How to find the perimeter and area

1 answer:

3 0

For Perimeter you add all the sides of the figure and the total is your perimeter. For Area you multiply the length times the width.

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Step-by-step explanation:

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The Rocky Mountain News (January 24, 1994) indicated that the 20-year mean snowfall in the Denver/Boulder region is 28.76 inches

ycow [4]

Answer:

The p-value of the test is 0.0007 < 0.05, indicating that the the snowfall for the 1993-1994 winters was higher than the previous 20-year average.

Step-by-step explanation:

20-year mean snowfall in the Denver/Boulder region is 28.76 inches. Test if the snowfall for the 1993-1994 winters has higher than the previous 20-year average.

At the null hypothesis, we test if the average was the same, that is, of 28.76 inches. So

$H_0: \mu = 28.76$

At the alternate hypothesis, we test if the average incresaed, that is, it was higher than 28.76 inches. So

$H_1: \mu > 28.76$

The test statistic is:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, $\sigma$ is the standard deviation and n is the size of the sample.

28.76 is tested at the null hypothesis:

This means that $\mu = 28.76$

Standard deviation of 7.5 inches. However, for the winter of 1993-1994, the average snowfall for a sample of 32 different locations was 33 inches.

This means that $\sigma = 7.5, X = 33, n = 32$ .

Value of the test statistic:

$z = \frac{X - \mu}{\frac{\sigma}{\sqrt{n}}}$

$z = \frac{33 - 28.76}{\frac{7.5}{\sqrt{32}}}$

$z = 3.2$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample mean above 33, which is 1 subtracted by the p-value of z = 3.2. In this question, we consider the standard level $\alpha = 0.05$ .