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ANTONII [103]
3 years ago
7

The running trail in the local park is 3.416 miles long. If the park board were planning to extend the trail by 1.34 miles, what

would the new length of the running trail be?
Mathematics
1 answer:
morpeh [17]3 years ago
7 0

Answer:

4.756 miles

Step-by-step explanation:

It is given that,

The running trail in the local park is 3.416 miles long. If the park board were planning to extend the trail by 1.34 miles.

We need to find the new length of the running trial. It is equal to the sum of length of running trail and the extended length. So,

l = 3.416 +1.34

l = 4.756 miles

So, the new length of the running trail be 4.756 miles

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Toll Brothers is a luxury home builder that would like to test the hypothesis that the average size of new homes exceeds 2,400 s
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Answer:

Since the pvalue of the test is 0.09 > 0.01, we do not reject the null hypothesis that the average size of new homes is of 2400 square feet.

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Toll Brothers is a luxury home builder that would like to test the hypothesis that the average size of new homes exceeds 2,400 square feet. Test to see if the average size of new homes exceeds 2,400 square feet.

At the null hypothesis, we test if the average is of 2400 square feet, that is:

H_o: \mu = 2400

At the alternate hypothesis, we test if the average is greater than 2400 square feet, that is:

H_a: \mu > 2400

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t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

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2400 is tested at the null hypothesis:

This means that \mu = 2400

A random sample of 36 newly constructed homes had an average of 2,510 square feet with a sample standard deviation of 480 square feet.

This means that n = 36, X = 2510, s = 480

Test statistic:

t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}

t = \frac{2510 - 2400}{\frac{480}{\sqrt{36}}}

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Pvalue of the test and decision:

The pvalue of the test is the probability of finding a sample mean of at least 2510, which is the pvalue of t = 1.375 with 36 - 1 = 35 degrees of freedom, using a one-tailed test.

With the help of a calculator, this pvalue is of 0.09

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