lnC = lnC0 – kt, C is the concentration of

Get college assignment help at Smashing Essays Question lnC = lnC0 – kt, C is the concentration of a reactant at time t and C0 is the initial concentration.Assume that when t = 0, C0 = 0.3 mole/L. The experiment was repeated n times to give a geometric mean of the concentration at time t= 450s of 0.22 mole/L.(a) Compute the mean of the constant k.(b) Compute the standard deviation of the rate constant k.I believe in order to solve the problem, we should use the fact that lnC can be equal to y and make the the equation into a linear equation.

Based on these results, t (92) = -.67, p =

Question Based on these results, t (92) = -.67, p = .001, d = .56. The reading scores of males (M = 31.54, SD = 5.16, n = 29) on and females were compared (M = 32.46, SD = 4.96, n = 65). What can be concluded based on these results?a. Males (M = 31.54, SD = 5.16, n = 29) on average score statistically significantly higher than females (M = 32.46, SD = 4.96, n = 65) in terms of reading achievement.b. Males (M = 31.54, SD = 5.16, n = 29) on average do not statistically significantly differ from females (M = 32.46, SD = 4.96, n = 65) in terms of reading achievementc. Males (M = 31.54, SD = 5.16, n = 29) on average score statistically significantly lower than females (M = 32.46, SD = 4.96, n = 65) in terms of reading achievement.

A sample of 25 was selected out of a specific

Question A sample of 25 was selected out of a specific population with mean equal to 18.4 and sample standard deviation of 3.6. Construct a 95% confidence interval for the mean of the population. Interpret the confidence interval.

A group of students were randomly selected to participate in

Question A group of students were randomly selected to participate in a study that compared the female grades to the male grades for a specific test. There were 15 females with a mean grade of 94.3 and sample standard deviation of 3.6. There were 12 males with a mean grade of 90.6 and sample standard deviation of 5.1. Construct a 90% confidence interval for the difference between the females’ and males’ test grades. Interpret the confidence interval

Assuming the population standard deviation how large should a sample

Question Assuming the population standard deviation  how large should a sample be to estimate a population mean with a margin of error of 0.2 for a 95% confidence interval?

A sample size of n = 120 produced the sample

Question A sample size of n = 120 produced the sample mean of  = 24.3. Assuming the population standard deviation  compute a 99% confidence interval for the population mean. Interpret the confidence interval.

Capstone Project Part 2 Template In the automated packaging process,

Question Capstone Project Part 2 Template In the automated packaging process, equipment is set to fill boxes with a mean weight of 574 grams. This is the standard for the population. Population standard deviation is not known. 1. Calculate a 95% confidence interval for the mean weight of the cereal boxes, using the sample data given in the Excel spreadsheet labeled Cereal Weights. 2. Based on the confidence interval for the mean weight of cereal boxes, does this sample of 400 boxes provide evidence that the packaging process is meeting the company standard? Explain. 3. The company engineer decided to make adjustments to the filling equipment to correct the issue. A random sample of 30 boxes was selected and weighed after the adjustment. The data for this sample can be found in the Excel spreadsheet labeled Weights after Adjustment. Find the descriptive statistics and the 95% confidence interval for the mean weight of packages after the equipment adjustment. Using the descriptive statistics and the confidence interval, does it appear that the equipment adjustment had any effect on the weight of the packages? Explain. 4. Using the data from the Weights after Adjustment, conduct a hypothesis test, using this sample data, to determine whether the mean weight of cereal boxes is meeting the company standard mean weight 574 grams after the equipment adjustment, with the level of significance of .05. Consider carefully whether a one-tail test or a two-tail test would be more useful to management. Use the company standard mean weight as the population mean; population standard deviation is not known. Null hypothesis: ____________________________________________________ Alternative hypothesis: ______________________________________________ Critical value: ______________________________________________________ Test statistic: ______________________________________________________ (Must show calculation of test statistic.) p-value: ___________________________________________________________ Decision: __________________________________________________________ What conclusions can you draw about the population mean weight based on the hypothesis test? Was the equipment adjustment enough to meet the company standard? C

Assuming the population standard deviation of 4, how large should

Question Assuming the population standard deviation of 4, how large should a sample be to estimate a population mean with a margin of error of 0.2 for a 95% confidence interval?

Driving under the influence of alcohol (DUI) is a serious

Question Driving under the influence of alcohol (DUI) is a serious offense. The following data give the ages of a random sample of 50 drivers arrested while driving under the influence of alcohol. This distribution is based on the age distribution of DUI arrests given in the Statistical Abstract of the United States (112th Edition).46 16 41 26 22 33 30 22 36 34 63 21 26 18 27 24 31 38 26 55 31 47 27 43 35 22 64 40 58 20 49 37 53 25 29 32 23 49 39 40 24 56 30 51 21 45 27 34 47 35(b) Make a frequency table using seven classes. (Give relative frequencies to 2 decimal places.) Class LimitsClass BoundariesMidpointFrequencyRelative FrequencyCumulative Frequency −                              (c) Make a histogram showing class boundaries. 

Suppose certain coins has weights that are normally distributed with

Question Suppose certain coins has weights that are normally distributed with a mean of 5.996 g and a standard deviation of 0.057 g. A vending machine is configured too accept those coins with weights between 5.936 g and 6.056 g.If 280 different coins were inserted into the vending machine, what is the expected number of rejected coins? (Round to the nearest integer)If 280 different coins are inserted into the vending machine, what is the probability that the mean falls between the limits of 5.963 g and 6.056 g ? ( round 4 decimal places please.)Which result is more important to the owner of the vending machine? why? A Part (a) because the average result is more important.B. Part (b) because rejected coins could mean lost sales.C. Part (b) because the average result is more important.D. Part (a) because rejected coins could mean lost sales.

I am having some trouble with this question. I’m not

Get college assignment help at Smashing Essays Question I am having some trouble with this question. I’m not sure how to find how many people were surveyed, computing the test statistic, and finding the P-value. ATTACHMENT PREVIEW Download attachment 65460965_723355858098823_999751876930961408_n.jpg ) Charles Stevens, the owner of Wilmot Orchards, historically had on average of 185 apples per tree for his Macintosh variety. He applies a new fertilizer to his crop and from a random sample of 36 trees, the average yield is 199 apples per tree with a standard deviation of 48 apples per tree. From these data, test the claim at 95% confidence that the fertilizer actually improved his crop yield. a) What are the Null and Alternative Hypotheses? (1) b) Prepare the PDF and state the Decision/Rejection Rule for this problem (1, 1) C ) Conduct the test (3) d) State the Decision and Interpretation (1, 1) e) What is the Pvalue? (2) f ) Using this same information, prepare the 95% confidence interval for this new crop. (4) g) Does the 95% confidence interval contain the older average yield (i.e., 185 apples per tree) before the application of this new fertilizer? (1) h) If the decision of part i) is to reject the Null Hypothesis, why w/could your confidence interval of part I) include the older average yield of 185 apples per tree? Explain. (2) Page 11 Assignment #2; HSLC 3800 Summer 2019

I don’t quite understand the problem 7.10 shown in the

Question I don’t quite understand the problem 7.10 shown in the picture, so im wondering could u plz solve the question for me? thanks! ATTACHMENT PREVIEW Download attachment IMG_0630.jpg (c) Cavendish made changes to his appa He considered this change as potentially important. Omit the first six observations from 5.50 to 5.55 and construct a box plot, a histogram (use 6 bins) and a QQ-plot of the remaining data. Does the removal of the first six observations appear to have changed the shape of the distribution? Problem 7.10. The concentration of a reactant in a first-order chemical reaction that proceeds at a rate k can be described as follows: In C = In Co – kt, where C is the concentration of the reactant at time t, Co is the initial concentration and t is the elapsed time since the reaction started. Consider an initial concentration of Co = 0.3 mol/L. The experiment was repeated n times to give a geometric mean of the concentration at time t = 450 seconds of 0.22 mol/L. The geometric standard deviation of the concentration at time t = 450 seconds is 1.17. (a) Compute the mean of the rate constant k. (b) Compute the standard deviation of the rate constant k. Hint: Use the fact that k is a linear function of y = In C. Problem 7.11. Carbon monoxide is a gas that is highly toxic. The authors of [16] observed that it was possible to have higher mean concentrations of carbon monoxide at urban intersections, compared to highways with much

A person with a cough is a persona non grata

Question A person with a cough is a persona non grata style=”color:rgb(0,0,0);”> on airplanes, elevators, or at the theater. In theaters especially, the irritation level rises with each muffled explosion. According to Dr. Brian Carlin, a Pittsburgh pulmonologist, in any large audience you’ll hear about 14 coughs per minute.(a) Let r = number of coughs in a given time interval. Explain why the Poisson distribution would be a good choice for the probability distribution of r.Coughs are a rare occurrence. It is reasonable to assume the events are independent.Coughs are a common occurrence. It is reasonable to assume the events are dependent.    Coughs are a common occurrence. It is reasonable to assume the events are independent.Coughs are a rare occurrence. It is reasonable to assume the events are dependent.(b) Find the probability of six or fewer coughs (in a large auditorium) in a 1-minute period. (Use 4 decimal places.) (c) Find the probability of at least five coughs (in a large auditorium) in a 27-second period. (Use 4 decimal places.)

A sample of 100 flounder of a certain species have

Question  A sample of 100 flounder of a certain species have sample mean weight 21.5 grams. Scientists want to perform a hypothesis test to determine how strong the evidence is that the mean weight differs from 20 grams. State the appropriate null and alternate hypotheses.

I don’t quite understand the question, could u plz solve

Question I don’t quite understand the question, could u plz solve this for me? thx! (by using R program) src=”/qa/attachment/8318645/” alt=”屏幕快照 2019-07-01 下午4.22.00.png” /> ATTACHMENT PREVIEW Download attachment 屏幕快照 2019-07-01 下午4.22.00.png Part (II) Use R for your computations to the following problems. Please attach the R commands, output and graphs that you used to answer the question. The R output alone is not an answer to the question. Please write a sentence or two to properly answer each question. Question 5: Assume that the distribution of the duration of human pregnancies can be approximated with a normal distribution with a mean of 266 days and a standard deviation of 16 days. (a) What percentage of pregnancies have a duration between 260 and 280 days? (b) Find a value :1: such that 10% of the pregnancies of a duration that is longer than :1: days.

“According to the February 2008 Federal Trade Commision report on

Question “According to the February 2008 Federal Trade Commision report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. In that year, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints. Does this data proivde enough evidence to show that Alaska had a lower proportion of identity theft than 23%? Test at the 5% level

I do not understand what the last part is asking.

Question I do not understand what the last part is asking. ATTACHMENT PREVIEW Download attachment Screen Shot 2019-07-01 at 4.23.05 PM.png Consider a finite population with five elements labeled A, B, C, D, and E. Ten possible simple random samples of size 2 can be selected. (a) List the 10 samples beginning with AB, AC, and so on. ( Enter your answers as a comma-separated list.) (b) Using simple random sampling, what is the probability that each sample of size 2 is selected? (Enter your probability as a fraction.) (c) Assume random number 1 corresponds to A, random number 2 corresponds to B, and so on. List the simple random sample of size 2 that will be selected by using the random digits 9, 0, 3, 6, 3, 2, 1. Need Help? mamun Mann |Back to a Tufees

Future Electronics makes compact disc players. Its research department found

Question Future Electronics makes compact disc players. Its research department found that the life of the laser beam device is normally distributed, with mean 5020 hours and standard deviation 500 hours.(a) Find the probability that the laser beam device will wear out in 5000 hours or less. (Round your answer to four decimal places.) (b) Future Electronics wants to place a guarantee on the players so that no more than 12% fail during the guarantee period. Because the laser pickup is the part most likely to wear out first, the guarantee period will be based on the life of the laser beam device. How many playing hours should the guarantee cover? (Round your answer down to the nearest hour.)

Academic advising: In 2014, the Community College Survey of Student

Question Academic advising: In 2014, the Community College Survey of Student Engagement reported that 32% of the students surveyed rarely or never use academic advising services. Suppose that in reality, 42% of community college students rarely or never use academic advising services at their college. In a simulation we select random samples from this population. For each sample we calculate the proportion who rarely or never use academic advising services.If we repeatedly obtain random samples of 200 students, what will be the standard deviation of the sampling distribution of sample proportions?0.0350.420.0012

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I followed the video and for some reason got the

Question I followed the video and for some reason got the wrong answer ATTACHMENT PREVIEW Download attachment Screen Shot 2019-07-01 at 4.51.04 PM.png You may need to use the appropriate appendix table or technology to answer this question. A population has a mean of 800 and a standard deviation of 200. Suppose a sample of size 400 is selected and x is used to estimate . (Round your answers to four decimal places.) (a) What is the probability that the sample mean will be within 15 of the population mean? 9544 X (b) What is the probability that the sample mean will be within 110 of the population mean? Need Help? Read It Watch It Talk to a Tutor

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