article contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range: 415 420 422 423 425 429 432 435 437 438 446 447 449 451 456 463 464 (c) Calculate a two-sided 95% confidence interval for true average degree of polymerization. (Round your answers to two decimal places.)

article contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range: 415 420 422 423 425 429 432 435 437 438 446 447 449 451 456 463 464 (c) Calculate a two-sided 95% confidence interval for true average degree of polymerization. (Round your answers to two decimal places.)

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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You wish to test the following claim (Ha) at a significance level of a 0.05. %3D H,:4 = 52.5 H:u> 52.5 You believe the population is normally distributed, but you do not know the standard deviation. You obtain a sample of size n = SD = 11.4. 102 with a mean of M 55.2 and a standard deviation of %3D What is the critical value for this test? (Report answer accurate to three decimal places.) critical value = What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic = The test statistic is. O in the critical region O not in the critical region This test statistic leads to a decision to... O reject the null Oaccept the null O fail to reject the null As such, the final conclusion is that.. OThere is sufficient evidence to warrant rejection of the claim that the population mean is -greater than 52.5. O There is not sufficient evidence to warrant rejection of the claim that the population mean is greater than 52.5. O The sample data support…
Managers of an outdoor coffee stand in Coast City are examining the relationship between (hot) coffee sales and daily temperature, hoping to be able to predict a day's total coffee sales from the maximum temperature that day. The bivariate data values for the coffee sales (denoted by y , in dollars) and the maximum temperature (denoted by x , in degrees Fahrenheit) for each of fifteen randomly selected days during the past year are given below. These data are plotted in the scatter plot in Figure 1. Temperature, x(in degrees Fahrenheit) Coffee sales, y(in dollars) 53.1 2215.5 76.6 1982.7 70.1 1937.3 39.7 2251.1 72.5 1603.9 48.4 2024.2 46.7 2135.3 83.9 1536.5 58.4 1965.6 38.5 1944.3 68.2 1746.1 75.4 1472.1 65.6 1819.5 53.9 1627.4 45.1 1782.3 Send data to calculator Send data to Excel Coffee sales(in dollars) y 1200 1400 1600 1800 2000 2200 2400 x 40 50 60 70 80 90…
Hello! I need answer for letter F only using Anderson-Darling test to support the normal distribution of the data.
The average income of a large sample of households is $63,000 and the s.d. is $7,000. Assume that the data is approximately bell-shaped. What percent of these households have incomes between $42,000 and $77,000? Don’t forget to properly justify your answer using the empirical rule.
Managers of an outdoor coffee stand in Coast City are examining the relationship between (hot) coffee sales and daily temperature, hoping to be able to predict a day's total coffee sales from the maximum temperature that day. The bivariate data values for the coffee sales (denoted by y, in dollars) and the maximum temperature (denoted by x, in degrees Fahrenheit) for each of fifteen randomly selected days during the past year are given below. These data are plotted in the scatter plot in Figure 1. Also given is the product of the temperature and the coffee sales for each of the fifteen days. (These products, written in the column labelled "xy", may aid in calculations.) Temperature, X 10 Coffee sales, y (in dollars) (in degrees Fahrenheit) 37.2 51.3 59.7 53.7 63.1 44.5 47.5 75.9 82.2 40.0 74.8 70.6 69.0 48.3 74.7 Send data to calculator V 2000.4 2205.9 1944.8 1579.3 1846.3 1797.3 2016.5 1563.3 1556.5 2249.4 1653.9 1923.0 1747.9 2154.6 1979.7 74,414.88 113,162.67 116,104.56 84,808.41…
In a sample of seven cars, each car was tested for nitrogen oxide emissions (in grams per mile). The results appear below. Assuming that nitrogen oxide emissions are normally distributed, 4. Car 1 2 3 4 5 6 7 Nitrogen oxide emission (g/mi) 0.13 0.18 0.15 0.09 0.12 0.18 0.08 a. Construct a 95% confidence interval for the mean amount of nitrogen oxide emissions for all cars. b. What effect would increasing the level of confidence have on the length of the interval?
Managers of an outdoor coffee stand in Coast City are examining the relationship between (hot) coffee sales and daily temperature, hoping to be able to predict a day's total coffee sales from the maximum temperature that day. The bivariate data values for the coffee sales (denoted by y, in dollars) and the maximum temperature (denoted by x, in degrees Fahrenheit) for each of sixteen randomly selected days during the past year are given below. These data are plotted in the scatter plot in Figure 1. Temperature, x (in degrees Fahrenheit) Coffee sales, y (in dollars) 71.5 1970.9 58.7 1953.9 53.7 1791.1 2400+ 83.0 1570.3 2200+ 62.8 1852.7 74.6 1633.6 2000- ** 40.4 1973.9 1800- 51.5 2250.9 44.6 1808.5 1600- 45.5 1977.3 1400- 45.6 2190.5 1200 69.4 1789.1 55.0 1598.2 50 70 80 90 40.6 2272.0 74.0 1937.0 Figure 1 76.6 1547.1 Send data to Excel Continue Submit Assignment O 2021 McGraw-Hill Education. All Rights Reserved. Terms of Use Privacy Accessibility here to search 99+ 5 4.
Women are recommended to consume 1790 calories per day. You suspect that the average calorie intake is smaller for women at your college. The data for the 15 women who participated in the study is shown below: 1959, 1908, 1789, 1544, 1805, 1656, 1899, 1811, 1686, 1688, 1687, 1943, 1612, 1789, 1467 Assuming that the distribution is normal, what can be concluded at the α = 0.05 level of significance? For this study, we should use The null and alternative hypotheses would be: H0: H1: The test statistic = (please show your answer to 3 decimal places.) The p-value = (Please show your answer to 4 decimal places.) The p-value is α Based on this, we should the null hypothesis. Thus, the final conclusion is that ... The data suggest the population mean is not significantly less than 1790 at α = 0.05, so there is sufficient evidence to conclude that the population mean calorie intake for women at your college is equal to 1790. The data suggest the populaton…
A researcher conducted a repeated measures study comparing three treatment conditions. Refer to attached images and tale to answer a to d. Mean Std. Deviation N Treatment I 1.00 1.414 5 Treatment II 5.00 2.345 5 Treatment III 6.00 1.581 5 In APA format, report the F-ratio related to the treatment effect: Is this treatment effect significant? What is the partial η2 value for the treatment effect? Is this a weak, moderate, or strong effect?
You wish to test the following claim (Ha) at a significance level of a = 0.05. H.:µ = 51.2 H.:µ + 51.2 You believe the population is normally distributed and you know the standard deviation is o = 19.9. You obtain a sample mean of M = 38.6 for a sample of size n = 24. What is the critical value for this test? (Report answer accurate to three decimal places.) critical value = What is the test statistic for this sample? (Report answer accurate to three decimal places.) test statistic =
An article contained the following observations on degree of polymerization for paper specimens for which viscosity times concentration fell in a certain middle range: 415 421 422 423 425 427 430 435 437 439 446 446 448 453 456 463 464 Calculate a two-sided 95% confidence interval for true average degree of polymerization.(Round your answers to two decimal places.)
These are the numbers I am working with: Sample mean= 78 Population mean= 75 Standard deviation:=10 Significance level: .05 Critical value: Z score > 1.96 Z score= 1.34 How do I figure the confidence limits for the 95% confidence interval. Z-score= 1.34 P-value= .1802