The Vitamin A levels in livers (μmol/L) between normal rats and radiation exposure rats are shownin the below table.GroupMean(μ)3525Standard deviation(0)23Sample number(n)50Normal ratsRadiation exposure rats(1) What is the standard error of mean (SEM) for each group?(2) Assume that the central-limit theorem is applicable. What does it mean in this context?40

Group meanstandard deviationsample sizeNormal rats35250Radiation exposure rats25340We have to find…

Answered: The Vitamin A levels in livers (µmol/L)…

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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Q scores are normally distributed with a mean of 95 and a standard deviation of 18. Assume that many samples of size n are taken from a large population of people and the mean IQ score is computed for each sample. a. If the sample size is n=81, find the mean and standard deviation of the distribution of sample means. The mean of the distribution of sample is __ The standard deviation of the distribution of sample means is__ (type an integer or decimal rounded to the nearest tenth as needed) b. If the sample size is n=169, find the mean and standard deviation of the distribution of sample means. The mean of distribution of sample means is __ The standard deviation of the distribution of sample means is __ (type an iteger or decimal rounded to the nearest tenth as needed) c. Why is the standard deviation in part A. different from the standard deviation in part B? Choose the correct answer below A. With smaller sample sizes (as in part a), the means tend to be further apart, so they have…
The tin and zinc contents of a brass sample are analyzed with the following results yielded: a) Zn: 33.27, 33.37, and 33.34%; b) Sn: 0.022, 0.025 and 0.026% Which of the following is the correct mean value, the calculated standard deviation "s" and the coefficient of variation "cov" for Tin? (Answers in that sequence) mean-33.33%;s-0.052% and cov -0.16% mean-0.024%;s-0.032% and cov-5.3% mean-0.024%;s-0.0034% and cov - 7,5% mean-0.024%s-0.0021% mean-0.024%;s-0.0021% and cov-8.8% and cov can not be calculated from the data set supplied
What is the interquartile range of these ages?
If a Type I error is made, what conclusion will be drawn regarding the mean breaking strength?
Use the 1odwing Wildlife: Coyotes A random sample of 46 adult coyotes in a region of northern Minnesota showed the average age to be z = 2.05 years, with sample standard deviation 8 = 0.82 years (based on information from the book Coyotes: Biology, Behavior and Management by M. Bekoff, Academic Press). However, it is thought that the overall population mean age of coyotes is µ=1.75. Do the sample data indicate that coyotes in this region of northern Minnesota tend to live longer than the average of 1.75 years? Use a = 0.01. Fl in the blanks a 007 • Ho u- 175 years * H u> 176 yean QUESTION 7 What is your test statistic? (t-statistic) 2.48 QUESTION 8 What is your p-value? (Hint You will need the degrees of freedom to find the p-value.) <0.01d QUESTION 9 Do you reject the null hypothesis? O Yes No
DO MALES AND FEMALES DIFFER ON ATTITUDES TOWARD CRIME ? MALES N=12 FEMALES N=13 CRIME ATTITUDE MALES 77, 97, 28, 55, 75, 43, 21, 70, 91, 22, 58, 45 FEMALES 32,31, 89, 21, 79, 67, 41, 24, 43, 87, 64, 28, 21 MEAN SD MALE FEMALE • DV= 1 •T-VALUE • df • SIGNIFICANT (Y/N). • TEST STATISTIC RAN • ACCEPT OR REJECT NULL HYPOTHESIS. • CONCLUSION:
The table below summarises data collected from male and female children about the time they spent on reading per week (X, in hours/week). Male Female All Sample size 20 22 ? 42 Sum values ? ? (i) of X Sum of X² values 8,648 17,352 (ii) 26000 Mean 19.9 27 ? 46.9 Standard Deviation 6.19 7.91 ? 14.1
D A small amount of the trace element selenium, 50-200 micrograms (ug) per day, is considered essential to good health. Suppose that random samples of 50 adults were selected from two regions of the United States and that each person's daily selenium intake was recorded. The means and standard deviations of the selenium daily intakes for the two groups are shown in the table. Sample Mean (ug) Sample Standard Deviation (ug) Find a 95% confidence interval for the difference (₁-₂) in the mean selenium intakes for the two regions. (Round your answers to three decimal places.) 49 Region Phase One-Tw_docx 1 167.4 22.8 ug to Interpret this interval, O The probability that the difference in the average daily selenium intake for the populations in Region 1 and Region 2 is exactly in the middle of the Interval is 95%. A 2 140.5 17.3 O The probability that the difference in the average daily selenium intake for the populations in Region 1 and Region 2 is within the interval is 95%. We are 95%…
10 points Q/ In the vehicle speed test, the results are given in the table below: Find 1. Arithmetic mean 2. Standard deviation 3. Skew coefficient 4. Kurtosis coefficient Speed range (km/hr Speed range (km/hr Speed range (km/hr Frequency(fi) Frequency(fi) Frequency(fi) 34-35.9 44-45.9 22 54-55.9 13 36-37.9 46-47.9 8 56-57.9 38-39.9 4 48-49.9 18 58-59.9 6. 40-41.9 10 50-51.9 17 60-61.9 42-43.9 7 52-53.9 16 62-63.9 3 add file
Need help with question iv) and v)
What is the test statistic? χ02 =
7. Ecological Applications study on the development following wildfires in the Pacific Northwest. One variable of interest to the researcher was tree diameter at breast height 110 years after the fire. The population of Douglas fir trees was shown to have an approximately normal distribution with mean tree diameter ? = 50 centimeters (cm) and ? = 14 cm. Find the diameter, d, such that 20.9% of the Douglas fir trees in the population have diameters that exceed d. published a of forests

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