A nutritionist daims that the mean tuna consumption by a person is 3.7 pounds per year. A sample of 80 people shows that the mean tuna consumption by a person is 3.3 pounds per year. Assume the population standard deviation is 1.04pounds. At a= 0.06, can you reject the claim?(a) Identify the null hypothesis and alternative hypothesis.OA Ho: H>3.7Hiu53.7O B. Ho: H= 3.7OC. Ho: H#3.3Hai H#3.7OD. Ho: H533O E. Ho: H>3.3Ha: H53.3OF. Ho: u53.7Hgiu>3.7H:u> 3.3(b) Identify the standardized test statistic.z= (Round to two decimal places as needed.)© Find the P-value(Round to three decimal places as needed.)(d) Decide whether to reject or fail to reject the null hypothesis.O A. Fail to reject Hn. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.7O B. Reject Ho. There is sufficient evidence to reject the claim that mean tuna consumptionsequal to 3.7 pounds.pounds.O C. Fail to reject Ho. Theresufficient evidence to reject the claim that mean tuna consumption is equal to 3.7O D. Reject Ho. There is not sufficient evidence to reject the claim that mean tuna consumptionequal to 3.7 pounds.pounds.O Time Remaining: 01:23:47NextMacBook Air80escF7FBF9F10F5F2F3F4@#$%&13479QWTYtab* 00

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Answered: A nutritionist dlaims 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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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.
The scores on a mathematics exam have a mean of 70 and a standard deviation of 6. Find the x-value that corresponds to the z-score 5.19. O 64.8 O 121.9 O 18.1 O 101.1
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
Claim: The standard deviation of pulse rates of adult males is less than 10 bpm. For a random sample of 173 adult males, the pulse rates have a standard deviation of 8.9 bpm. Find the value of the test statistic. The value of the test statistic is (Round to two decimal places as needed)
A nutritionist claims that the mean tuna consumption by a person is 3.8 pounds per year. A sample of 70 people shows that the mean tuna consumption by a person is 3.5 pounds per year. Assume the population standard deviation is 1.04 pounds. At α=0.09, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. H0: μ=3.8 Ha: μ≠3.8 B. H0: μ≠3.5 Ha: μ=3.5 C. H0: μ>3.5 Ha: μ≤3.5 D. H0: μ>3.8 Ha: μ≤3.8 E. H0: μ≤3.5 Ha: μ>3.5 F. H0: μ≤3.8 Ha:μ>3.8 b) Identify the standardized test statistic. z= (Round to two decimal places as needed.) (c) Find the P-value. (d) Decide whether to reject or fail to reject the null hypothesis. A. Reject H0. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.8pounds. B. Reject H0. There is sufficient evidence to reject the claim that mean tuna consumption is equal to…
Claim: The mean pulse rate (in beats per minute) of adult males is equal to 69 bpm. For a random sample of 161 adult males, the mean pulse rate is 69.2 bpm and the standard deviation is 11.3 bpm. Find the value of the test statistic. The value of the test statistic is (Round to two decimal places as needed.)
A nutritionist claims that the mean tuna consumption by a person is 3.9 pounds per year. A sample of 80 people shows that the mean tuna consumption by a person is 3.8 pounds per year. Assume the population standard deviation is 1.02 pounds. At a = 0.09, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. O A. Ho: u>3.8 Ha: μs 3.8 O B. Ho: H#3.8 Ha: μ= 3.8 O C. Ho: Hs3.9 Hai H> 3.9 O D. Ho: H>3.9 O E. Ho: Hs3.8 Ha: H> 3.8 F. Ho: H=3.9 Ha:us3.9 Ha: H#3.9 (b) Identify the standardized test statistic. (Round to two decimal places as needed.)
Claim: The standard deviation of pulse rates of adult males is more than 10 bpm. For a random sample of 157 adult males, the pulse rates have a standard deviation of 10.9 bpm. Find the value of the test statistic. The value of the test statistic is (Round to two decimal places as needed.)
Claim: The standard deviation of pulse rates of adult males is less than 12 bpm. For a random sample of 160 adult males, the pulse rates have a standard deviation of 10.7 bpm. Find the value of the test statistic. The value of the test statistic is :
A study determined that the average student who graduates takes 58 months to graduate from college with a bachelor's degree and a standard deviation of 10 months. The distribution was approximately normal. What percentage of college graduates will take longer than 4 years (48 months)? 34.1% 34.1% 13.6% 15.6% 83% 23 15.9% 47.7% O84.1% 97.7% 56°F Clou

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