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Anna11 [10]
3 years ago
9

A researcher at a large company has collected data on both the beginning salary and the current salary of 48 randomly selected e

mployees. The least‑squares regression equation for predicting their current salary from their beginning salary is y^ =−2532.7 + 2.12x.(a) The current salaries had a mean of $32,070 with a standard deviation of $15,300. The beginning salaries had a mean of $16,340 with a standard deviation of $5,970. What is the correlation between current and beginning salary?
(b) Mr. Joseph Keller started working for the company earning $22,000. What do you predict his current salary to be?(c) Kathy Jones started working for the company earning $19,000. She currently earns $40,000. What is the residual for Ms. Jones?
Mathematics
1 answer:
Nadya [2.5K]3 years ago
8 0

Answer:

a. 0.83

b. 44107.3

c. 2252.7

Step-by-step explanation:

a)

The correlation coefficient is calculated as

byx=rSy/Sx

rSy=byx*Sx

⇒r=byx*Sx/Sy

Sy=standard deviation of y=standard deviation of current salary=15300

Sx=standard deviation of x=standard deviation of beginning salary=5970

byx=2.12

r=2.12*5970/15300

r=12656.4/15300

r=0.83

There is a strong positive correlation between current and beginning salary.

b)

Beginning salary=$22000

As

y^ =−2532.7 + 2.12x.

Current salary=-2532.7+2.12*beginning salary

Current salary=-2532.7+2.12*22000

Current salary=-2532.7+46640

Current salary=$44107.3

The predicted current salary of Mr. Joseph Keller is $44107.3.

c)

Residual=y-y^=Observed-Predicted

Observed Current Salary of Kathy Jones=40000

Beginning salary of Kathy Jones=19000

y^ =−2532.7 + 2.12x.

Predicted Current salary=-2532.7+2.12*beginning salary

Predicted Current salary=-2532.7+2.12*19000

Predicted Current salary=-2532.7+40280

Predicted Current salary=$37747.3

Residual=Observed Current Salary-Predicted Current salary

Residual=40000-27747.3

Residual=$2252.7

The residual for Ms. Jones is $2252.7.

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And we can find this probability with the following difference:

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The other possible code would be:

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Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

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Solution to the problem

For this case we want to find this probability:

P(0.7

And we can find this probability with the following difference:

P(0.7

We can use the following commands on the ti 84

2nd>VARS>DISTR

And then we look for normalcdf and we input this:

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The other possible code would be:

normalcdf(-1000,1,4,0,1)-normalcdf(-1000,0.7,0,1)

And we got:

P(0.7

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