Calculate the test statistic and determine the P-value. (Round your test statistic to two decimal places and your P-value to four decimalplaces.)P-value=State the conclusion in the problem context.O Reject the null hypothesis. There is not sufficient evidence that average daily zinc intake falls below 15 mg/dayO Do not reject the null hypothesis. There is sufficient evidence that average daily zinc intake falls below 15 mg/dayO Reject the null hypothesis. There is sufficient elsdence that average daily zinc intake falls below 15 mg/dayO Do not reject the null hypothesis. There is not sufficient evidence that average daily zinc intake falls below 15 mg/day. The recommended daily dietary allowance for zinc among males older than age 50 years is 15 mg/day. An article reports the followingsummary data on intake for a sample of males age 65-74 years: n-118, x- 12.2, and s-6.96, Does this data indicate that averagedaily zinc intake in the population of all males age 65-74 falls below the recommended allowance? (Use a -0.05.)State the appropriate null and alternative hypotheses,ⒸHR=15OH₂ = 15H₂p=15H₂ <15ⒸH-15H₁ s 15Calculate the test statistic and determine the P-valde (Round your test statistic to two decimal places and your P-value to four decimal

The mean significance between the sample and population tested using hypothesis testing. If the…

Answered: The recommended daily dietary allowance…

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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Consider the following sodium content of 26 randomly picked canned food. Propose appropriate data analysis to extract significant information from the data. Perform the analysis, discuss the findings, and state conclusion in a video and submit the video for evaluation. Please show the step by step of the solution using SPSS/Minitab Sodium content (mg/kg): 12148, 10426, 10912, 9116, 13226, 11663, 11781, 10680, 8457, 10788, 12605, 10591, 11040, 10815, 12962, 11644, 10047, 10478, 10108, 12353, 11778, 11092, 11673, 8758, 11145, 11495.
here are the 25 data set 4;4;5;5;5;6;6;6;6;7;7;7;7;8;8;8;9;9;9;10;10;10;11; 12;12 mean :7.64 sd: 2.3431 9.) State whether it is appropriate to use the Empirical rule to approximate the proportion of data between a set of values for you data set? Why or why not?
For a normally distributed data set with mean 20 and standard deviation 4. According to the 68-95-99.7 rule, what percentage of the data lies between 8 and 32?
Consider the following measurements of blood hemoglobin concentrations (in g/dL) from three human populations at different geographic locations: population1 = [ 14.7 , 15.22, 15.28, 16.58, 15.10 ] population2 = [ 15.66, 15.91, 14.41, 14.73, 15.09] population3 = [ 17.12, 16.42, 16.43, 17.33] Perform ANOVA to check if any of these populations have different mean hemoglobin concentrations. (Assume that all the ANOVA requirements such as normality, equal variances and random samples are met.) After you perform ANOVA perform a Tukey-Kramer post-hoc test at a significance level of 0.05 to see which populations actually have different means. As usual, round all answers to two digits after the decimal point. (Make sure you round off to at least three digits any intermediate results in order to obtain the required precision of the final answers.) For any questions, which ask about differences in means or test statistics, which depend on differences in means provide absolute values. In…
Compare risk factors in men and women. Complete the following table to compare men and women using data collected at the first examination for each participant (n = 4434). Summarize your results in three to four sentences. Patient Characteristic* Men (n = 1944) Women (n = 2490) Age, years 49.79 (8.72) 50.03(8.64) Systolic blood pressure, mmHg 131.74 (19.44) 133.82(8,64) Hypertension 83.71(11.44) 82.60 (12.50) Use of anti-hypertensive medication 42 (2.19) 102 (4.16) Current smoker 1175 (60.44) 1006 (40.4) Total serum cholesterol, mg/dL 233.58 (4236) 239.68 (46.22) Serum Glucose Stroke * Mean (Standard deviation) or n (%
The summary statistics for the amount of rainfall received in San Diego, CA over the past 117 years is as follows: N Mean Median Std. Dev. Minimum Maximum 117 14.91 13.07 6.75 4.85 38.18 A California newspaper article reported that, “in 2008 only 10 inches of rain fell, an extremely dry year and a rare occurrence for the city”. Comment on the accuracy of this statement
The leader of two postpartum women’s support groups is interested in the depression levels of the women in her groups. She administers the Center for Epidemiologic Studies Depression Scale (CES-D) screening test to the members of her groups. The CES-D is a 20-question self-test that measures depressive feelings and behaviors during the previous week. The mean depression level from the screening test for the 10 women in the first group is μ₁ = 16; the mean depression level for the 14 women in the second group is μ₂ = 10. Without calculating the weighted mean for the combined group, you know that the weighted mean is:
9.85 Iron Deficiency? Iron is essential to most life forms and to normal human physiology. It is an integral part of many proteins and enzymes that maintain good health. Recommendations for iron are provided in Dietary Reference Intakes, developed by the Institute of Medicine of the National Academy of Sciences. The recommended dietary allowance (RDA) of iron for adult females under the age of 51 is 18 milligrams (mg) per day. The following iron intakes, in milligrams, were obtained during a 24-hour period for 45 randomly selected adult females under the age of 51. 15.0 18.1 16.0 12.6. 15.3 9.4 14.6 11.9 17.0 6.3 18.2 18.3 15.0 10.9 18.1 19.8 11.6 12.8 15.6 11.0 14.5 16.6 11.5 16.4 12.5 13,1 10.7 12.1 17.3 12.4 16.3 14.7 12.7 16.3 11.5 = 14.6 14.4 16.6 20.7 19.5 18.3 12.5 18.6 16.8 12.5 At the 1% significance level, do the data suggest that adult females un- der the age of 51 are, on average, getting less than the RDA of 18 mg of iron? Assume that the population standard deviation is…
Consider the following measurements of blood hemoglobin concentrations (in g/dL) from three human populations at different geographic locations: population1 = [ 14.7 , 15.22, 15.28, 16.58, 15.10 ] population2 = [ 15.66, 15.91, 14.41, 14.73, 15.09] population3 = [ 17.12, 16.42, 16.43, 17.33] Perform ANOVA to check if any of these populations have different mean hemoglobin concentrations. (Assume that all the ANOVA requirements such as normality, equal variances and random samples are met.) After you perform ANOVA perform a Tukey-Kramer post-hoc test at a significance level of 0.05 to see which populations actually have different means. As usual, round all answers to two digits after the decimal point. (Make sure you round off to at least three digits any intermediate results in order to obtain the required precision of the final answers.) For any questions, which ask about differences in means or test statistics, which depend on differences in means provide absolute values. In…
An MP of one city constituency complained in parliament that his constituency consists of mainly poorly paid workers. His colleagues demanded that he substantiates his claim. Mtuwetu hired a statistician who collected data on monthly salaries from a sample of 500 residents of the constituency. The statistician recorded the following data: Monthly salary (ksh 000) No. of residents 0-5 50 5 – 10 170 10 – 15 210 15 – 20 40 20 – 25 20 25 – 30 10 Suppose ksh 8,000 and 22,000 are the divided salary points of lower, middle, and upper-income groups. What proportion of the constituents fall in each of the income groups
Please provide Handwritten answer
Humerus bones from the same species of animal have approximately the same length-to-width ratios. It is known that Species A has a mean ratio of 5.9. Suppose that 100 fossil humerus bones were unearthed at a site where Species A is known to have flourished. (We assume that all bones are from the same species.) The length-to-width ratios of these bones have a sample mean 5.6 and sample standard deviation 0.25. Can we conclude that the mean ratio of these bones is significantly smaller than those of Species A? Test with a 0.01 significance level. What is the standard error?

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