During the nutrition research, the amount of consumed kilocalories per day was measured for 18 people - 10 women and 8 men.Results are as follows:Women: 1967, 1666, 1607, 1622, 1749, 1805, 1590, 1921, 1822, 1675;Men: 2076, 2071, 2167, 1885, 1841, 1868, 2299, 2119.Calculate a 90% confidence interval on the mean for women and men separately. Assume distribution to be normal.Round your answers to the nearest integer (e.g. 9876).Women:i

Confidence interval is the combined value of sample statistic and the margin of error. Margin of…

Answered: During the nutrition research, the…

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

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Joan’s finishing time for the Bolder Boulder 10K race was 1.67 standard deviations faster than the women’s average for her age group. There were 415 women who ran in her age group. Assuming a normal distribution, how many women ran faster than Joan?
The National Center for Education Statistics surveyed 4400 college graduates about the lengths of time required to earn their bachelor's degrees. The mean is 5.15 years, and the standard deviation is 1.68 years. Based enthuse sample data, construct the 99% confidence interval for the mean time required by all college graduates. left left: right:
Women are recommended to consume 1730 calories per day. You suspect that the average calorie intake is different for women at your college. The data for the 16 women who participated in the study is shown below: 1764, 1612, 1772, 1405, 1685, 1634, 1823, 1770, 1691, 1762, 1485, 1883, 1834, 1578, 1733, 1885 Assuming that the distribution is normal, what can be concluded at the a = 0.10 level of significance? a. For this study, we should use E-test for a population mean b. The null and alternative hypotheses would be: Ho: ?v|| Select an answer v H.: ?vSelect an answer v c. The test statistic ? (please show your answer to 4 decimal places.) d. The p-value = (Please show your answer to 4 decimal places.) e. The p-value is f. Based on this, we should fail to reject g. Thus, the final conclusion is that. O The data suggest that the population mean calorie intake for women at your college is not significantly different from 1730 at a = 0.10, so there is not enough evidence to conclude that the…
For each day of last year, the number of vehicles passing through a certain intersection was recorded by a city engineer. One objective of this study was to determine the percentage of days that more than 425 vehicles used the intersection. If the mean data was 375 vehicles per day and the standard deviation was 25 vehicles: (c) Suppose the relative frequency distribution for the data is bell-shaped, then what percentage of days between 325 and 425 vehicles used the intersection ?
According to the Center for Disease Control and Prevention (CDC), the mean life expectancy in 2015 for non‑Hispanic white males was 76.3 years. Assume that the standard deviation was 15 years, as suggested by the Bureau of Economic Research.The distribution of age at death, ????, is not normal because it is skewed to the left. Nevertheless, the distribution of the mean, x⎯⎯⎯, in all possible samples of size ???? is approximately normal if ???? is large enough, by the central limit theorem.Let x⎯⎯⎯ be the mean life expectancy in a sample of 100 non‑Hispanic white males. Determine the interval centered at the population mean ???? such that 95% of sample means x⎯⎯⎯ will fall in the interval. Give your answers precise to one decimal.
Reports on a student's test score such as the SAT or a child's height or weight usually give the percentile as well as the actual value of the variable. The percentile is just the cumulative proportion stated as a percent: the percent of all values of the variable that were lower than this one. The upper arm lengths of females in the United States are approximately Normal with mean 35.735.7 cm and standard deviation 2.12.1 cm, and those for males are approximately Normal with mean 39.139.1 and standard deviation 2.32.3 cm. Larry, a 6060‑year‑old male in the United States, has a upper arm length of 37.837.8 cm. What is his percentile? (Enter your answer rounded to two decimal places.)
The board of examiners that administers the real estate broker’s examination in a certain state found that the mean score on the test was 450 and the standard deviation was 50. If the board wants to set the passing scores so that only the top 5% of all applicants pass, what should the passing score be? Assume that the scores are normally distributed.
For the normal distribution of adult males in North America (mean=70.0 inches, standard deviation=4.0 inches), find the proportion of those who are a) under 65 inches b) over 76 inches c) between 67 and 73 inches
In the US women between the ages of 18 to 50 have shoe sizes that approximately follow a normal distribution. The approximate mean is shoe size 8.43 and the standard deviation is 1.35. Find the 90th percentile of this distribution, write the calculator function used to compute the result, and interpret it in a complete sentence.
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A parenting magazine reports that the average amount of wireless data used by teenagers each month is 5 Gb. For her science fair project, Ella sets out to prove the magazine wrong. She claims that the mean among teenagers in her area is less than reported. Ella collects information from a simple random sample of 25 teenagers at her high school, and calculates a mean of 4.7 Gb per month with a standard deviation of 0.9 Gb per month. Assume that the population distribution is approximately normal. Test Ella's claim at the 0.01 level of significance. Step 3 of 3: Draw a conclusion and interpret the decision. E Tables E Keypad Answer Keyboard Shortcuts We reject the null hypothesis and conclude that there is sufficient evidence at a 0.01 level of significance that the average amount of wireless data used by teenagers each month is less than 5 Gb. We fail to reject the null hypothesis and conclude that there is insufficient evidence at a 0.01 level of significance that the average amount of…
According to the records of an electric company serving the Boston area, the mean electric consumption for all households during winter is 1650 kilowatt-hours per month. Assume that the monthly electric consumptions during winter by all households in this area have a normal distribution with a mean of 1650 kilowatt-hours and a standard deviation of 320 kilowatt-hours. What percentage of the households in this area have a monthly electric consumption of 1838 to 1940 kilowatt-hours? Round your answer to two decimal places. P= Enter your answer in accordance to the question statement %

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