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lys-0071 [83]
2 years ago
6

Why was open trade a strength in the articles of confederation

Social Studies
1 answer:
Sladkaya [172]2 years ago
7 0

The Articles of Confederation gave the federal government the power to declare war and to manage its own department of international relations. However, the Articles of Confederation did not give the US government the power to collect taxes nationwide, regulate interstate or international trade, or direct the government of these states. According to the Articles of Confederation, each state would be responsible for managing its own government and each state would also be the only one with the power to create its own taxes and laws.

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An investigator theorizes that people who participate in a regular program of exercise will have levels of systolic blood pressu
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Answer:

a) H0: mean of exercise group = mean of non exercise group

   Ha: mean of exercise group ≠ mean of non exercise group

b) p value is 0.002

c) see the explanation

d) Null hypothesis is rejected. Ha is true

Explanation:

The complete question is:

An investigator theorizes that people who participate in a regular program of exercise will have levels of systolic blood pressure that are significantly different from that of people who do not participate in a regular program of exercise. To test this idea the investigator randomly assigns 21 subjects to an exercise program for 10 weeks. After ten weeks the mean systolic blood pressure of subjects in the exercise group is 137 and the standard deviation of blood pressure values in exercise group is 10. After ten weeks the mean systolic blood pressure of subjects in the non-exercise group is 127 and the standard deviation of blood pressure values in exercise group is 9.

a) State null and alternate hypothesis

b) Include the critical value of appropriate as part of decision rule to for rejecting the null hypothesis

c) show all of your work in reaching a decision as to whether the investgator should reject the null hypothesis or not.

d) State the conclusion te investigator is entitled ot draw on the basis of results

a) H0: mean of exercise group = mean of non exercise group

   Ha: mean of exercise group ≠ mean of non exercise group

b) at 5% significance value, the p vaue is 0.002

t=  (x1 — x2) /√ (σ1² / √n1 + σ2² / n2),

x1: mean of exercise group

x2: mean of non exercise group

σ1: standard deviation of exercise group

σ2: standard deviation of non exercise group

N1: sample size of exercise group

N2: sample size of non exercise group

t= 4.948

p value is 0.002

Since p value is less t statistic, null hypothesis is rejected.

Ha is true

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3 years ago
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