Respuesta :
Answer:
answer is
[tex]-0.245 \pm2.160(0.205)[/tex]
Step-by-step explanation:
After working this way for 6 months he takes a simple random sample of 15 days. He records how long he walked that day (in hours) as recorded by his fitness watch as well as his billable hours for that day as recorded by a work app on his computer.
Slope is -0.245
Sample size n = 15
Standard error is 0.205
Confidence level 95
Sognificance level is (100 - 95)% = 0.05
Degree of freedom is n -2 = 15 -2 = 13
Critical Value =2.16 = [using excel = TINV (0.05, 13)]
Marginal Error = Critical Value * standard error
= 2.16 * 0.205
= 0.4428
[tex]-0.245 \pm2.160(0.205)[/tex]
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Answer:
The correct answer is C. [-0.245±2.160*0.205]
Step-by-step explanation:
Hello!
To analyze the if walking while working (independent variable X) affects his working productivity (dependent variable Y) James took a random sample of 15 days (from a 6 month period in which he worked and trained at home) and recorded how long he walked and his billable hours for that day.
Using this information he estimated the linear regression between the working productivity and the time spent training.
^Y= 7.785 -0.245X
The interest is to estimate the change in the average number of billable hours every time he increases the time spent walking in 1 hour.
To estimate the range of values the average billable hours may take each time the training time increases sone hour, you have to calculate a confidence interval for the population slope of the regression. The statistic is a t-test:
[tex]t= \frac{b-\beta}{Sb} ~~t_{n-1}[/tex]
[b±[tex]t_{n-2;1-\alpha /2}[/tex]*Sb]
Using a 95% confidence level the value of t is [tex]t_{n-2;1-\alpha /2}= t_{13; 0.975}= 2.160[/tex]
b= -0.245
Sb= 0.205
[-0.245±2.160*0.205]
[-0.68780.1978]
Using a 95% confidence level you'd expect the interval [-0.68780.1978] to contain the value of the average billable hours when the time spent walking increases in 1 hour.
I hope this helps!
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