Elementary Statistics (Text Only)
2nd Edition
ISBN: 9780077836351
Author: Author
Publisher: McGraw Hill
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Question
Chapter 13.3, Problem 21E
a.
To determine
To find:The regression equation for the data.
a.
Expert Solution
Answer to Problem 21E
The regression equation for the data is
Explanation of Solution
Given information: The data is shown below.
| 31.3 | 50 | 19 | 4.0 | 37.4 | 60 | 30 | 5.0 | 24.0 | 60 | 24 | 4.0 | 51.0 | 80 | 34 | 7.5 |
| 56.9 | 90 | 38 | 8.0 | 43.3 | 60 | 26 | 7.0 | 36.2 | 50 | 21 | 7.0 | 34.3 | 60 | 22 | 2.5 |
| 43.1 | 70 | 28 | 6.5 | 36.3 | 70 | 25 | 7.5 | 26.5 | 50 | 17 | 2.0 | 31.5 | 60 | 24 | 5.0 |
| 41.5 | 70 | 25 | 5.5 | 38.4 | 70 | 31 | 5.5 | 47.1 | 80 | 34 | 8.5 | 33.2 | 60 | 23 | 4.0 |
| 39.0 | 60 | 26 | 6.5 | 41.5 | 60 | 27 | 7.5 | 38.1 | 70 | 27 | 5.5 | 39.2 | 60 | 29 | 6.5 |
| 40.9 | 70 | 29 | 5.0 | 36.l | 60 | 23 | 6.0 | 33.5 | 60 | 24 | 2.5 | 46.7 | 70 | 27 | 7.5 |
| 35.9 | 60 | 23 | 5.5 | 38.5 | 60 | 23 | 6.0 | 43.6 | 70 | 27 | 10.0 | 30.4 | 70 | 32 | 4.0 |
| 43.5 | 70 | 28 | 5.5 | 42.4 | 60 | 24 | 9.0 | 41.0 | 80 | 32 | 6.5 | 43.2 | 60 | 25 | 5.5 |
| 47.9 | 80 | 34 | 6.5 | 46.5 | 70 | 31 | 5.5 | 50.2 | 50 | 26 | 9.0 | 30.6 | 60 | 26 | 3.5 |
| 33.8 | 70 | 26 | 4.5 | 43.1 | 80 | 32 | 6.0 | 34.4 | 50 | 22 | 4.0 | 43.3 | 70 | 28 | 7.5 |
| 41.1 | 70 | 26 | 8.0 | 50.8 | 60 | 26 | 10.0 | 47.9 | 80 | 34 | 6.5 | 32.4 | 60 | 22 | 5.5 |
| 38.7 | 70 | 26 | 8.0 | 44.2 | 70 | 28 | 4.5 | 46.2 | 50 | 21 | 10.0 | 35.5 | 60 | 22 | 4.5 |
Calculation:
The MINITAB is shown below,
Figure-1
From Figure-1 it is clear that the regression equation is
b.
To determine
To find: The value of variable
b.
Expert Solution
Answer to Problem 21E
The value of variable
Explanation of Solution
Given information: The data is shown below.
| 31.3 | 50 | 19 | 4.0 | 37.4 | 60 | 30 | 5.0 | 24.0 | 60 | 24 | 4.0 | 51.0 | 80 | 34 | 7.5 |
| 56.9 | 90 | 38 | 8.0 | 43.3 | 60 | 26 | 7.0 | 36.2 | 50 | 21 | 7.0 | 34.3 | 60 | 22 | 2.5 |
| 43.1 | 70 | 28 | 6.5 | 36.3 | 70 | 25 | 7.5 | 26.5 | 50 | 17 | 2.0 | 31.5 | 60 | 24 | 5.0 |
| 41.5 | 70 | 25 | 5.5 | 38.4 | 70 | 31 | 5.5 | 47.1 | 80 | 34 | 8.5 | 33.2 | 60 | 23 | 4.0 |
| 39.0 | 60 | 26 | 6.5 | 41.5 | 60 | 27 | 7.5 | 38.1 | 70 | 27 | 5.5 | 39.2 | 60 | 29 | 6.5 |
| 40.9 | 70 | 29 | 5.0 | 36.l | 60 | 23 | 6.0 | 33.5 | 60 | 24 | 2.5 | 46.7 | 70 | 27 | 7.5 |
| 35.9 | 60 | 23 | 5.5 | 38.5 | 60 | 23 | 6.0 | 43.6 | 70 | 27 | 10.0 | 30.4 | 70 | 32 | 4.0 |
| 43.5 | 70 | 28 | 5.5 | 42.4 | 60 | 24 | 9.0 | 41.0 | 80 | 32 | 6.5 | 43.2 | 60 | 25 | 5.5 |
| 47.9 | 80 | 34 | 6.5 | 46.5 | 70 | 31 | 5.5 | 50.2 | 50 | 26 | 9.0 | 30.6 | 60 | 26 | 3.5 |
| 33.8 | 70 | 26 | 4.5 | 43.1 | 80 | 32 | 6.0 | 34.4 | 50 | 22 | 4.0 | 43.3 | 70 | 28 | 7.5 |
| 41.1 | 70 | 26 | 8.0 | 50.8 | 60 | 26 | 10.0 | 47.9 | 80 | 34 | 6.5 | 32.4 | 60 | 22 | 5.5 |
| 38.7 | 70 | 26 | 8.0 | 44.2 | 70 | 28 | 4.5 | 46.2 | 50 | 21 | 10.0 | 35.5 | 60 | 22 | 4.5 |
Calculation:
From Figure-1 it is clear that the regression equation is
Substitute the values in above equation.
Thus, the value of variable
c.
To determine
To find: The confidence interval.
c.
Expert Solution
Answer to Problem 21E
The confidence intervalis
Explanation of Solution
Given information: The data is shown below.
| 31.3 | 50 | 19 | 4.0 | 37.4 | 60 | 30 | 5.0 | 24.0 | 60 | 24 | 4.0 | 51.0 | 80 | 34 | 7.5 |
| 56.9 | 90 | 38 | 8.0 | 43.3 | 60 | 26 | 7.0 | 36.2 | 50 | 21 | 7.0 | 34.3 | 60 | 22 | 2.5 |
| 43.1 | 70 | 28 | 6.5 | 36.3 | 70 | 25 | 7.5 | 26.5 | 50 | 17 | 2.0 | 31.5 | 60 | 24 | 5.0 |
| 41.5 | 70 | 25 | 5.5 | 38.4 | 70 | 31 | 5.5 | 47.1 | 80 | 34 | 8.5 | 33.2 | 60 | 23 | 4.0 |
| 39.0 | 60 | 26 | 6.5 | 41.5 | 60 | 27 | 7.5 | 38.1 | 70 | 27 | 5.5 | 39.2 | 60 | 29 | 6.5 |
| 40.9 | 70 | 29 | 5.0 | 36.l | 60 | 23 | 6.0 | 33.5 | 60 | 24 | 2.5 | 46.7 | 70 | 27 | 7.5 |
| 35.9 | 60 | 23 | 5.5 | 38.5 | 60 | 23 | 6.0 | 43.6 | 70 | 27 | 10.0 | 30.4 | 70 | 32 | 4.0 |
| 43.5 | 70 | 28 | 5.5 | 42.4 | 60 | 24 | 9.0 | 41.0 | 80 | 32 | 6.5 | 43.2 | 60 | 25 | 5.5 |
| 47.9 | 80 | 34 | 6.5 | 46.5 | 70 | 31 | 5.5 | 50.2 | 50 | 26 | 9.0 | 30.6 | 60 | 26 | 3.5 |
| 33.8 | 70 | 26 | 4.5 | 43.1 | 80 | 32 | 6.0 | 34.4 | 50 | 22 | 4.0 | 43.3 | 70 | 28 | 7.5 |
| 41.1 | 70 | 26 | 8.0 | 50.8 | 60 | 26 | 10.0 | 47.9 | 80 | 34 | 6.5 | 32.4 | 60 | 22 | 5.5 |
| 38.7 | 70 | 26 | 8.0 | 44.2 | 70 | 28 | 4.5 | 46.2 | 50 | 21 | 10.0 | 35.5 | 60 | 22 | 4.5 |
Calculation:
The MINITAB output is shown below.
Figure-2
From Figure-2 it is clear that the confidence interval is
d.
To determine
To find: The prediction interval.
d.
Expert Solution
Answer to Problem 21E
The prediction intervalis
Explanation of Solution
Given information: The data is shown below.
| 31.3 | 50 | 19 | 4.0 | 37.4 | 60 | 30 | 5.0 | 24.0 | 60 | 24 | 4.0 | 51.0 | 80 | 34 | 7.5 |
| 56.9 | 90 | 38 | 8.0 | 43.3 | 60 | 26 | 7.0 | 36.2 | 50 | 21 | 7.0 | 34.3 | 60 | 22 | 2.5 |
| 43.1 | 70 | 28 | 6.5 | 36.3 | 70 | 25 | 7.5 | 26.5 | 50 | 17 | 2.0 | 31.5 | 60 | 24 | 5.0 |
| 41.5 | 70 | 25 | 5.5 | 38.4 | 70 | 31 | 5.5 | 47.1 | 80 | 34 | 8.5 | 33.2 | 60 | 23 | 4.0 |
| 39.0 | 60 | 26 | 6.5 | 41.5 | 60 | 27 | 7.5 | 38.1 | 70 | 27 | 5.5 | 39.2 | 60 | 29 | 6.5 |
| 40.9 | 70 | 29 | 5.0 | 36.l | 60 | 23 | 6.0 | 33.5 | 60 | 24 | 2.5 | 46.7 | 70 | 27 | 7.5 |
| 35.9 | 60 | 23 | 5.5 | 38.5 | 60 | 23 | 6.0 | 43.6 | 70 | 27 | 10.0 | 30.4 | 70 | 32 | 4.0 |
| 43.5 | 70 | 28 | 5.5 | 42.4 | 60 | 24 | 9.0 | 41.0 | 80 | 32 | 6.5 | 43.2 | 60 | 25 | 5.5 |
| 47.9 | 80 | 34 | 6.5 | 46.5 | 70 | 31 | 5.5 | 50.2 | 50 | 26 | 9.0 | 30.6 | 60 | 26 | 3.5 |
| 33.8 | 70 | 26 | 4.5 | 43.1 | 80 | 32 | 6.0 | 34.4 | 50 | 22 | 4.0 | 43.3 | 70 | 28 | 7.5 |
| 41.1 | 70 | 26 | 8.0 | 50.8 | 60 | 26 | 10.0 | 47.9 | 80 | 34 | 6.5 | 32.4 | 60 | 22 | 5.5 |
| 38.7 | 70 | 26 | 8.0 | 44.2 | 70 | 28 | 4.5 | 46.2 | 50 | 21 | 10.0 | 35.5 | 60 | 22 | 4.5 |
Calculation:
From Figure-12 it is clear that the
e.
To determine
To find: The percentage of variation in variable
e.
Expert Solution
Answer to Problem 21E
The percentage of variation in variable
Explanation of Solution
Given information: The data is shown below.
| 31.3 | 50 | 19 | 4.0 | 37.4 | 60 | 30 | 5.0 | 24.0 | 60 | 24 | 4.0 | 51.0 | 80 | 34 | 7.5 |
| 56.9 | 90 | 38 | 8.0 | 43.3 | 60 | 26 | 7.0 | 36.2 | 50 | 21 | 7.0 | 34.3 | 60 | 22 | 2.5 |
| 43.1 | 70 | 28 | 6.5 | 36.3 | 70 | 25 | 7.5 | 26.5 | 50 | 17 | 2.0 | 31.5 | 60 | 24 | 5.0 |
| 41.5 | 70 | 25 | 5.5 | 38.4 | 70 | 31 | 5.5 | 47.1 | 80 | 34 | 8.5 | 33.2 | 60 | 23 | 4.0 |
| 39.0 | 60 | 26 | 6.5 | 41.5 | 60 | 27 | 7.5 | 38.1 | 70 | 27 | 5.5 | 39.2 | 60 | 29 | 6.5 |
| 40.9 | 70 | 29 | 5.0 | 36.l | 60 | 23 | 6.0 | 33.5 | 60 | 24 | 2.5 | 46.7 | 70 | 27 | 7.5 |
| 35.9 | 60 | 23 | 5.5 | 38.5 | 60 | 23 | 6.0 | 43.6 | 70 | 27 | 10.0 | 30.4 | 70 | 32 | 4.0 |
| 43.5 | 70 | 28 | 5.5 | 42.4 | 60 | 24 | 9.0 | 41.0 | 80 | 32 | 6.5 | 43.2 | 60 | 25 | 5.5 |
| 47.9 | 80 | 34 | 6.5 | 46.5 | 70 | 31 | 5.5 | 50.2 | 50 | 26 | 9.0 | 30.6 | 60 | 26 | 3.5 |
| 33.8 | 70 | 26 | 4.5 | 43.1 | 80 | 32 | 6.0 | 34.4 | 50 | 22 | 4.0 | 43.3 | 70 | 28 | 7.5 |
| 41.1 | 70 | 26 | 8.0 | 50.8 | 60 | 26 | 10.0 | 47.9 | 80 | 34 | 6.5 | 32.4 | 60 | 22 | 5.5 |
| 38.7 | 70 | 26 | 8.0 | 44.2 | 70 | 28 | 4.5 | 46.2 | 50 | 21 | 10.0 | 35.5 | 60 | 22 | 4.5 |
Calculation:
From Figure-1 it is clear that the percentage of variation in variable
f.
To determine
To find:Whether the given model is useful for prediction.
f.
Expert Solution
Answer to Problem 21E
The model is useful in prediction.
Explanation of Solution
Given information: The data is shown below.
| 31.3 | 50 | 19 | 4.0 | 37.4 | 60 | 30 | 5.0 | 24.0 | 60 | 24 | 4.0 | 51.0 | 80 | 34 | 7.5 |
| 56.9 | 90 | 38 | 8.0 | 43.3 | 60 | 26 | 7.0 | 36.2 | 50 | 21 | 7.0 | 34.3 | 60 | 22 | 2.5 |
| 43.1 | 70 | 28 | 6.5 | 36.3 | 70 | 25 | 7.5 | 26.5 | 50 | 17 | 2.0 | 31.5 | 60 | 24 | 5.0 |
| 41.5 | 70 | 25 | 5.5 | 38.4 | 70 | 31 | 5.5 | 47.1 | 80 | 34 | 8.5 | 33.2 | 60 | 23 | 4.0 |
| 39.0 | 60 | 26 | 6.5 | 41.5 | 60 | 27 | 7.5 | 38.1 | 70 | 27 | 5.5 | 39.2 | 60 | 29 | 6.5 |
| 40.9 | 70 | 29 | 5.0 | 36.l | 60 | 23 | 6.0 | 33.5 | 60 | 24 | 2.5 | 46.7 | 70 | 27 | 7.5 |
| 35.9 | 60 | 23 | 5.5 | 38.5 | 60 | 23 | 6.0 | 43.6 | 70 | 27 | 10.0 | 30.4 | 70 | 32 | 4.0 |
| 43.5 | 70 | 28 | 5.5 | 42.4 | 60 | 24 | 9.0 | 41.0 | 80 | 32 | 6.5 | 43.2 | 60 | 25 | 5.5 |
| 47.9 | 80 | 34 | 6.5 | 46.5 | 70 | 31 | 5.5 | 50.2 | 50 | 26 | 9.0 | 30.6 | 60 | 26 | 3.5 |
| 33.8 | 70 | 26 | 4.5 | 43.1 | 80 | 32 | 6.0 | 34.4 | 50 | 22 | 4.0 | 43.3 | 70 | 28 | 7.5 |
| 41.1 | 70 | 26 | 8.0 | 50.8 | 60 | 26 | 10.0 | 47.9 | 80 | 34 | 6.5 | 32.4 | 60 | 22 | 5.5 |
| 38.7 | 70 | 26 | 8.0 | 44.2 | 70 | 28 | 4.5 | 46.2 | 50 | 21 | 10.0 | 35.5 | 60 | 22 | 4.5 |
Calculation:
The null hypothesis is, the model is not useful for prediction and the alternative hypothesis is, the model is useful in prediction.
From Figure-1 it is clear that the p value is less than the level of significance of
Hence, the null hypothesis is rejected.
Thus, the model is useful in prediction.
g.
To determine
To explain:The test for the hypothesis
g.
Expert Solution
Explanation of Solution
Given information: The data is shown below.
| 31.3 | 50 | 19 | 4.0 | 37.4 | 60 | 30 | 5.0 | 24.0 | 60 | 24 | 4.0 | 51.0 | 80 | 34 | 7.5 |
| 56.9 | 90 | 38 | 8.0 | 43.3 | 60 | 26 | 7.0 | 36.2 | 50 | 21 | 7.0 | 34.3 | 60 | 22 | 2.5 |
| 43.1 | 70 | 28 | 6.5 | 36.3 | 70 | 25 | 7.5 | 26.5 | 50 | 17 | 2.0 | 31.5 | 60 | 24 | 5.0 |
| 41.5 | 70 | 25 | 5.5 | 38.4 | 70 | 31 | 5.5 | 47.1 | 80 | 34 | 8.5 | 33.2 | 60 | 23 | 4.0 |
| 39.0 | 60 | 26 | 6.5 | 41.5 | 60 | 27 | 7.5 | 38.1 | 70 | 27 | 5.5 | 39.2 | 60 | 29 | 6.5 |
| 40.9 | 70 | 29 | 5.0 | 36.l | 60 | 23 | 6.0 | 33.5 | 60 | 24 | 2.5 | 46.7 | 70 | 27 | 7.5 |
| 35.9 | 60 | 23 | 5.5 | 38.5 | 60 | 23 | 6.0 | 43.6 | 70 | 27 | 10.0 | 30.4 | 70 | 32 | 4.0 |
| 43.5 | 70 | 28 | 5.5 | 42.4 | 60 | 24 | 9.0 | 41.0 | 80 | 32 | 6.5 | 43.2 | 60 | 25 | 5.5 |
| 47.9 | 80 | 34 | 6.5 | 46.5 | 70 | 31 | 5.5 | 50.2 | 50 | 26 | 9.0 | 30.6 | 60 | 26 | 3.5 |
| 33.8 | 70 | 26 | 4.5 | 43.1 | 80 | 32 | 6.0 | 34.4 | 50 | 22 | 4.0 | 43.3 | 70 | 28 | 7.5 |
| 41.1 | 70 | 26 | 8.0 | 50.8 | 60 | 26 | 10.0 | 47.9 | 80 | 34 | 6.5 | 32.4 | 60 | 22 | 5.5 |
| 38.7 | 70 | 26 | 8.0 | 44.2 | 70 | 28 | 4.5 | 46.2 | 50 | 21 | 10.0 | 35.5 | 60 | 22 | 4.5 |
The null hypothesis is, there is no relationship between
For the variable
From Figure-1 it is clear that the p value is
Hence, the null hypothesis is not rejected.
Thus, there is no linear relationship between
For the variable
From Figure-1 it is clear that the p value is
Hence, the null hypothesis is rejected.
Thus, there is alinear relationship between
For the variable
From Figure-1 it is clear that the p value is
Hence, the null hypothesis is rejected.
Thus, there is a linear relationship between
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Chapter 13 Solutions
Elementary Statistics (Text Only)
Ch. 13.1 - Prob. 7ECh. 13.1 - Prob. 8ECh. 13.1 - In Exercises 9 and 10, determine whether the...Ch. 13.1 - Prob. 10ECh. 13.1 - Prob. 11ECh. 13.1 - Prob. 12ECh. 13.1 - Prob. 13ECh. 13.1 - Prob. 14ECh. 13.1 - Prob. 15ECh. 13.1 - Prob. 16E
Ch. 13.1 - Prob. 17ECh. 13.1 - Prob. 18ECh. 13.1 - Prob. 19ECh. 13.1 - Prob. 20ECh. 13.1 - Prob. 21ECh. 13.1 - Prob. 22ECh. 13.1 - Prob. 23ECh. 13.1 - Prob. 24ECh. 13.1 - Prob. 25ECh. 13.1 - Prob. 26ECh. 13.1 - Calculator display: The following TI-84 Plus...Ch. 13.1 - Prob. 28ECh. 13.1 - Prob. 29ECh. 13.1 - Prob. 30ECh. 13.1 - Confidence interval for the conditional mean: In...Ch. 13.2 - Prob. 3ECh. 13.2 - Prob. 4ECh. 13.2 - Prob. 5ECh. 13.2 - Prob. 6ECh. 13.2 - Prob. 7ECh. 13.2 - Prob. 8ECh. 13.2 - Prob. 9ECh. 13.2 - Prob. 10ECh. 13.2 - Prob. 11ECh. 13.2 - Prob. 12ECh. 13.2 - Prob. 13ECh. 13.2 - Prob. 14ECh. 13.2 - Prob. 15ECh. 13.2 - Prob. 16ECh. 13.2 - Prob. 17ECh. 13.2 - Dry up: Use the data in Exercise 26 in Section...Ch. 13.2 - Prob. 19ECh. 13.2 - Prob. 20ECh. 13.2 - Prob. 21ECh. 13.3 - Prob. 7ECh. 13.3 - Prob. 8ECh. 13.3 - Prob. 9ECh. 13.3 - In Exercises 9 and 10, determine whether the...Ch. 13.3 - Prob. 11ECh. 13.3 - Prob. 12ECh. 13.3 - Prob. 13ECh. 13.3 - For the following data set: Construct the multiple...Ch. 13.3 - Engine emissions: In a laboratory test of a new...Ch. 13.3 - Prob. 16ECh. 13.3 - Prob. 17ECh. 13.3 - Prob. 18ECh. 13.3 - Prob. 19ECh. 13.3 - Prob. 20ECh. 13.3 - Prob. 21ECh. 13.3 - Prob. 22ECh. 13.3 - Prob. 23ECh. 13 - A confidence interval for 1 is to be constructed...Ch. 13 - A confidence interval for a mean response and a...Ch. 13 - Prob. 3CQCh. 13 - Prob. 4CQCh. 13 - Prob. 5CQCh. 13 - Prob. 6CQCh. 13 - Construct a 95% confidence interval for 1.Ch. 13 - Prob. 8CQCh. 13 - Prob. 9CQCh. 13 - Prob. 10CQCh. 13 - Prob. 11CQCh. 13 - Prob. 12CQCh. 13 - Prob. 13CQCh. 13 - Prob. 14CQCh. 13 - Prob. 15CQCh. 13 - Prob. 1RECh. 13 - Prob. 2RECh. 13 - Prob. 3RECh. 13 - Prob. 4RECh. 13 - Prob. 5RECh. 13 - Prob. 6RECh. 13 - Prob. 7RECh. 13 - Prob. 8RECh. 13 - Prob. 9RECh. 13 - Prob. 10RECh. 13 - Air pollution: Following are measurements of...Ch. 13 - Icy lakes: Following are data on maximum ice...Ch. 13 - Prob. 13RECh. 13 - Prob. 14RECh. 13 - Prob. 15RECh. 13 - Prob. 1WAICh. 13 - Prob. 2WAICh. 13 - Prob. 1CSCh. 13 - Prob. 2CSCh. 13 - Prob. 3CSCh. 13 - Prob. 4CSCh. 13 - Prob. 5CSCh. 13 - Prob. 6CSCh. 13 - Prob. 7CS
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- Please conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points 2. Two-Sample T-Test: Compare the average sales revenue of two different regions to determine if there is a significant difference. (Hints: The null can be about maintaining status-quo or no difference among groups; if alternative hypothesis is non-directional use the two-tailed p-value from excel file to make a decision about rejecting or not rejecting null) H0 = H1=arrow_forwardPlease conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points 3. Paired T-Test: A company implemented a training program to improve employee performance. To evaluate the effectiveness of the program, the company recorded the test scores of 25 employees before and after the training. Determine if the training program is effective in terms of scores of participants before and after the training. (Hints: The null can be about maintaining status-quo or no difference among groups; if alternative hypothesis is non-directional, use the two-tailed p-value from excel file to make a decision about rejecting or not rejecting the null) H0 = H1= Conclusion:arrow_forwardPlease conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points. The data for the following questions is provided in Microsoft Excel file on 4 separate sheets. Please conduct these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points. 1. One Sample T-Test: Determine whether the average satisfaction rating of customers for a product is significantly different from a hypothetical mean of 75. (Hints: The null can be about maintaining status-quo or no difference; If your alternative hypothesis is non-directional (e.g., μ≠75), you should use the two-tailed p-value from excel file to…arrow_forward
- Please conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points. 1. One Sample T-Test: Determine whether the average satisfaction rating of customers for a product is significantly different from a hypothetical mean of 75. (Hints: The null can be about maintaining status-quo or no difference; If your alternative hypothesis is non-directional (e.g., μ≠75), you should use the two-tailed p-value from excel file to make a decision about rejecting or not rejecting null. If alternative is directional (e.g., μ < 75), you should use the lower-tailed p-value. For alternative hypothesis μ > 75, you should use the upper-tailed p-value.) H0 = H1= Conclusion: The p value from one sample t-test is _______. Since the two-tailed p-value is _______ 2. Two-Sample T-Test:…arrow_forwardPlease conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points. What is one sample T-test? Give an example of business application of this test? What is Two-Sample T-Test. Give an example of business application of this test? .What is paired T-test. Give an example of business application of this test? What is one way ANOVA test. Give an example of business application of this test? 1. One Sample T-Test: Determine whether the average satisfaction rating of customers for a product is significantly different from a hypothetical mean of 75. (Hints: The null can be about maintaining status-quo or no difference; If your alternative hypothesis is non-directional (e.g., μ≠75), you should use the two-tailed p-value from excel file to make a decision about rejecting or not…arrow_forwardThe data for the following questions is provided in Microsoft Excel file on 4 separate sheets. Please conduct a step by step of these statistical tests on separate sheets of Microsoft Excel. If the calculations in Microsoft Excel are incorrect, the null and alternative hypotheses, as well as the conclusions drawn from them, will be meaningless and will not receive any points. What is one sample T-test? Give an example of business application of this test? What is Two-Sample T-Test. Give an example of business application of this test? .What is paired T-test. Give an example of business application of this test? What is one way ANOVA test. Give an example of business application of this test? 1. One Sample T-Test: Determine whether the average satisfaction rating of customers for a product is significantly different from a hypothetical mean of 75. (Hints: The null can be about maintaining status-quo or no difference; If your alternative hypothesis is non-directional (e.g., μ≠75), you…arrow_forward
- What is one sample T-test? Give an example of business application of this test? What is Two-Sample T-Test. Give an example of business application of this test? .What is paired T-test. Give an example of business application of this test? What is one way ANOVA test. Give an example of business application of this test? 1. One Sample T-Test: Determine whether the average satisfaction rating of customers for a product is significantly different from a hypothetical mean of 75. (Hints: The null can be about maintaining status-quo or no difference; If your alternative hypothesis is non-directional (e.g., μ≠75), you should use the two-tailed p-value from excel file to make a decision about rejecting or not rejecting null. If alternative is directional (e.g., μ < 75), you should use the lower-tailed p-value. For alternative hypothesis μ > 75, you should use the upper-tailed p-value.) H0 = H1= Conclusion: The p value from one sample t-test is _______. Since the two-tailed p-value…arrow_forward4. Dynamic regression (adapted from Q10.4 in Hyndman & Athanasopoulos) This exercise concerns aus_accommodation: the total quarterly takings from accommodation and the room occupancy level for hotels, motels, and guest houses in Australia, between January 1998 and June 2016. Total quarterly takings are in millions of Australian dollars. a. Perform inflation adjustment for Takings (using the CPI column), creating a new column in the tsibble called Adj Takings. b. For each state, fit a dynamic regression model of Adj Takings with seasonal dummy variables, a piecewise linear time trend with one knot at 2008 Q1, and ARIMA errors. c. What model was fitted for the state of Victoria? Does the time series exhibit constant seasonality? d. Check that the residuals of the model in c) look like white noise.arrow_forwardce- 216 Answer the following, using the figures and tables from the age versus bone loss data in 2010 Questions 2 and 12: a. For what ages is it reasonable to use the regression line to predict bone loss? b. Interpret the slope in the context of this wolf X problem. y min ball bas oft c. Using the data from the study, can you say that age causes bone loss? srls to sqota bri vo X 1931s aqsini-Y ST.0 0 Isups Iq nsalst ever tom vam noboslios tsb a ti segood insvla villemari aixs-Yediarrow_forward
- 120 110 110 100 90 80 Total Score Scatterplot of Total Score vs. Putts grit bas 70- 20 25 30 35 40 45 50 Puttsarrow_forward10 15 Answer the following, using the figures and tables from the temperature versus coffee sales data from Questions 1 and 11: a. How many coffees should the manager prepare to make if the temperature is 32°F? b. As the temperature drops, how much more coffee will consumers purchase?ov (Hint: Use the slope.) 21 bru sug c. For what temperature values does the voy marw regression line make the best predictions? al X al 1090391-Yrit,vewolf 30-X Inlog arts bauoxs 268 PART 4 Statistical Studies and the Hunt forarrow_forward18 Using the results from the rainfall versus corn production data in Question 14, answer DOV 15 the following: a. Find and interpret the slope in the con- text of this problem. 79 b. Find the Y-intercept in the context of this problem. alb to sig c. Can the Y-intercept be interpreted here? (.ob or grinisiques xs as 101 gniwollol edt 958 orb sz) asiques sich ed: flow wo PEMAIarrow_forward
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