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Vladimir [108]
4 years ago
12

What is the value of the interquartile range of the data below? 6 12 14 24

Mathematics
1 answer:
Strike441 [17]4 years ago
7 0

Answer:

14

Step-by-step explanation:

The interquartile range is the value of quartile 3 minus quartile 1.

The "box" part of this diagram has 3 lines, the left most, the middle, and the rightmost.

The leftmost line is Quartile 1. The right most is Quartile 3.

Hence

interquartile range = quartile 3 - quartile 1

Looking at the box plot, we can see that each small line in the number line is 2 units.

The leftmost line (quartile 1 ) is at 1 unit left of 30, so that is 30 -2 = 28

The rightmost line (quartile 3) is at  1 unit right of 40, so that is 40 + 2 = 42

Hence,

Interquartile range = 42 - 28 = 14

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Assume that a random sample of 4 Goldfinch birds are going to be used to test the above hypothesis. Compute the largest value of
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This question is incomplete, the complete question is;

Assume that body masses of Goldfinch birds follow a normal distribution, with a standard deviation equal to 0.04 oz. An ornithologist who would like to make some inference about the average body mass of the Goldfinch birds. In particular, at a significance level of 0.01, she would like to test the null hypothesis H₀: Average body mass of the Goldfinch birds is 0.5 oz, against the alternative claim that average body size is less than 0.5 oz.

      Assume that a random sample of 4 Goldfinch birds are going to be used to test the above hypothesis. Compute the largest value of the sample mean, that will allow the experimenter reject the null and prove the alternative at significance level 0.01. For this problem, make sure that your solution includes all steps of your computation. You will not earn points if you just use a formula

Answer: the largest value of the sample mean that will enable the experimenter to reject the null hypothesis is 0.4534 oz.

Step-by-step explanation:

Let us consider testing a hypothesis about a population mean with population standard deviation that is known,

Let the average body mass of the Goldfinch bird be μ^μ

Now our hypotheses are will be;

H₀ : μ = 0.5    and

H₁ : μ < 0.5

For the test, we shall use the z statistic which will be

z = (x"-μ)/(σ/√n)

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sample mean x" = ?

Now in order to be able to reject the null hypothesis, the p-value should be ≤ α = 0.01.

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For a p-value = 0.01 at the lower tail (which is the percentile for z),

the corresponding z will be -2.33

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z = (x"-μ)/(σ/√n)

z(σ/√n) = (x"-μ)

-2.33(0.04/√n) = (x"-0.5)

-0.0466 = x" - 0.5

x" = 0.5 - 0.0466

x"= 0.4534

therefore the largest value of the sample mean that will enable the experimenter to reject the null hypothesis is 0.4534 oz.

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Answer:

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Step-by-step explanation:

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