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# Yes I have a test that is due today and I noticed you answered

### Customer Question

yes I have a test that is due today and I noticed you answered most of the questions for someone else....how does this work
Submitted: 1 year ago.
Category: Calculus and Above
Customer: replied 1 year ago.
Given the linear correlation coefficient r and the sample size n, determine the critical values of r and use your finding to state whether or not the given r represents a significant linear correlation. Use a significance level of 0.05.
r = 0.75, n = 9
Critical values: r = ±0.666, no significant linear correlation
Critical values: r = 0.666, no significant linear correlation
Critical values: r = -0.666, no significant linear correlation
Critical values: r = ±0.666, significant linear correlation
5 points
Question 2
Construct a scatterplot for the given data.
Choose A, B, C, or D.
5 points
Question 3
Find the value of the linear correlation coefficient r.
-0.054
0.214
0.109
-0.078
5 points
Question 4
Suppose you will perform a test to determine whether there is sufficient evidence to support a claim of a linear correlation between two variables. Find the critical values of r given the number of pairs of data n and the significance level alpha.
n = 11, = 0.01
r = 0.765
r = ± 0.602
r = 0.735
r = ± 0.735
5 points
Question 5
Use the given data to find the best predicted value of the response variable.
Six pairs of data yield r = 0.789 and the regression equation What is the best predicted value of y for x = 5?
22.0
18.0
18.5
19.0
5 points
Question 6
Use the given data to find the equation of the regression line. Round the final values to three significant digits, if necessary.
Choose A, B, C, or D.
5 points
Question 7
Is the data point, P, an outlier, an influential point, both, or neither?
Neither
Outlier
Both
Influential point
5 points
Question 8
Use the given information to find the coefficient of determination.
A regression equation is obtained for a collection of paired data. It is found that the total variation is 24.488, the explained variation is 15.405, and the unexplained variation is 9.083. Find the coefficient of determination.
0.629
0.590
1.590
0.371
5 points
Question 9
Use the computer display to answer the question.
82.7%
17.0%
91.1%
83.0%
5 points
Question 10
Find the explained variation for the paired data.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is = 44.8447 + 3.52427x. Find the explained variation.
498.103
511.724
87.4757
599.2
5 points
Question 11
Find the unexplained variation for the paired data.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is = 44.8447 + 3.52427x. Find the unexplained variation.
511.724
87.4757
96.103
599.2
5 points
Question 12
Find the total variation for the paired data.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is = 44.8447 + 3.52427x. Find the total variation.
599.2
498.103
511.724
87.4757
5 points
Question 13
Find the standard error of estimate for the paired data.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is = 44.8447 + 3.52427x. Find the the standard error of estimate.
4.1097
7.1720
5.3999
13.060
5 points
Question 14
Construct the indicated prediction interval for an individual y.
The paired data below consists of test scores and hours of preparation for 5 randomly selected students. The equation of the regression line is = 44.845 + 3.524x and the standard error of estimate is Se = 5.40. Find the 99% prediction interval for the test score of a person who spent 7 hours preparing for the test.
58 < y < 82
62 < y < 78
35 < y < 104
32 < y < 107
5 points
Question 15
Solve the problem.
1.936 < B1 < 4.874
0.686 < B1 < 6.124
0.322 < B1 < 6.488
0.134 < B1 < 6.676
5 points
Question 16
Use computer software to find the multiple regression equation. Can the equation be used for prediction?
An anti-smoking group used data in the table to relate the carbon monoxide of various brands of cigarettes to their tar and nicotine content.
CO = 1.37 - 5.53TAR + 1.33NIC; Yes, because the R2 is high
CO = 1.37 + 5.50TAR - 1.38NIC; Yes, because the P-value is high
CO = 1.3 + 5.5TAR - 1.3NIC; Yes, because the adjusted R2 is high
CO = 1.25 + 1.55TAR - 5.79NIC; Yes, because the P-value is too low
5 points
Question 17
Use computer software to obtain the multiple regression equation and identify R2, adjusted R2, and the P-value.
An anti-smoking group used data in the table to relate the carbon monoxide of various brands of cigarettes to their tar and nicotine content.
0.976, 0.921, 0.002
0.943, 0.934, 0.000
0.931, 0.902, 0.000
0.861, 0.900, 0.015
5 points
Question
Expert:  Mr. Glenn replied 1 year ago.
Hi there!
Thank you for requesting me (I am trigeoscal or aka Mr. Glenn)
What time or how many hours/minutes do you need the answers?
I am now working on these.
By the way, you forgot to include the actual data for Q2, Q3 Q6, Q7 and Q9
kindly sent them (the complete data) so that I can now answer those questions
or better, kindly attach and send the actual questions in word or excel format/file including the complete data.
Regards,
Mr. Glenn G. (Trigeoscal)
*************************************************
You can request me in future by writing "For Mr. Glenn G. only".
Expert:  Mr. Glenn replied 1 year ago.
Hi there!
By the way, you forgot to include the actual data for Q2, Q3 Q6, Q7 and Q9
kindly sent them (the complete data) so that I can now answer those questions
or better, kindly attach and send the actual questions in word or excel format/file including the complete data.