A nutritionist claims that the mean tuna consumption by a person is 3.3 pounds per year. A sample of 90 people shows that the mean tuna consumption by a person is3.1 pounds per year. Assume the population standard deviation is 1.08 pounds. At a =0.04, can you reject the claim?(a) Identify the null hypothesis and alternative hypothesis.O A. H,: us3.3О в. Но: и> 3.3O C. H,: µ>3.1Ha: H53.1Ha: u> 3.3Ha: u53.3O F. Ho:H#3.1O E. Ho: H= 3.3Ha: H#3.3O D. Ho: us3.1Ha:=3.1Ha: µ> 3.1

Givensample size(n)=90Mean(x)=3.1population standard deviation(σ)=1.08α=0.04

Answered: A nutritionist claims that the mean…

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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A standard I.Q. test produces normaly distributed results 12 with a mean of 104 and a standard deviation of 16 for 52 000 students in grade 12 in the province. Approximately how many of these people would have LQ.s below 76? A 2080 B 4170 C 23 915 D 49915 e 2021 EducAde Safware ie Alrights reserved.
A nutritionist claims that the mean tuna consumption by a person is 3.4 pounds per year. A sample of 90 people shows that the mean tuna consumption by a person is 3.2 pounds per year. Assume the population standard deviation is 1.11 pounds. At a = 0.04, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. H0: με3.4 Haiµ> 3.4 B. H0: μ>3.4 Hai us3.4 C. H : μ>3.2 Hai us3.2 D. Ho: H#3.2 Haiu= 3.2 E. Ho: µs3.2 Ha: µ> 3.2 F H: μ=3.4 Ha: u#3.4 (b) Identify the standardized test statistic. (Round to two decimal places as needed.)
pls. answer using 7 steps:-Parameter of interest-Null hypothesisAlternative hypothesis -Test statistic -Reject H0 if: -Computations: -Conclusion
We want to obtain a sample to estimate a population mean. Based on previous evidence, researchers believe the population standard deviation is approximately σ=26.1. We would like to be 99% confident that the estimate is within 3 of the true population mean. How large of a sample size is required?
A random sample is obtained from a normal population with a mean of u=95 and a standard deviation of o=40. The sample mean is M=86. a.is the a representative sample mean or an extreme value for a sample of n=16 scores?
A nutritionist claims that the mean tuna consumption by a person is 3.1 pounds per year. A sample of 50 people shows that the mean tuna consumption by a person is 2.9 pounds per year. Assume the population standard deviation is 1.08 pounds. At a= 0.09, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. OB. Ho Ha p> 2.9 OC. H, p#2.9 Ha H=2.9 µ32.9 us2.9 A. Ho H= 3.1 Ha H#3.1 OD. H, H>3.1 Ha µ53.1 OE H, s3.1 Ha p>31 OF Ho p> 2.9 Hps2.9 s. every ti s hir wing. es (b) Identify the standardized test statistic. Home z= - 1.31 (Round to two decimal places as needed.) (c) Find the P-value. ments (Round to three decimal places as needed.) book t: 0 ter Contents Clear all Check answer ltimedia Library View an example Get more help- Help me solve this zudy Plan 11:17. OK Innls for Success 68F ^@口a a 12/4/2 99+ A6 A Type here to search 10/05/17 97 PrtSc Insert Delete F10 F11 F12 F6 F7 FB F9 F1 F2 F3 F4 F5 Backspace Num Lock
A nutritionist claims that the mean tuna consumption by a person is 3.4 pounds per year. A sample of 90 people shows that the mean tuna consumption by a person is 3.3 pounds per year. Assume the population standard deviation is 1.02 pounds. At α=0.03, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. H0: μ=3.4 Ha: μ≠3.4 Your answer is correct. B. H0: μ>3.4 Ha: μ≤3.4 C. H0: μ≤3.3 Ha: μ>3.3 D. H0: μ≤3.4 Ha: μ>3.4 E. H0: μ>3.3 Ha: μ≤3.3 F. H0: μ≠3.3 Ha: μ=3.3 (b) Identify the standardized test statistic. z=negative 1.34−1.34 (Round to two decimal places as needed.)
nutritionist claims that the mean tuna consumption by a person is 3.4 pounds per year. A sample of 90 people shows that the mean tuna consumption by a person is 3.3 pounds per year. Assume the population standard deviation is 1.02 pounds. At α=0.03, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. H0: μ=3.4 Ha: μ≠3.4 Your answer is correct. B. H0: μ>3.4 Ha: μ≤3.4 C. H0: μ≤3.3 Ha: μ>3.3 D. H0: μ≤3.4 Ha: μ>3.4 E. H0: μ>3.3 Ha: μ≤3.3 F. H0: μ≠3.3 Ha: μ=3.3 (b) Identify the standardized test statistic. z=negative 0.93−0.93 (Round to two decimal places as needed.) (c) Find the P-value. 0.3520.352 (Round to three decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis.
5.58 The concentration of mercury in a lake has been monitored for a number of years. Measurements taken on a weekly basis yielded an average of 1.20 mg/m (milligrams per cubic meter) with a standard deviation of .32 mg/m. Following an accident at a smelter on the shore of the lake, 15 measurements produced the following mercury concentrations. 1.60 1.77 1.61 1.08 1.07 1.79 1.34 1.07 1.45 1.59 1.43 2.07 1.16 0.85 2.11 ES ABOUT POPULATION CENTRAL VALUES a. Give a point estimate of the mean mercury concentration after the accident. b. Construct a 95% confidence interval on the mean mereury concentration after the accident. Interpret this interval. c. Is there sufficient evidence that the mean mercury concentration has increased since the accident? Use a = .05. d. Assuming that the standard deviation of the mercury concentration is .32 mg/m. calculate the power of the test to detect mercury concentrations of 1.28, 1.32. 1.36, and 1.40.
Claim: The standard deviation of pulse rates of adult males is more than 11 bpm. For a random sample of 170 adult males, the pulse rates have a standard deviation of 12.2 bpm. Find the value of the test statistic. The value of the test statistic is (Round to two decimal places as needed.)
15
A car battery's lifetime has an unknown distribution that is symmetric around its average of 800 days. The standard deviation of the lifetime is 50 days. Use Chebyshev's theorem to find an upper bound for the percentage of batteries that last shorter than 550 days? A. 5.56% B. 2.00% O C. 1.39% O D. 3.13%

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