Problem 1. During your spring break you decided to take a fishing trip. You caught and released two different types of bass. For each fish, you recorded its weight as in the following table. Florida Bass Weight in lbs. 6.8 7.2 12.6 10.5 7.4 13.1 11.4 6.9 6.8 12.9Northern Bass Weight in lbs. 6.5 7.2 6.7 6.2 8.3 6.4 8.2 12.2 10.8 6.8 Assume that weights of both the Florida bass and the Northern bass are normally distributed. Use a 0.01 significance level to test the claim of a biologist that the mean weight of the Florida bass is different from the mean weight of the Northern bass.Use the following steps: Step 1. Write the Claim, Null Hypothesis and Alternative Hypothesis in symbols. Step 2. Identify Step 3. Check and identify the requirements for this test. Step 4. State the test statistic and P-value (use Excel but do not round the results) Step 5. State your decision to either reject or fail to reject the null hypothesis and why. Step 6. Consider the claim and your decision and state the conclusion in the context of the problem. Problem 2. COURSE LEARNING OUTCOMEPerform a hypothesis test for two population proportions. Give the null hypothesis, alternative hypothesis, reject or fail to reject the null hypothesis, and give the conclusion. For the Claim state the null hypothesis and alternative hypothesis in symbols For the P-value = 0.073, state whether you reject or fail to reject the null hypothesis if the significance level is 0.05 and why. State the conclusion of the problem. Problem 3. ECHINACEA TO PREVENT COMMON COLDSRhinovirus typically cause common colds. In a study to prevent common colds, subjects are divided into two groups. In group #1, 45 subjects were treated with echinacea and 38 of them developed rhinovirus infections. In group #2, 103 subjects were given a placebo and 88 of them developed rhinovirus infections. Use a 0.05 significance level to test the claim that people taking echinacea develop less rhinovirus infections than people who do not take echinacea.Use the following steps: Step 1. Write the Claim, Null Hypothesis and Alternative Hypothesis in symbols. Step 2. Identify Step 3. Check and identify the requirements for this test. Step 4. State the type of test (right-tail, left-tail, or two-tail test) and use the given test statistic to find the P-value (use the provided Standard Normal Distribution table, aka z-table, without rounding) Test statistic Step 5. State your decision to either reject or fail to reject the null hypothesis and why. Step 6. Consider the claim and your decision and state the conclusion in the context of the problem. Step 7. Considering your conclusion from part 6, does echinacea seem to be effective in preventing the common colds? Why or why not?

Problem 1 For Florida bass : sample size (n1) = 10sample mean (x̄1) = Σx1 /n1 = 95.6/10 = 9.56sample standard deviation (s1) = √ (x1-x̄1)2 /(n1-1)= √ 69.744/(10-1)= 2.7838For Northern bass:sample size (n2) = 10sample mean (x̄2) = Σx2 /n2 = 79.3/10 = 7.93sample standard deviation (s2) = √ (x2-x̄2)2 /(n2-1) = √ 37.381/(10-1) = 2.0380 Note : As per answering guidelines I have to answer only the first question in case of multiple questions. Please repost the questions seperately. 1) The null hypothesis and alternative hypothesis H0: μ1 = μ2 H1: μ1 ≠ μ2 2) This is an independent t distribution test. 3) population standard deviation is unknown and sample size (n) > 30, so we use t- distribution test. 4) pooled sample variance ( sp2 ) = ( (n1-1)*s12 + (n2-1)*s22) /(n1+n2-2) = ((10-1)*2.78382 + (10-1)*2.03802)/(10+10-2) = 5.9515 Standard error (S.E.) = √sp2 * (1/n1 + 1/n2) = √ 5.9515*(1/10 +1/10) = 1.0910 Test statistic (t) = (x̄1-x̄2)/Standard error (S.E.) = (9.56-7.93)/1.0910 = 1.4940 degree of freedom(df) = n1+n2-2 = 10+10-2 = 18 significance level α = 0.01 For a two tailed test, P-value = 0.152499078 (By using excel) 5) Since p-value > significance level α, we fail to reject the null hypothesis. The result is not significant at 0.01 significance level. 6) Conclusion : There is no significant evidence to conclude that the mean weight of the Florida bass is different from the mean weight of the Northern bass.

During your spring break you decided to take a fishing trip. You caught and released two different types of bass. For each fish, you recorded its weight as in the following table. Florida Bass Weight in lbs. 6.8 7.2 12.6 10.5 7.4 13.1 11.4 6.9 6.8 12.9 Northern Bass Weight in lbs. 6.5 7.2 6.7 6.2 8.3 6.4 8.2 12.2 10.8 6.8 Assume that weights of both the Florida bass and the Northern bass are normally distributed. Use a 0.01 significance level to test the claim of a biologist that the mean weight of the Florida bass is different from the mean weight of the Northern bass. Use the following steps: Step 1. Write the Claim, Null Hypothesis and Alternative Hypothesis in symbols. Step 2. Identify Step 3. Check and identify the requirements for this test. Step 4. State the test statistic and P-value (use Excel but do not round the results) Step 5. State your decision to either reject or fail to reject the null hypothesis and why. Step 6. Consider the claim and your decision and state the conclusion in the context of the problem.

Functions and Change: A Modeling Approach to College Algebra (MindTap Course List)

6th Edition

ISBN:9781337111348

Author:Bruce Crauder, Benny Evans, Alan Noell

Publisher:Bruce Crauder, Benny Evans, Alan Noell

Chapter3: Straight Lines And Linear Functions

Section3.3: Modeling Data With Linear Functions

Problem 17SBE

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Problem 1.

During your spring break you decided to take a fishing trip. You caught and released two different types of bass. For each fish, you recorded its weight as in the following table.

Florida Bass Weight in lbs.

6.8

7.2

12.6

10.5

7.4

13.1

11.4

6.9

6.8

12.9

Northern Bass Weight in lbs.

6.5

7.2

6.7

6.2

8.3

6.4

8.2

12.2

10.8

6.8

Assume that weights of both the Florida bass and the Northern bass are normally distributed. Use a 0.01 significance level to test the claim of a biologist that the mean weight of the Florida bass is different from the mean weight of the Northern bass.

Use the following steps:

Step 1. Write the Claim, Null Hypothesis and Alternative Hypothesis in symbols.

Step 2. Identify

Step 3. Check and identify the requirements for this test.

Step 4. State the test statistic and P-value (use Excel but do not round the results)

Step 5. State your decision to either reject or fail to reject the null hypothesis and why.

Step 6. Consider the claim and your decision and state the conclusion in the context of the problem.

Problem 2. COURSE LEARNING OUTCOME

Perform a hypothesis test for two population proportions. Give the null hypothesis, alternative hypothesis, reject or fail to reject the null hypothesis, and give the conclusion.

For the Claim state the null hypothesis and alternative hypothesis in symbols

For the P-value = 0.073, state whether you reject or fail to reject the null hypothesis if the significance level is 0.05 and why.

State the conclusion of the problem.

Problem 3. ECHINACEA TO PREVENT COMMON COLDS

Rhinovirus typically cause common colds. In a study to prevent common colds, subjects are divided into two groups. In group #1, 45 subjects were treated with echinacea and 38 of them developed rhinovirus infections. In group #2, 103 subjects were given a placebo and 88 of them developed rhinovirus infections. Use a 0.05 significance level to test the claim that people taking echinacea develop less rhinovirus infections than people who do not take echinacea.

Use the following steps:

Step 1. Write the Claim, Null Hypothesis and Alternative Hypothesis in symbols.

Step 2. Identify

Step 3. Check and identify the requirements for this test.

Step 4. State the type of test (right-tail, left-tail, or two-tail test) and use the given test statistic to find the P-value (use the provided Standard Normal Distribution table, aka z-table, without rounding)

Test statistic

Step 5. State your decision to either reject or fail to reject the null hypothesis and why.

Step 6. Consider the claim and your decision and state the conclusion in the context of the problem.

Step 7. Considering your conclusion from part 6, does echinacea seem to be effective in preventing the common colds? Why or why not?

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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