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1 year ago

When calculating correlation and regression both sets of data must be __________

Mathematics

1 answer:

Misha Larkins [42]1 year ago

7 0

When calculating correlation and regression both sets of data must be Statistical.

According to the statement

we have to find the type of data when we calculate the correlation and regression both sets.

so, The difference between these two statistical measurements is that correlation measures the degree of a relationship between two variables (x and y), whereas regression is how one variable affects another.

And when we calculate both then data sets must be a statistical data. because correlation summarizing direct relationship between two variables and regression predict or explain numeric response. So, without statistical data this is not possible to calculate correlation and regression both sets.

so, When calculating correlation and regression both sets of data must be Statistical.

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expeople1 [14]

Answer:

The sample mean is of 1925 calories.

The margin of error is of 75 calories.

The sample standard deviation is of 109.7992 calories.

Step-by-step explanation:

Sample mean:

The sample mean is the mean value of the two bounds of the confidence interval. So

$M = \frac{1850 + 2000}{2} = 1925$

The sample mean is of 1925 calories.

The margin of error

Difference between the bounds and the sample mean. So

2000 - 1925 = 1925 - 1850 = 75 calories.

The margin of error is of 75 calories.

Sample standard deviation:

Here, I am going to expand on the t-distribution.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 18 - 1 = 17

99% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 17 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.99}{2} = 0.995$ . So we have T = 2.898

The margin of error is:

$M = T\frac{s}{\sqrt{n}}$

In which s is the standard deviation of the sample and n is the size of the sample.

Since $M = 75, T = 2.898, n = 18$

$M = T\frac{s}{\sqrt{n}}$

$75 = 2.898\frac{s}{\sqrt{18}}$

$s = \frac{75\sqrt{18}}{2.898}$

$s = 109.7992$

The sample standard deviation is of 109.7992 calories.

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2 years ago

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2m = 200cm

200cm-65m= 135cm

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Sorry, but I have no idea what your "<span>eaqkesbnnn" means.
</span>
If you were supposed to add 0.16 and 0.45, the correct answer would be 0.61.

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