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#50.If α = .05 and β = .25, calculate the

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Question #50.If α = .05 and β = .25, calculate the following:  The statistical decision is to reject the null, and H0 (the null) is really true (i.e., a Type I error)  .05.99.20.95.01#51.If α = .05 and β = .25, calculate the following:  The statistical decision is to fail reject the null, and H0 (the null) is really true (i.e., a correct decision)  .05.75.20.01.95#52.If α = .05 and β = .25, calculate the following:  The statistical decision is to reject the null, and H0 (the null) is really false (i.e., Power)  .05.25.75.95.01#53.If α = .05 and β = .25, calculate the following:  The statistical decision is to fail to reject the null, and H0 (the null) is really false (i.e., a Type II error)  .05.25.75.95.01

A random sample of 49 style=”color:rgb(0,0,0);background-color:transparent;”> cans of soda is

Question A random sample of 49 style=”color:rgb(0,0,0);background-color:transparent;”> cans of soda is obtained and the contents are measured. The sample mean is 12.04 oz and the standard deviation is 0.11 oz. Test the claim that the contents of all such cans have a mean different from 12.00 ​oz, as indicated by the label. Use a 0.05 significance level. State what is the Null and alternative hypotheses? What is the Z score? (round to two decimal places). Find the P value? (round to four decimal places). State the conclusion choose the correct answer:A.The​ P-value is greater than the significance level. There is sufficient evidence to support the claim that the contents of all such cans have a mean different from 12.00 oz.B.The​ P-value is less than or equal to the significance level. There is not sufficient evidence to support the claim that the contents of all such cans have a mean different from 12.00 oz.C.The​ P-value is less than or equal to the significance level. There is sufficient evidence to support the claim that the contents of all such cans have a mean different from 12.00 oz.D.The​ P-value is greater than the significance level. There is not sufficient evidence to support the claim that the contents of all such cans have a mean different from 12.00 oz.

Looking to see if I got this correct

Question Looking to see if I got this correct

Assist for a better understanding below: ATTACHMENT PREVIEW Download attachment

Question Assist for a better understanding below: ATTACHMENT PREVIEW Download attachment C1.jpg An animal sanctuary houses a variety of animals. To understand the workload of the sanctuary’s veterinarians, the director looked at which types of animals are treated or not treated by various doctors. The Venn diagram shows this Information for three of the doctors. All ammals housed- (a) Select all the doctors who treat cats. Treated by Or. Cook Treated by Dr. Young Dr. Cook Dr. Young ) Dr. Watson Rabbits Koalas (b) How many types of animals are treated by Dr. Cook or Dr. Young for both)? Dogs Chits (c) Which types of animals are treated only by one of the three doctors [Dr. Cook, Dr. Young, or Or! Watson)? Choose all that apply. Llamas Dogs Elks Frogs Horses Iquanas Koalas Lamas Pigs Rabbits Horses I Snakes Turtles Zebras Treated by Or. Watson Turtles X

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Question alt=”Capture15.2.PNG” />I tried to enter the function in excel but cant get any of the options Attachment 1 Attachment 2 ATTACHMENT PREVIEW Download attachment Capture15.2.PNG ATTACHMENT PREVIEW Download attachment Capture15.PNG

Please check to see if I did this correct and

Question Please check to see if I did this correct and if wrong give correct answer: src=”/qa/attachment/9312643/” alt=”Screen Shot 2019-08-09 at 2.33.03 PM.png” /> ATTACHMENT PREVIEW Download attachment Screen Shot 2019-08-09 at 2.33.03 PM.png An important application of regression in manufacturing is the estimation of cost of production. Based on DATA from Ajax Widgets relating cost (Y) to volume (X), what is the cost of producing 600 widgets? 0 None of the answers are correct.

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Question />Assist with above to get a better understanding of the entire selection. Please read carefully. Attachment 1 Attachment 2 ATTACHMENT PREVIEW Download attachment C2.jpg A college radio station surveyed 252 Incoming freshmen to gather Information about the genres of music that they like. The table below gives the results for two of the genres. Number of freshmen Like classical 76 Like rap 195 Like both classical and rap 48 Construct a Venn diagram Illustrating these results. Then answer the questions All freshmen In the survey How many freshmen like rap but not classical? freshmen Like rap How many freshmen like classical or rap (or both)? freshmen X ATTACHMENT PREVIEW Download attachment C2C.jpg Like classical Like rap How many freshmen like classical or rap (or both)? freshmen X ? OO O

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Question />Please help me – i cannot get any of the options listed. Missing a step Attachment 1 Attachment 2 ATTACHMENT PREVIEW Download attachment Capture21.1.PNG ATTACHMENT PREVIEW Download attachment Capture21.PNG

You are performing a Segmentation Analysis and want to target

Question You are performing a Segmentation Analysis and want to target customers that are most likely to respond. A cell in your analytic sample has a response rate of 4%, and the overall response rate for all cells is 1.5% If your targeted group must have a response rate index greater than 1.0, would this cell be part of the Target group –  Yes or No, and explain why. 

Aardvark Auto Rental rents vehicles. Each day Aardvark Automotive Rental

Question Aardvark Auto Rental rents vehicles. Each day Aardvark Automotive Rental keeps track of whether its SUVs were returned in “failing” condition. Aardvark must use more of its resources to restore each SUV with a “failing” returned condition. Aardvark rents an average of five SUVs each day. The condition of returned SUVs is recorded for many days in which five SUVs were rented.Download the file titled SUV Failures. It contains a scatter plot of the number of failures versus frequency. To compare the results to the binomial distribution, complete the following:Explain why SUV rental condition scenario can be a binomial experiment.Using the SUV Failure scatter plot, construct a frequency distribution for the number of failures.Compute the mean number of failures. The formula for the mean is as follows: ∑(x⋅f)∑fThe terms x represent the total number of failures (0, 1, 2, 3, 4, 5) and f is the corresponding frequency (number of days where xᵢ failures occurred). Explain what the numerical result means. From the frequency distribution, construct the corresponding relative frequency distribution.Explain why the relative frequency distribution table is a probability distribution.Then, use Excel to put together a scatter plot of the probability distribution:Select the two columns of the probability distribution. Click on INSERT, and then go to the Charts area and select Scatter. Then choose the first Scatter chart (the one without lines connecting). Using the frequency distribution, what is the SUV failure average? In part 3, note that the numerator in the formula for the mean is the total number of failures. The total number of trials is the denominator of the formula for the mean multiplied by 5. What does this average mean?The Binomial Distribution is uniquely determined by n, the number of trials, and p, the probability of “success” on each trial. Using Excel, construct the Binomial Probability Distribution for five trials, n, and probability of success, p, as the SUV failure average in part 5. Here is an explanation of the BINOM.DIST function in Excel: https://support.office.com/en-ie/article/BINOM-DIST-function-c5ae37b6-f39c-4be2-94c2-509a1480770c?ui=en-US

:The CEO says he’s heard you did some analysis on

Question :The CEO says he’s heard you did some analysis on the company’s return policy. He doesn’t want to see equations or calculations but wants you to explain to him what your EMV result means for the company and whether the policy is a good idea or not. How would you respond.

Imagine that you work in web media, and your company

Question Imagine that you work in web media, and your company publishes news and analysis. Success is measured by the amount of time visitors stay on the site consuming content (longer time is better). Two websites are created and tested: Deep-Analysis (DA) vs. Just-Headlines (JH). An analysis of the duration of time spent by unique visitors is as follows: Deep-Analysis: 1000 unique visitors; Duration= 2:40 (hh:mm); std.deviation = 25 minutes. (or in decimal numbers, mean=2.66, sd=0.42) Just-Headlines: 1000 unique visitors; Duration= 2:15 (hh:mm); std.deviation =30 minutes. (or in decimal numbers, SLS mean=2.25, sd=0.50) A.Test the null hypothesis claim at 90% confidence (or alpha=.10) that the difference in time spent between these two sites is not significantly different from zero. (i.e. Note that this means that the null hypothesis is claiming that the same amount of time is spent on these sites). Do you retain or reject the null hypothesis?B.Based on your analysis of comparing the time visitors spend on the DA and JH websites, which site would be more attractive for advertisers to use?  Explain.

If you calculate a 90% confidence interval to estimate the

Question If you calculate a 90% confidence interval to estimate the true difference in time spent between the sites, and also calculate a traditional hypothesis test (using an alpha=0.10), should the conclusion based on Confidence Intervals be the same as the conclusion your would reach in the hypothesis test? Explain or demonstrate via calculations.

Result probability 0 heads, 6 tails 1/641 heads, 5 tails

Question Result probability 0 heads, 6 tails 1/641 heads, 5 tails 3/322 heads, 4 tails 15/643 heads, 3 tails 5/164 heads, 2 tails 15/645 heads, 1 tail 3/32a. What is the probability that all six tosses are the same​ (all heads or​ tails)?The probability is ? .​ (Simplify your​ answer.)b. What is the probability that the six tosses are not all the​ same?The probability is ?(Simplify your​ answer.)c. What is the probability of getting one head and five tails when you toss six coins at​ once?The probability is ?(Simplify your​ answer.)(Show work)

A statistical test like the z test can me used

Question A statistical test like the z test can me used to test the null hypothesis. It is used when the population standard deviation is know and the variables is normally distributed, or when the sample size is greater than or equal to 30. Is this true or false

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Question alt=”Capture1.PNG” />I cannot get the solutions listed. Attachment 1 Attachment 2 ATTACHMENT PREVIEW Download attachment Capture.PNG ATTACHMENT PREVIEW Download attachment Capture1.PNG

If my chi square statistic is 9.94781 and df (r-1)*(c-1)

Question If my chi square statistic is 9.94781 and df (r-1)*(c-1) = 1 and alpha is 0.05….the chi square distribution is 3.84, I am should reject the null hypothesis because 9.94781 is way beyond p value of >0.01 (because when p value is less than alpha level, we are to reject null hypothesis) Is this correct?

How many years is a randomly selected 19​-year-old expected to

Question How many years is a randomly selected 19​-year-old expected to live beyond his or her 19th ​birthday? Use the data in the table for people between the ages of 16 and 21 years. Show workage interval probability of dying during interval # of surviving to the beg of interval # of deaths during the interval Expected remaining lifetime from the beg16-17 0.000647 98,972 64 61.317-18 0.000718 98,875 71 60.318-19 0.000779 98,812 77 59.219-20 0.000831 98,723 82 58.720-21 0.000872 98,657 86 57.7

Describe why hypothesis testing is important to businesses.(need some detailed

Question Describe why hypothesis testing is important to businesses.(need some detailed explanation)

Altitude- 1 8 13 23 28 32 34temperature 57 36

Question Altitude- 1 8 13 23 28 32 34temperature 57 36 23 -7 -31 -43 -56 />A)construct a scatterplot of the above numbersB) characterize the correlation in wordsC) Estimate the correlation coefficient r=?D) from the data it seems that as the aircraft gain altitude, the outside temperature appears to ? in a ? pattern.

#45.In the general population, patients who are recovering from bronchial

Question #45.In the general population, patients who are recovering from bronchial pneumonia normally take a mean (average) of 4.75 days to leave the hospital, with a population standard deviation of 1.82 days. Medical personnel at a hospital are trying a new medication that is designed to ease breathing during recovery. What is not currently known is whether the new medication will impact duration of hospital stay. The first sample of 25 patients (n=25) to whom this medication was administered were released from the hospital in a mean (average) of 2.60 days. Based on this result, should the hospital conclude that there is a significant difference in duration of hospital stay because of the new medication? The best conclusion for this problem is:There is a significant difference in the duration of hospital stay when patients administered the new medication are compared to the general population of patients not receiving this medication.There is no significant difference in the duration of hospital stay when patients administered the new medication are compared to the general population of patients not receiving this medication.Duration of hospital stay is significantly decreased for patients administered the new medication compared to the general population of patients not receiving this medication.Duration of hospital stay is significantly increased for patients administered the new medication compared to the general population of patients not receiving this medication.#46.Whenever we reject the null hypothesis, as researchers what type of error are we most concerned that we may have made? Standard errorNull hypothesis errorAlternative hypothesis errorType II statistical errorType I statistical error#47.In the general population, patients who are recovering from bronchial pneumonia normally take a mean (average) of 4.75 days to leave the hospital, with a population standard deviation of 1.82 days. Medical personnel at a hospital are trying a new medication that is designed to ease breathing during recovery. What is not currently known is whether the new medication will impact duration of hospital stay. The first sample of 25 patients (n=25) to whom this medication was administered were released from the hospital in a mean (average) of 2.60 days. Based on this result, should the hospital conclude that there is a significant difference in duration of hospital stay because of the new medication? Calculate the 99% confidence interval.  The mean that you will use in this calculation is ____:4.603.204.752.601.82#48.In the general population, patients who are recovering from bronchial pneumonia normally take a mean (average) of 4.75 days to leave the hospital, with a population standard deviation of 1.82 days. Medical personnel at a hospital are trying a new medication that is designed to ease breathing during recovery. What is not currently known is whether the new medication will impact duration of hospital stay. The first sample of 25 patients (n=25) to whom this medication was administered were released from the hospital in a mean (average) of 2.60 days. Based on this result, should the hospital conclude that there is a significant difference in duration of hospital stay because of the new medication? What is the critical value you will use for the 99% confidence interval? /- 1.65 /- 1.70 /- 1.96 /- 2.33 /- 2.58

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