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maks197457 [2]
1 year ago
12

Which of the following statements about least squares regression analysis is true?

Mathematics
1 answer:
Lemur [1.5K]1 year ago
7 0

The statements about least squares regression analysis is true are I and III.

<h3>What is meant by regression analysis?</h3>

The statistical technique of regression analysis demonstrates the link between two or more variables. The method evaluates the relationship between a dependent variable and independent factors, and is frequently presented as a graph.

You can anticipate the effects of the independent variable on the dependent one by creating a regression analysis. For instance, we can claim that a linear regression model can adequately account for age and height.

Here,

From the given information in the question:

The statements I and III are correct.

And the statements are:

A point with a large residual is an outlier and the removal of an influential point from the data set could change the value of the correlation coefficient.

So, the statements  I and III are correct for the given information in the question.

To know more about regression analysis, visit:

brainly.com/question/28298210

#SPJ1

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Triangle rst is dilated by a scale factor of 5 to form triangle r's't'. side r's' measures 40. what is the measure of side rs?
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Answer: 8

Step-by-step explanation:

We know that scale factor = (image)/(preimage).

Substituting in what we have,

5=\frac{R'S'}{RS}\\\\5=\frac{40}{RS}\\\\RS=\boxed{8}

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What is 100.34567 in a fraction and word form? Thank you! ❤
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3 years ago
Polly's Polls asked 1850 second-year college students if they still had their original major. According to the colleges, 65% of
suter [353]

Answer:

The probability that Polly's Sample will give a result within 1% of the value 65% is 0.6424

Step-by-step explanation:

The variable that assigns the value 1 if a person had its original major and 0 otherwise is a Bernoulli variable with paramenter 0.65. Since she asked the question to 1850 people, then the number of students that will have their original major is a Binomial random variable with parameters n = 1850, p = 0.65.

Since the sample is large enough, we can use the Central Limit Theorem to approximate that random variable to a Normal random variable, which we will denote X.

The parameters of X are determined with the mean and standard deviation of the Binomal that we are approximating. The mean is np = 1850*0.65 = 1202.5, and the standard deviation is √np(1-p) = √(1202.5*0.35) = 20.5152.

We want to know the probability that X is between 0.64*1850 = 1184 and 0.66*1850 = 1221 (that is, the percentage is between 64 and 66). In order to calculate this, we standarize X so that we can work with a standard normal random variable W ≈ N(0,1). The standarization is obtained by substracting the mean from X and dividing the result by the standard deviation, in other words

W = \frac{X-\lambda}{\sigma} = \frac{X-1202.5}{20.5152}

The values of the cummulative function of the standard normal variable W, which we will denote \phi are tabulated and they can be found in the attached file.

Now, we are ready to compute the probability that X is between 1184 and 1221. Remember that, since the standard random variable is symmetric through 0, then \phi(-z) = 1-\phi(z) for each positive value z.

P(1184 < X < 1221) = P(\frac{1184-1202.5}{20.5152} < \frac{X-1202.5}{20.5152} < \frac{1221-1202.5}{20.5152})\\ = P(-0.9018 < W < 0.9018) = \phi(0.9018) - \phi(-0.9018) = \phi(0.9018)-(1-\phi(0.9018))\\ = 2\phi(0.9018)-1 = 2*0.8212-1 = 0.6424

Therefore, the probability that Polly's Sample will give a result within 1% of the value 65% is 0.6424.

Download pdf
4 0
3 years ago
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