Question

2)Given: Mean is 235.97 and Standard deviation is 87.04 and N is 30. Use a 0.01...

2)Given: Mean is 235.97 and Standard deviation is 87.04 and N is 30.

Use a 0.01 significance level to test the claim of a Forest Ranger that the mean
weight of the population is greater than 220 pounds.

a) Hypothesis

B) Diagram

C) Conclusion

D)Check that the requirements for performing the hypothesis test have been met. Any
assumptions that you had to make?

Homework Answers

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
Given: Female Mean is 178.8 and Standard Deviation is 63.29 N=20 Male Mean is 231.9 and...
Given: Female Mean is 178.8 and Standard Deviation is 63.29 N=20 Male Mean is 231.9 and Standard Deviation is 98.31 N=20 4) Use a 0.05 significance level to test the claim that the mean weight of male bears is greater than that of female bears. Take samples of size 20 from each gender. a) Hypothesis b)Diagram c) Conclusion Check that the requirements for performing the hypothesis test have been met. Any assumptions that you had to make?
There are 16 female bears out of a total of 30 bears. Use a 0.05 significance...
There are 16 female bears out of a total of 30 bears. Use a 0.05 significance level to test the claim of a Forest Ranger that the proportion of female bears is not equal to 0.50. a)Hypotheses b)Diagram c)Conclusion Check that the requirements for performing the hypothesis test have been met. Any assumptions that you had to make?
Given the sample mean = 22.325, sample standard deviation = 5.8239, and N = 40 for...
Given the sample mean = 22.325, sample standard deviation = 5.8239, and N = 40 for the low income group, Test the claim that the mean nickel diameter drawn by children in the low income group is greater than 21.21 mm. Test at the 0.01 significance level. a) Identify the correct alternative hypothesis: μ>21.21μ>21.21 p<21.21p<21.21 p=21.21p=21.21 μ<21.21μ<21.21 p>21.21p>21.21 μ=21.21μ=21.21 Give all answers correct to 3 decimal places. b) The test statistic value is:      c) Using the Traditional method, the critical...
A data set lists earthquake depths. The summary statistics are n=500 x̄=6.39sequals=4.18 km. Use a 0.01...
A data set lists earthquake depths. The summary statistics are n=500 x̄=6.39sequals=4.18 km. Use a 0.01 significance level to test the claim of a seismologist that these earthquakes are from a population with a mean equal to 6.00 Assume that a simple random sample has been selected. Identify the null and alternative​ hypotheses, test​ statistic, P-value, and state the final conclusion that addresses the original claim. What are the null and alternative​ hypotheses? Determine the​ P-value State the final conclusion...
Given the sample mean = 21.15, sample standard deviation = 4.7152, and N = 40 for...
Given the sample mean = 21.15, sample standard deviation = 4.7152, and N = 40 for the low income group, Test the claim that the mean nickel diameter drawn by children in the low income group is greater than 21.21 mm. Test at the 0.1 significance level. a) Identify the correct alternative hypothesis: p=21.21p=21.21 μ>21.21μ>21.21 μ=21.21μ=21.21 μ<21.21μ<21.21 p<21.21p<21.21 p>21.21p>21.21 Give all answers correct to 3 decimal places. b) The test statistic value is:      c) Using the Traditional method, the critical...
A sample mean, sample standard deviation, and sample size are given. Use the one-mean t-test to...
A sample mean, sample standard deviation, and sample size are given. Use the one-mean t-test to perform the required hypothesis test about the mean, μ, of the population from which the sample was drawn. Use the critical-value approach. , , n = 18, H0: μ = 10, Ha: μ < 10, α = 0.01 Group of answer choices Test statistic: t = -4.43. Critical value: t = -2.33. Reject H0. There is sufficient evidence to support the claim that the...
Children of a certain age are known to have a mean weight of 85 pounds. A...
Children of a certain age are known to have a mean weight of 85 pounds. A complaint is made that the children living in a municipal home are overfed. As one bit of evidence, 25 randomly selected children are weighed and found to have a mean weight of 87 pounds. It is known that the population standard deviation is 11.6 pounds. Is there convincing evidence at 0.05 significance level that mean weight of the children is greater than 85 pounds?...
Salmon: Assume that the weights of Chinook Salmon in the Columbia River are normally distributed. You...
Salmon: Assume that the weights of Chinook Salmon in the Columbia River are normally distributed. You randomly catch and weigh 30 such salmon. The mean weight from your sample is 23.8 pounds with a standard deviation of 2.5 pounds. Test the claim that the mean weight of Columbia River salmon is greater than 23 pounds. Test this claim at the 0.01 significance level. (a) What type of test is this? This is a right-tailed test. This is a left-tailed test.    ...
At the 0.01 significance level, test the claim that the three brands have the same mean...
At the 0.01 significance level, test the claim that the three brands have the same mean level if the following sample results have been obtained. Use Anova. Brand A Brand B Brand C 32 27 22 34 24 25 37 33 32 33 30 22 36 21 39 Claim: Null Hypothesis: Alternative Hypothesis: Calculator Screen Name in Ti 183: test statistics: Pvalue/alpha conversion decision: Conclusion:
Suppose a growing health and wellness company is rolling out three new nutritional diet plans. The...
Suppose a growing health and wellness company is rolling out three new nutritional diet plans. The research department wants to test the effectiveness of these three new nutritional plans, a low-fat plan and a low-carb plan, as well as its standard low-calorie plan. Marcia, a nutrition researcher with the company, randomly assigns 24 new weight loss program participants to each of the three plans, assigning 8 to each plan. After administering the plans, she follows up with the participants a...