Question

In: Statistics and Probability

A researcher believes that there is a positive relationship between the time spent in studying and...

A researcher believes that there is a positive relationship between the time spent in studying and the score a student gets at the end of the semester. She selected a sample of students and recorded the following data:

Time Spent in minutes

Score out of 100

120

86

75

83

60

78

45

75

180

91

30

72

90

84

After analyzing the data using Microsoft Excel, the researcher got the outputs below:

Answer the following (Please, use at least 2 decimals when you type numbers ):

a.    The correlation coefficient (r) = Answer

b.    The relationship between the time spent in studying and the score a student gets is: Answerweak and positivestrong and negativeweak and negativestrong and positive

c.    The y-intercept equals: Answer

d.    The regression (time spent) coefficient equals = Answer

e.    The regression model will be: AnswerScore = 70.704 + 0.123 MinuteMinute = 0.123 + 70.704 ScoreScore = 0.123 + 70.704 MinuteMinute = 70.704 + 0.123 Score

f.     From the regression model, the researcher can predict that if a student spends 2.5 hours, the score he gets will be: Answer

g.    From the regression model, if a student did not study at all, his score would be expected to be: Answer

h.    The coefficient of determination for this model is:  Answer95708890%

Solutions

Expert Solution

## Q ) A researcher believes that there is a positive relationship between the time spent in studying and the score a student gets at the end of the semester.

## a) The correlation coefficient (r) = 0.9507

## b) The relationship between the time spent in studying and the score a student gets is : = postive strong

## c) The y-intercept equals = 70.7037

## d) The regression (time spent) coefficient equals : 0.123

## e) The regression model will be:  Answer

Score = 70.704 + 0.123 Minute

## f) From the regression model, the researcher can predict that if a student spends 2.5 hours, the score

he gets will be :

2.5 hours into the Minute is 150 min :

predict y when x = 150  

that is score =   70.704 + 0.123 *150 =

score = 89.2222

## g) From the regression model, if a student did not study at all, his score would be expected to be

Answer : x = 0 then y hat = 70.704 that is score will be 70.704

## h) The coefficient of determination for this model is : = 0.9038 %

variatiion explained by model is : 90.38 %

## MIcrosoft Excel Output :


Related Solutions

A teacher is interested in the relationship between the time spent studying for an exam and...
A teacher is interested in the relationship between the time spent studying for an exam and exam score. The table lists scores for 5 students. The value of b0 = 67.91 and the value of b1 = 0.75 Hours studied 5 18 3 15 17 Exam score 63 87 79 72 82 . Step 4 of 6 : Calculate the Coefficient of Determination (R2). Round answer to four decimal places.
A researcher measures the relationship between the number of interruptions during a class and time spent...
A researcher measures the relationship between the number of interruptions during a class and time spent "on task" (in minutes). Answer the following questions based on the results provided. Number of Interruptions Time Spent "On Task" 8 15 4 38 7 18 2 32 Part (a) Compute the Pearson correlation coefficient. (Round your answer to three decimal places.) Part (b) Multiply each measurement of interruptions times 3 and recalculate the correlation coefficient. (Round your answer to three decimal places.) Part...
A researcher measures the relationship between the number of interruptions during a class and time spent...
A researcher measures the relationship between the number of interruptions during a class and time spent "on task" (in minutes). Answer the following questions based on the results provided. Number of Interruptions Time Spent "On Task" 9 15 3 37 7 17 2 32 Part (a) Compute the Pearson correlation coefficient. (Round your answer to three decimal places.) Part (b) Multiply each measurement of interruptions times 3 and recalculate the correlation coefficient. (Round your answer to three decimal places.) Part...
A researcher measures the relationship between the number of interruptions during a class and time spent...
A researcher measures the relationship between the number of interruptions during a class and time spent "on task" (in minutes). Answer the following questions based on the results provided. Number of Interruptions - 9, 4, 6,3 Time Spent "On Task"- 18, 38, 17, 32 Part (a) Compute the Pearson correlation coefficient. (Round your answer to three decimal places.) Part (b) Multiply each measurement of interruptions times 3 and recalculate the correlation coefficient. (Round your answer to three decimal places.) Part...
A researcher measures the relationship between the number of interruptions during a class and time spent...
A researcher measures the relationship between the number of interruptions during a class and time spent "on task" (in minutes). Answer the following questions based on the results provided. Number of Interruptions Time Spent "On Task" 11 15 6 37 9 20 4 32 Part (a) Compute the Pearson correlation coefficient. (Round your answer to three decimal places.) Part (b) Multiply each measurement of interruptions times 3 and recalculate the correlation coefficient. (Round your answer to three decimal places.) Part...
In studying the relationship between average daily temperature and time spent on a telephone, suppose you...
In studying the relationship between average daily temperature and time spent on a telephone, suppose you computed r=0.607 using n=31 data points. Using the critical values table below, determine if the value of r is significant or not.
Homer is studying the relationship between the average daily temperature and time spent watching television and...
Homer is studying the relationship between the average daily temperature and time spent watching television and has collected the data shown in the table. The line of best fit for the data is yˆ=−0.6x+94.5. Assume the line of best fit is significant and there is a strong linear relationship between the variables. Temperature (Degrees) 40506070 Minutes Watching Television 70655952 (a) According to the line of best fit, what would be the predicted number of minutes spent watching television for an...
A researcher measures the relationship between sleep medication use (times used per week) and time spent...
A researcher measures the relationship between sleep medication use (times used per week) and time spent working (in hours per week). Answer the following questions based on the results provided. Sleep Medication Use Time Spent Working 9 16 4 39 7 18 2 31 Part A) Compute the Pearson correlation coefficient. (Round your answer to three decimal places.) Part B) Multiply each measurement of drug use times 3 and recalculate the correlation coefficient. (Round your answer to three decimal places.)...
A researcher measures the relationship between sleep medication use (times used per week) and time spent...
A researcher measures the relationship between sleep medication use (times used per week) and time spent working (in hours per week). Answer the following questions based on the results provided. Sleep Medication Use Time Spent Working 8 16 3 39 7 19 2 30 A) Compute the Pearson correlation coefficient. (Round your answer to three decimal places.) _____ B) Multiply each measurement of drug use times 3 and recalculate the correlation coefficient. (Round your answer to three decimal places.) _____...
A researcher measures the relationship between sleep medication use (times used per week) and time spent...
A researcher measures the relationship between sleep medication use (times used per week) and time spent working (in hours per week). Answer the following questions based on the results provided. Sleep Medication Use 9,4, 7, 3 Time Spent Working 16, 38, 18, 31 Compute the Pearson correlation coefficient. (Round your answer to three decimal places.) Multiply each measurement of drug use times 3 and recalculate the correlation coefficient. (Round your answer to three decimal places.) Divide each measurement in half...
ADVERTISEMENT
ADVERTISEMENT
ADVERTISEMENT