MATLAB: An Introduction with Applications
6th Edition
ISBN: 9781119256830
Author: Amos Gilat
Publisher: John Wiley & Sons Inc
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- A nutritionist claims that the mean tuna consumption by a person is 3.3 pounds per year. A sample of 50 people shows that the mean tuna consumption by a person is 3.1 pounds per year. Assume the population standard deviation is 1.21 pounds. At α = 0.1, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. Ho: μ#3.1 H₂:μ = 3.1 D. H₂:μ≤3.1 Ha:μ>3.1 (b) Identify the standardized test statistic. Z= (Round to two decimal places as needed.) B. Ho: μ = 3.3 H₂:μ#3.3 E. Ho:μ≤3.3 Ha:μ>3.3 C. Ho: μ>3.3 H₂:μ≤3.3 F. Ho: μ>3.1 Hg:μ≤3.1arrow_forwardA 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.)arrow_forward5) A nutritionist claims that the mean tuna consumption by a person is 3.3 pounds per year. A sample of 60 people shows that the mean tuna consumption by a person is 3.2 pounds per year. Assume the population standard deviation is 1.17 pounds. At α=0.03, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. A. H0: μ>3.2 Ha: μ≤3.2 B. H0: μ≤3.2 Ha: μ>3.2 C. H0: μ≠3.2 Ha: μ=3.2 D. H0: μ≤3.3 Ha: μ>3.3 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= (Round to two decimal places as needed.) (c) Find the P-value. _____ (Round to three decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis. A. Fail to reject H0. There is not sufficient evidence to reject the claim that mean tuna consumption is equal to 3.3 pounds. B. Reject H0. There is…arrow_forward
- Answer the questionarrow_forward15arrow_forwardA 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 1.17 pounds. At a = 0.07, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. O A. Ho: HS3.8 H: > 3.8 O B. Ho: H#3.5 Ha: H= 3.5 OC. Ho: H= 3.8 Ha:H43.8 O D. Ho: H>3.8 ΟΕ. H :μs 3.5 H:u>3.5 OF. Ho: H>3.5 H:Hs3.8 H: us3.5 (b) Identify the standardized test statistic, z. z= (Round to two decimal places as needed.) (c) Find the P-value. Hint: Keep in mind that P-value depends on the type of test. Left-tailed test: P-value=Area Right-tailed test: P-value=1-Area Two-tailed test: P-value=2'Area (where Area = the number from the table) P-value = (Round to three decimal places as needed.) (d) Decide whether to reject or fail to reject the null hypothesis Compare P with a and then State your conclusion. O B. Fail to reject Ho. There is…arrow_forward
- 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…arrow_forwardA nutritionist claims that the mean tuna consumption by a person is 3.3 pounds per year. A sample of 80 people shows that the mean tuna consumption by a person is 3.1 pounds per year. Assume the population standard deviation is 1.02 pounds. At a = 0.06, can you reject the claim? (a) Identify the null hypothesis and alternative hypothesis. OA. Ho: H>3.1 Ha: HS3.1 D. Ho: H=3.3 H₂:μ*3.3 OB. Ho: ≤3.3 H:H>3.3 OE. Ho: >3.3 Ha: 53.3 (b) Identify the standardized test statistic. z = (Round to two decimal places as needed.) OC. Ho: * 3.1 Ha: 3.1 OF. Ho: $3.1 H₂:μ>3.1arrow_forwardA 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.)arrow_forward
- A population has parameters u = 214.3 and o = 73.5. You intend to draw a random sample of size n = 237. What is the mean of the distribution of sample means? What is the standard deviation of the distribution of sample means? (Report answer accurate to 2 decimal places.) = 20arrow_forwardFind the p-value for the hypothesis test. A random sample of size 54 is taken. The sample has a mean of 375 and a standard deviation of 83. H0: µ = 400 Ha: µ< 400 The p-value for the hypothesis test is . rounded to 4 decimal placesarrow_forwardAssume all populations are normally distributed. A travel analyst claims that the mean price of a round trip flight from New York City to Los Angeles is less than $507. In a random sample of 55 round trip flights from NYC to LA, the mean price is $502. Assume the population standard deviation is $111. At α = 0.05, is there enough evidence to support the travel analyst’s claim?arrow_forward
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