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2 years ago

Below is a data set. Use desmos to find the slope, y-intercept, and correlation

Mathematics

1 answer:

Debora [2.8K]2 years ago

3 0

The slope of the least squares regression line is 1.0 and the y-intercept is 6.3

<h3>How to determine the slope?</h3>

To do this, we enter the data values in a graphing calculator, and we interpret the results:

Using a graphing calculator, we have the following calculation summary

Sum of X = 83
Sum of Y = 123
Mean X = 13.8333
Mean Y = 20.5
Sum of squares (SSX) = 356.8333
Sum of products (SP) = 365.5
R = 0.8728

The slope (b) is calculated using:

b = SP/SSX

This gives

b = 365.5/356.83

b = 1.0

Hence, the slope is 1.0

<h3>How to determine the y-intercept?</h3>

The y-intercept (a) is calculated using:

a = MY - bMX

a = 20.5 - (1.02*13.83)

a = 6.3

Hence, the y-intercept is 6.3

<h3>How to determine the correlation coefficient?</h3>

In (a), we have:

R = 0.8728

Approximate

R = 0.9

This means that the correlation coefficient is 0.9

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Answer:

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Step-by-step explanation:

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3 years ago

Read 2 more answers

Researchers believe that 24.5% of the adults in the United States are obese. A Gallup poll examined the rate of obesity in Ameri

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Answer:

The p-value of the test is 0 < 0.05(standard significance level), which means that this is enough evidence to state that more than 24.5% of all adults in the U.S. are obese

Step-by-step explanation:

Researchers believe that 24.5% of the adults in the United States are obese. Test if there is enough evidence to state that more than 24.5% of all adults in the U.S. are obese.

At the null hypothesis, we test if the proportion is of 24.5% or less, that is:

$H_0: p \leq 0.245$

At the alternative hypothesis, we test if this proportion is greater than 24.5%, that is:

$H_1: p > 0.245$

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.

0.245 is tested at the null hypothesis:

This means that $\mu = 0.245, \sigma = \sqrt{0.245*0.755}$

Survey 86,664 randomly sampled U.S. adults. Of the adults surveyed, 23,053 said that they were obese.

This means that $n = 86664, X = \frac{23053}{86664} = 0.266$

Value of the test statistic:

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

$z = \frac{0.266 - 0.245}{\frac{\sqrt{0.245*0.755}}{\sqrt{86664}}}$

$z = 14.37$

P-value of the test and decision:

The p-value of the test is the probability of finding a sample proportion of 0.266 or higher, which is 1 subtracted by the p-value of z = 14.37.

The p-value of z = 14.37 is 1.

1 - 1 = 0

The p-value of the test is 0 < 0.05(standard significance level), which means that this is enough evidence to state that more than 24.5% of all adults in the U.S. are obese

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Answer:

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