Using this information: Lycopene is a bright red pigment which has been shown to have powerful antioxidant and anticarcinogenic properties. A biological research company conducts experiments in which seeds from cherry tomatoes are exposed to a series of pulsations of ultra-violet radiation with the aim of triggering a genetic mutation to increase lycopene production. It is well established from past experiments that the lycopene content for untreated cherry tomatoes is 12.7 mg/100g. A random sample of 36 treated seeds from a population of seeds of an identical variant was selected and were planted and grown in identical conditions. The mean lycopene content of the treated seeds was determined via mass spectrometry to be 13.1 mg/100g and a standard deviation of 1.2. A histogram indicates that the distribution of the lycopene content appears to be roughly normally distributed. set of hypothesis: H0:μ>12.7 vs. HA:μ=12.7answer the following questions in the images please (: Which of the following is the correct conclusion and summarizing statement to the hypothesis test at the a0.01 level ?Select one:O a. Fail to reject Ho. The collected sample does not provide sufficient evidence at an a = 0.01 level to suggest that the mean lycopene contentof all seeds treated with the ultraviolet radiation is any larger than 12.7 mg/100g, the average lycopene content of untreated seeds.O b. Reject Họ. The collected sample provides sufficient evidence at an a = 0.01 level to suggest that the mean lycopene content of all seedstreated with the ultraviolet radiation is greater than 12.7 mg/100g, the average lycopene content of untreated seeds.O c. Fail to reject Ho. The collected sample provides sufficient evidence at an a = 0.01 level to suggest that the mean lycopene content of allseeds treated with the ultraviolet radiation is 12.7 mg/100g, the average lycopene content of untreated seeds.O d. Fail to reject Ho. The collected sample does not provide sufficient evidence at an a0.01 level to suggest that the mean lycopene contentof the 36 sampled seeds treated with the ultraviolet radiation is any larger than 12.7 mg/100g, the average lycopene content of untreatedseeds.O e. Reject Họ. The collected sample provides sufficient evidence at an a = 0.01 level to suggest that the mean lycopene content of the 36sampled seeds treated with the ultraviolet radiation is greater than 12.7 mg/100g, the average lycopene content of untreated seeds.Suppose that the experiment were repeated with more seeds, but coincidentally resulted in the same sample mean and sample standard deviation.How would this change the P-value of our hypothesis test? (Think about it: would this provide more orevidence against the Ho in favor of thealternative?)Select one:O a. It would not change the P-value relative to the first test.Ob.It would increase the P-value relative to the first test.Oc.It would decrease the P-value relative to the first test.O d. It is not possible to know how this would affect the P-value without more information.

We have given that Sample size n =36 Sample mean = 13.1 Standard deviation=1.2

Answered: bright red pigment which has been shown…

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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Using this information: Lycopene is a bright red pigment which has been shown to have powerful antioxidant and anticarcinogenic properties. A biological research company conducts experiments in which seeds from cherry tomatoes are exposed to a series of pulsations of ultra-violet radiation with the aim of triggering a genetic mutation to increase lycopene production. It is well established from past experiments that the lycopene content for untreated cherry tomatoes is 12.7 mg/100g. A random sample of 36 treated seeds from a population of seeds of an identical variant was selected and were planted and grown in identical conditions. The mean lycopene content of the treated seeds was determined via mass spectrometry to be 13.1 mg/100g and a standard deviation of 1.2. A histogram indicates that the distribution of the lycopene content appears to be roughly normally distributed.

set of hypothesis: H0:μ>12.7 vs. HA:μ=12.7

answer the following questions in the images please (:

Which of the following is the correct conclusion and summarizing statement to the hypothesis test at the a
0.01 level ?
Select one:
O a. Fail to reject Ho. The collected sample does not provide sufficient evidence at an a = 0.01 level to suggest that the mean lycopene content
of all seeds treated with the ultraviolet radiation is any larger than 12.7 mg/100g, the average lycopene content of untreated seeds.
O b. Reject Họ. The collected sample provides sufficient evidence at an a = 0.01 level to suggest that the mean lycopene content of all seeds
treated with the ultraviolet radiation is greater than 12.7 mg/100g, the average lycopene content of untreated seeds.
O c. Fail to reject Ho. The collected sample provides sufficient evidence at an a = 0.01 level to suggest that the mean lycopene content of all
seeds treated with the ultraviolet radiation is 12.7 mg/100g, the average lycopene content of untreated seeds.
O d. Fail to reject Ho. The collected sample does not provide sufficient evidence at an a
0.01 level to suggest that the mean lycopene content
of the 36 sampled seeds treated with the ultraviolet radiation is any larger than 12.7 mg/100g, the average lycopene content of untreated
seeds.
O e. Reject Họ. The collected sample provides sufficient evidence at an a = 0.01 level to suggest that the mean lycopene content of the 36
sampled seeds treated with the ultraviolet radiation is greater than 12.7 mg/100g, the average lycopene content of untreated seeds.
Suppose that the experiment were repeated with more seeds, but coincidentally resulted in the same sample mean and sample standard deviation.
How would this change the P-value of our hypothesis test? (Think about it: would this provide more or
evidence against the Ho in favor of the
alternative?)
Select one:
O a. It would not change the P-value relative to the first test.
Ob.
It would increase the P-value relative to the first test.
Oc.
It would decrease the P-value relative to the first test.
O d. It is not possible to know how this would affect the P-value without more information.

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

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