Group 1Group 2Group 3Group 4Mean57.6Mean29.93333333 Mean30.8Mean97.86666667Standard Error5.315565642 Standard Error2.411217173 Standard Error11.16209572 Standard Error0.16523192Median65Median32Median7Median98Mode#N/AMode22Mode100Mode98Standard Deviation20.58709721 Standard Deviation9.338603954 Standard Deviation43.23061085 Standard Deviation0.639940473Sample Variance423.8285714 Sample Variance87.20952381 Sample Variance1868.885714 Sample Variance0.40952381Kurtosis-0.443312906 Kurtosis-0.963469672 Kurtosis-0.739206836 Kurtosis5.639430877Skewness-0.320329658 Skewness0.128440214 Skewness1.16990546Skewness-1.783484308Range76Range31Range98Range3Minimum18Minimum14MinimumMinimum96Maximum94Maximum45Maximum100Maximum99Sum864Sum449Sum462Sum1468Count15Count15Count15Count15Largest(1)Smallest(1)Largest(1)Smallest(1)Largest(1)9445100Largest(1)99Smallest(1)18142Smallest(1)96Confidence Level(95.0%) 11.40075443 Confidence Level(95.0%) 5.171546494 Confidence Level(95.0%) 23.94031432 Confidence Level(95.0%) 0.354387222Analyze the result between the groups in terms of mean and standard deviation.

Answered: Image /qna-images/answer/29c3e655-18c3-410e-92ef-23f059b5da30.jpg

Answered: Group 1 Group 2 Group 3 Group 4 Мean…

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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7
How many households out of the city population of 16,245 households had discarded paper goods weighing between 13.66 and 14.44 pounds. mean = 9.42806451612903 standard deviation= 4.16806903387879
Compute the mean and standard deviation of the provided sample data on cost of living adjustments. Then compute the z-scores for the smallest and largest comparative salaries and housing adjustments and interpret your results. "The z-scores show that the comparative salary amount for __ (St. Louis/New York/Columbus) is the farthest away from the __ (variance/standard deviation/mean) comparative salary amount. The z-scores also show that the housing amount for ___(St. Louis/New York/Columbus) is the fate these away from the ___ (variance/standard deviation/mean)."
An airline is charged by an airport based upon the total taxi-time that an airplane spends to take off and land. Suppose taxi time, a, at airport A is normally distributed with mean of 27 minutes and standard deviation 8 minutes; taxi time, b, at airport B is normally distributed with mean of 39 minutes and standard deviation 12 minutes; taxi time, c, at airport C is normally distributed with mean of 41 minutes and standard deviation 10 minutes. JJ Airlines has a plane flying to airport A, airport B, and airport C on Sunday. Airport A charges $131 per minute of taxi time; airport B charges $150 per minute of taxi time; airport c charges $200 per minute of taxi time. Let X = total amount of taxi time in minutes for the JJ plane. Let W = total taxi charges for the JJ plane. X is defined as X = a + b + c. W = 131a + 150b + 200c. Note that a, b, and c are independent of one another. d) Calculate the standard deviation of W. e) If we pick a value k such that the probability that X > k…
The mean of the commute time to work for a resident of a certain city is 27.3 minutes. Assume that the standard deviation of the commute time is 7.1 minutes. What minimum percentage of commuters in the city has a commute time within 2.5 standard deviations of the mean? What are the commute times within 2.5 standard deviations of the mean?
According to the Census Bureau, the mean number of people in a typicalAmerican household in 2010 was 2.6 with a standard deviation of 1.5 - this is based on thecensus information. Suppose, instead the Census Bureau has estimated this mean using arandom sample of 225 homes - the sample mean came out to be 2.4 and standard deviation of1.4. What is the mean and standard error of the sampling distribution of the sample mean for225 homes ?
Suppose that the annual employer participation in the Canada/Quebec Pension Plan (CPP/QPP) per employee is normally distributed with a standard deviation of $625, but the mean is unknown. If 73.89% of such employer contributions are greater than $1,700, what is the mean annual employer contribution per employee? Suppose the mean annual CPP/QPP employer contribution per employee is $2,259 and the standard deviation is $625. If such employer contributions are normally distributed, 31.56% of the contributions are greater than what value?
We have to compare the result of all variables using range,interquatile range,variance and standard deviation of male and female
21)
Suppose that a category of world class runners are known to run a marathon (26 miles) in an average of 146 minutes with a standard deviation of 16 minutesConsider 49 of the races Find the probabillity that the average of the sample will be between 144 and 146 min in the 49 marathons B find the 80th percentile for the average of these 49 marathons C find the median
9) Steel rods are manufactured with a mean length of 25 centimeters (cm). Because of variability in the manufacturing process, the lengths of the rods are approximately normally distributed, with a standard deviation of 0.07 cm. What proportion of rods has a length less than 24.9 cm? b. C. d. 25 Any rods that are shorter than 24.85 cm or longer than 25.15 cm are discarded. What proportion of rods will be discarded? Using the results of part (b), if 5,000 rods are manufactured in a day, how many should the plant manager expect to discard? If an order comes in for 10,000 steel rods, how many rods should the plant manager manufacture if the order states that all rods must be between 24.9 cm and 25.1 cm?
SAMPLE 1 MEAN 5.61 MEDIAN 5.6 MODE 5.6 STANDARD ERROR 0.39992384 STANDARD DEVIATION 2.1904731 SAMPLE VARIANCE 4.79817241 KUTOSIS -0.07445435 SKEWNESS -0.69709258 RANGE 7.9 MINIMUM 1.1 MAXIMUM 9 SAMPLE COUNT 30 SUM 168.3 SAMPLE 2 MEAN 7.326666667 MEDIAN 7.3 MODE 8.7 STANDARD ERROR 0.248717785 STANDARD DEVIATION 1.362283411 SAMPLE VARIANCE 1.855816092 KUTOSIS -0.118922156 SKEWNESS -0.015286932 RANGE 5.6 MINIMUM 4.6 MAXIMUM 10.2 SAMPLE COUNT 30 SUM 219.8 AND THIS IS MY t-TEST-ASSUMING EQUAL VARIANCE Variable 1 Variable 2 Mean 5.61 7.326666667 Variance 4.798172414 1.855816092 Observations 30 30 Pooled Variance 3.326994253 Hypothesized Mean Difference 0 df 58 t Stat -3.645067529 P(T<=t) one-tail 0.000286392 t Critical one-tail 1.671552762 P(T<=t) two-tail 0.000572783 t Critical two-tail 2.001717484 IS IT RIGHT THAT MY T-STAT CAME OUT AS A NEGATIVE, AND ARE MY CALCULATIONS…

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