch. 13 - Prob. 79SECh. 13 - A dealer has in stock three cars (models A, B, and...Ch. 13 - In the hope of attracting more riders, a city...Ch. 13 - Prob. 84SECh. 13 - Prob. 85SECh. 13 - Prob. 86SECh. 13 - Prob. 87SECh. 13 - Prob. 88SECh. 13 - Prob. 89SECh. 13 - Prob. 90SECh. 13 - Prob. 92SECh. 13 - Prob. 94SE
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, statistics and related others by exploring similar questions and additional content below.Similar questions
- Let a random sample of size 17 from the normal distribution N(μ, σ2) yieldx = 4.7 and s2 = 5.76. Determine a 90% confidence interval for μ.arrow_forwardGiven a random sample of n = 17 subjects drawn from a normally distributed population with sample mean of M = 130 and sample standard deviation of sx = 40, what are the 95 percent confidence limits for the population mean (μ)?arrow_forwardA random sample of size n1=23, taken from a normal population with a standard deviation σ1=5, has a mean x1=77. A second random sample of size n2=37, taken from a different normal population with a standard deviation σ2=3, has a mean x2=38. Find a 96% confidence interval for μ1−μ2.arrow_forward
- Consider the following results for independent random samples taken from two populations. Sample 1Sample 2n 1 = 20n 2 = 40x 1 = 22.4x 2 = 20.7s 1 = 2.2s 2 = 4.7 What is the point estimate of the difference between the two population means (to 1 decimal)?What is the degrees of freedom for the t distribution (round down your answer to nearest whole number)?At 95% confidence, what is the margin of error (to 1 decimal)?What is the 95% confidence interval for the difference between the two population means (to 1 decimal)? Use z-table. If your answer is negative value, entre minus (-) sign.(, )arrow_forwardThe following data, drawn from a normal population, are the waiting times (in minutes) of 12randomly selected customers at a travel desk.1.8 2.4 2.7 2.1 1.3 1.7 1.5 1.5 2.3 2.8 2.5 1.9a. Calculate the point estimates of the population variance and the population standard deviation.b. Compute a 90% confidence intervals for the population variance and the population standarddeviationarrow_forwardCollege-aged adults need at least 7 hours of sleep each night to stay healthy. Sleep deprivation can lead to decreased immune system function, lack of concentration, depression, and poor memory. A simple random sample of 34 college students reports the number of hours of sleep they had last night. Is there evidence that the true population means hours of sleep for college students in the population is less than the 7 hours that are recommended? Conduct a two-tailed, one-sample t-test, with alpha set at .05. Our test statistic is 7 hours of sleep. Reported hours of sleep (n = 34): 4, 4, 6, 5, 6, 5, 5, 6, 4, 3, 6, 8, 9, 10, 8, 9, 7, 5, 4, 7, 8, 7, 5, 7, 7, 6, 7, 4, 5, 3, 7, 9, 10, 5arrow_forward
- A random sample of 24 army recruits has a mean height of 68 inches with a standard deviation of 2.3 inches. If a 98% confidence interval is constructed, how many degrees of freedom would be allowed to find the t-critical value? a) 67 b) 97 c) 23arrow_forwardThe NAEP considers that a national average of 283 is an acceptable performance. Using α = .05, run a two-tail t-test for one sample to test Ho: µ=283 for the 2019 scores. Report the t-obt, df, and p-values. Would you reject the null hypothesis that the 2019 scores come from a population with average 283? If this is the case, does it come from a population from larger or smaller average?arrow_forwardConstruct a 90% confidence interval for the true average cost of a 75 000 kilometre car service in Randburg, if it was found that a sample of 85 different service centres in the area charged an average of R2 180 for such a service. Assume a population standard deviation of R181.25.arrow_forward
- For a population data set sigma = 12.7 How large a sample should be selected so that the margin of error of estimate for a 96 confidence interval for u is 3.20?arrow_forwardAn article in Engineering Horizons (Spring 1990, p. 26) reported that 117 of 484 new engineering graduates were c08.indd 302 9/24/2013 6:57:11 PMSection 8-7/Tolerance and Prediction Intervals 303planning to continue studying for an advanced degree. Consider this as a random sample of the 1990 graduating class.(a) Find a 90% confidence interval on the proportion of such graduates planning to continue their education.(b) Find a 95% confidence interval on the proportion of such graduates planning to continue their education.(c) Compare your answers to parts (a) and (b) and explain why they are the same or different.(d) Could you use either of these confidence intervals to determine whether the proportion is actually 0.25? Explain your answer. Hint: Use the normal approximation to the binomial.arrow_forwardA sports statiscian claims that the mean winning times for Boston Marathon women's division open champions is at least 2.68 hours. The mean winning time of a sample of 30 randomly selected Boston Marathon women's open division is 2.60 hours. Assume the population standard deviation is 0.32 hour. Alpha = 0.05, can you reject the claim?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
MATLAB: An Introduction with ApplicationsStatisticsISBN:9781119256830Author:Amos GilatPublisher:John Wiley & Sons Inc
Probability and Statistics for Engineering and th...StatisticsISBN:9781305251809Author:Jay L. DevorePublisher:Cengage Learning
Statistics for The Behavioral Sciences (MindTap C...StatisticsISBN:9781305504912Author:Frederick J Gravetter, Larry B. WallnauPublisher:Cengage Learning
Elementary Statistics: Picturing the World (7th E...StatisticsISBN:9780134683416Author:Ron Larson, Betsy FarberPublisher:PEARSON
The Basic Practice of StatisticsStatisticsISBN:9781319042578Author:David S. Moore, William I. Notz, Michael A. FlignerPublisher:W. H. Freeman
Introduction to the Practice of StatisticsStatisticsISBN:9781319013387Author:David S. Moore, George P. McCabe, Bruce A. CraigPublisher:W. H. Freeman
MATLAB: An Introduction with Applications
Statistics
ISBN:9781119256830
Author:Amos Gilat
Publisher:John Wiley & Sons Inc
Probability and Statistics for Engineering and th...
Statistics
ISBN:9781305251809
Author:Jay L. Devore
Publisher:Cengage Learning
Statistics for The Behavioral Sciences (MindTap C...
Statistics
ISBN:9781305504912
Author:Frederick J Gravetter, Larry B. Wallnau
Publisher:Cengage Learning
Elementary Statistics: Picturing the World (7th E...
Statistics
ISBN:9780134683416
Author:Ron Larson, Betsy Farber
Publisher:PEARSON
The Basic Practice of Statistics
Statistics
ISBN:9781319042578
Author:David S. Moore, William I. Notz, Michael A. Fligner
Publisher:W. H. Freeman
Introduction to the Practice of Statistics
Statistics
ISBN:9781319013387
Author:David S. Moore, George P. McCabe, Bruce A. Craig
Publisher:W. H. Freeman
Statistics 4.1 Point Estimators; Author: Dr. Jack L. Jackson II;https://www.youtube.com/watch?v=2MrI0J8XCEE;License: Standard YouTube License, CC-BY
Statistics 101: Point Estimators; Author: Brandon Foltz;https://www.youtube.com/watch?v=4v41z3HwLaM;License: Standard YouTube License, CC-BY
Central limit theorem; Author: 365 Data Science;https://www.youtube.com/watch?v=b5xQmk9veZ4;License: Standard YouTube License, CC-BY
Point Estimate Definition & Example; Author: Prof. Essa;https://www.youtube.com/watch?v=OTVwtvQmSn0;License: Standard Youtube License
Point Estimation; Author: Vamsidhar Ambatipudi;https://www.youtube.com/watch?v=flqhlM2bZWc;License: Standard Youtube License