What is an equation of the line that passes through the points (2,5) and

11) Which of the following is NOT an assumption of Simple Linear Regression?

makvit [3.9K]

Answer:

11) c. the independent variables are not col-linear

12) b. Bar graph

13) a. Units of distance from the mean

14) d. None of the above

15) b. The difference between the predicted Y and observed Y

16) c. There is no spatial auto-correlation

17) a. It has a low Adjusted R²

18) c. A point that can influence the regression in a way that reflects the value of most other points

19) c. Rejecting the Null hypothesis incorrectly

20) c. The variable is not normally distributed

Step-by-step explanation:

11) With simple linear regression there is a col linear relationship between the dependent variable and the independent variables

Therefore, the correct option is c. the independent variables are not col linear

12) The bar graph which is also known as a bar chart, is used for the frequencies of data that are categorized

Therefore, the correct option is b. Bar graph

13) The z-value is a measure of the distance from the mean, therefore, the correct option is a. Units of distance from the mean

14) The higher the R² value the increased correlation between the factors of the dependent variable and the independent variable

Therefore, the correct option is d. None of the above

15) In an OLS residual error = Observed value of Y - Predicted value of Y

Therefore, the best option is b. The difference between the predicted Y and observed Y

16) A Moran's I value of zero indicates perfect randomness or no auto-correlation

The correct option is therefore, c. There is no spatial auto-correlation

17) The adjusted R² value increases when the added variable improves the model to the observation correlation more than would be expected by chance alone, therefore, the correct option is a. It has a low Adjusted R²

18) A leverage point is a point that is far from the other points in the observation, therefore, the correct option is c. A point that can influence the regression in a way that reflects the value of most other points

19) A type 1 error occurs when an accurate (statistically acceptable) null hypothesis is rejected therefore, the correct option is c. Rejecting the Null hypothesis incorrectly

20) The Quantile-Quanttile (QQ) plot is a technique used for checking the commonness of the sources of two sets of data, therefore sources of data that are from the source (hence normally distributed) align with the diagonal lines in a QQ-Plot

The correct option is therefore c. The variable is not normally distributed

8 0

4 years ago

Walter bought 500 marbles at the swap meet. Two-fifths of them

MrRissso [65]

Step-by-step explanation:

2/5 of marbles are blue

= 3/5 of marble are not blue.

Number of non-blue marbles = 500 * (3/5) = 300.

300 marbles are not blue.

4 0

3 years ago

Estimate the answer 102.3 × 4.7

ddd [48]

The estimated answer would be 480.81

8 0

3 years ago

Read 2 more answers

Please a need the answer and the solution please

Elis [28]

Answer:

Step-by-step explanation:

To find the first table y = 4x + 2, to find each y value we have to substitute each x value into the equation.

4(-2) + 2 = -6 so -2 is going to have the y as -6

4(-1) + 2 = -2 so -1 is going to have the y as -2

4(0) + 2 = 2 so 0 is going to have the y as 2

4(1) + 2 = 6 so 1 is going to have the y as 6

4(2) + 2 = 10 so 2 is going to have the y as 10

So these are the points you would graph (-2, -6), (-1, -2), (0, 2), (1,6), and (2, 10)

For the next talbe y = 3x + 1 we do the same thing

3(-2) + 1 = -5 so -2 is going to have the y as -5

3(-1) + 1 = -2 so -1 is going to have the y as -2

3(0) + 1 = 1 so 0 is going to have the y as 1

3(1) + 1 = 4 so 1 is going to have the y as 4

3(2) + 1 = 7 so 2 is going to have the y as 7

The ooints to graph are (-2, -5), (-1, -2),(0,1), (1, 4) and (2,7)

For the next table y = 6x - 5

6(-2) - 5 = -17 so -2 is going to have the y as -17

6(-1) - 5 = -11 so -1 is going to have the y as -11

6(0) - 5 = -5 so 0 is going to have the y as -5

6(1) - 5 = 1 so 1 is going to have the y as 1

6(2) - 5 = 7 so 2 is going to have the y as 7

The points to graph (-2, -17), (-1, -11), (0, -5), (1, 1) and (2, 7)

For the next table y = 5 + 2x

5 + 2(-4) = -3 so -4 is going to have the y as -3

5 + 2(-2) = 1 so -2 is going to have the y as 1

5 + 2(0) = 5 so 0 is going to have the y as 5

5 + 2(3) = 11 so 3 is going to have the y as 11

5 + 2(5) = 15 so 5 is going to have the y as 15

(-4, -3), (-2, 1), (0, 5), (3, 11) and (5, 15)

For the next table y = 8x - 10

8(-4) - 10 = -42 so -4 is going to have the y as -42

8(-2) - 10 = -26 so -2 is going to have the y as -26

8(0) - 10 = -10 so 0 is going to have the y as -10

8(2) - 10 = 6 so 2 is going to have the y as 6

8(4) - 10 = 22 so 4 is going to have the y as 22

(-4, -42), (-2, -26), (0, -10), (2, 6) and (4, 22)

For the last table

2(-2) = -4 so -2 is going to have the y as -4

2(-1) = -2 so -1 is going to have the y as -2

2(0) = 0 so 0 is going to have the y as 0

2(-1) = -2 so -1 is going to have the y as -2

2(-2) = -4 so -2 is going to have the y as -4

(-2, -4), (-1, -2), (0, 0), (-1, -2) and (-2, -4)

7 0

3 years ago

The TI-83/84 Plus calculator can be used to generate random data from a normally distributed population. The command randNorm(10

postnew [5]

Answer:

99% confidence interval for the mean value from the random number generator is a lower limit of 85.87 and an upper limit of 110.93.

Step-by-step explanation:

Confidence interval = mean +/- margin of error (E)

mean = 98.4

sd = 16.3

n = 15

degree of freedom (df) = n - 1 = 15 - 1 = 14

confidence level (C) = 99% = 0.99

significance level = 1 - C = 1 - 0.99 = 0.01 = 1%

t-value corresponding to 14 df and 1% significance level is 2.977.

E = t×sd/√n = 2.977×16.3/√15 = 12.53

Lower limit = mean - E = 98.4 - 12.53 = 85.87

Upper limit = mean + E = 98.4 + 12.53 = 110.93

99% confidence interval for the mean is between 85.87 and 110.93

3 0

3 years ago