In 2006 the mean weight of men was believed to be 86.8 kg. To testfor an increase in the weight a random sample of 100 men showed amean weight of 88.7 kg with a sample standard deviation of 11.64 kg.The null and alternative hypotheses are:Họ: µ = 86.8 ; H;: µ > 88.7Họ: u = 86.8 ; H1: µ < 86.8С.А.В.%3DHọ: µ = 86.8 ; H;: µ > 86.8Họ: µ = 86.8 ; H;: 86.8 < µ< 88 .7D.А. АВ. ВС. СD. D

Given : In 2006 the mean weight of men was believed to be 86.8 To test for an increase in the…

Answered: In 2006 the mean weight of men was…

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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In a sample of 14 randomly selected high school seniors, the mean score on a standardized test was 1197 and the standard deviation was 166.9. Further research suggests that the population mean score on this test for high school seniors is 1018. Does the t-value for the original sample fall between -to.95 and to.95? Assume that the population of test scores for high school seniors is normally distributed. fall between -to.95 and to.95 because to.95 = The t-value of t= (Round to two decimal places as needed.) C
A recent study showed that the average number of sticks of gum a person chews in a week is 10. A college student believes that the guys in his dormitory chew more gum in a week. He conducts a study and samples 10 of the guys in his dorm and finds that on average they chew 11 sticks of gum in a week with a standard deviation of 3.7. Test the college student's claim at a=0.10. The correct hypotheses would be: Ο Η: μ 10 (claim) Ο Η0: μ Σ 10 HA:µ < 10 (claim) Ο Η: μ = 10 HA:µ + 10 (claim) Since the level of significance is 0.10 the critical value is 1.383 The test statistic is: (round to 3 places) The p-value is: (round to 3 places) The decision can be made to: O reject Ho O do not reject Ho The final conclusion is that: O There is enough evidence to reject the claim that more gum in a week. O There is not enough evidence to reject the claim that more gum in a week. O There is enough evidence to support the claim that more gum in a week. O There is not enough evidence to support the claim…
the mean and standard deviation of the body mass index (BMI) for a sample of 10,215 women who drink at least one caffeinated carbonated beverage a day are 25.7 and 5.3 Approximately 68% of women in the sample have BMI between
The 2015 general Social Survey asked a large number of people how much time they spent watching TV each day. The mean number of hours was 4.2 with a standard deviation of 2.8. Assume that in a sample of 39 teenagers, the sample standard deviation of daily TV time is 2.8 hours, and that the population of TV watching times is normally distributed. Under 10% significance level can you conclude that the population standard deviation of TV watching times for teenagers is less than 2.8? Procedure: Select an answer Assumptions: (select everything that applies) Simple random sample Population standard deviation is unknown O Sample size is greater than 30 Normal population Population standard deviation is known The number of positive and negative responses are both greater than 10
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3.3.15 Use z scores to compare the given values. Based on sample data, newborn males have weights with a mean of 3293.7 g and a standard deviation of 551.3 g. Newborn females have weights with a mean of 3000.6 g and a standard deviation of 590.5 g Who has the weight that is more extreme relative to the group from which they came; a male who weighs 1700 g or a female who weighs 1700 g? V has the weight that is more extreme. Since the z score for the male is z= and the z score for the female isz= the (Round to two decimal places.)
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A recent study showed that the average number of sticks of gum a person chews in a week is 14. A college student believes that the guys in his dormitory chew less gum in a week. He conducts a study and samples 25 of the guys in his dorm and finds that on average they chew 17 sticks of gum in a week with a standard deviation of 3.1. Test the college student's claim at a=0.10. The correct hypotheses would be: Ο Η: μ 14 HA: 14 (claim) Ho: μ = 14 HA: The p-value is: 14 (claim) Since the level of significance is 0.10 the critical value is -1.318 The test statistic is: The decision can be made to: O reject Ho do not reject Ho (round to 3 places) (round to 3 places) The final conclusion is that: O There is enough evidence to reject the claim that less gum in a week. There is not enough evidence to reject the claim that less gum in a week. O There is enough evidence to support the claim that less gum in a week. O There is not enough evidence to support the claim that less gum in a week.

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