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Sedaia [141]
2 years ago
11

A population mean of 360, a standard deviation of 4, and a margin of error of 2.5%

Mathematics
1 answer:
Ad libitum [116K]2 years ago
3 0

Answer:

(351.04, 368.96)

Step-by-step explanation:

Given that population mean= 360

Std devitaion = 4

margin of error = 2.5%= 0.025

Since population std deviation is known we can use Z critical value

For two tailed critical value for 2.5% would be

2.24

Margin of error = ±Critical value*std deviation

= ±2.24(4)

= ±8.96

Confidence interval =(Mean - margin of error, mean + margin of error)

=(360-8.96, 360+8,.96)\\= (351.04, 368.96)

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Answer:

1. If it is true that 60 percent of the trees in a forested region are classified as softwood, 0.015 is the probability of obtaining a population proportion greater than 0.6.

Step-by-step explanation:

Hello!

The historical information indicates that 60% of the forest trees are classified as softwood.

A botanist thinks that the proportion might be greater than 60%, so he tested his belief obtaining:

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You need to interpret this p-value. Little reminder:

The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis. It represents the % of size n samples from a population with proportion p=p₀, which will produce a measure that provides evidence as (or stronger) than the current sample that p is not equal to p₀.

The correct answer is:

1. If it is true that 60 percent of the trees in a forested region are classified as softwood, 0.015 is the probability of obtaining a population proportion greater than 0.6.

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Let us say weight of Javier is x pounds.

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