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Use the Computer Display to Answer the Question =31.55+10.90= 31.55 + 10.90

question 23

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Use the computer display to answer the question.
-A collection of paired data consists of the number of years that students have studied Spanish and their scores on a Spanish language proficiency test. A computer program was used to obtain the least squares linear Regression line and the computer output is shown below. Along with the paired sample data, the program was Also given an x value of 2 (years of study) to be used for predicting test score.
The regression equation is Score =31.55+10.90= 31.55 + 10.90 Years.

 Use the computer display to answer the question. -A collection of paired data consists of the number of years that students have studied Spanish and their scores on a Spanish language proficiency test. A computer program was used to obtain the least squares linear Regression line and the computer output is shown below. Along with the paired sample data, the program was Also given an x value of 2 (years of study)  to be used for predicting test score. The regression equation is Score  = 31.55 + 10.90  Years.       \mathrm { S } = 5.651 \quad \mathrm { R } - \mathrm { Sq } = 83.0 \% \quad \mathrm { R } - \mathrm { Sq } (  Adj  )  = 82.7 \%   Predicted values   \begin{array} { l c c c }  \text { Fit } & \text { StDev Fit } & 95.0 \% \text { CI } & 95.0 \% \text { PI } \\ 53.35 & 3.168 & ( 42.72,63.98 )  & ( 31.61,75.09 )  \end{array}   For a person who studies for 2 years, obtain the 95% prediction interval and write a statement interpreting the Interval.  A)   ( 42.72,63.98 )  ; We can be  95 \%  confident that the mean test score of all individuals who study 2 years will lie in the interval  ( 42.72,63.98 )   B)   ( 42.72,63.98 )  ; We can be  95 \%  confident that the test score of an individual who studies 2 years will lie in the interval  ( 42.72,63.98 )   C)   ( 31.61,75.09 )  ; We can be  95 \%  confident that the test score of an individual who studies 2 years will lie in the interval (31.61, 75.09)  D)   ( 31.61,75.09 )  ; We can be  95 \%  confident that the mean test score of all individuals who study 2 years will lie in the interval  ( 31.61,75.09 )


S=5.651RSq=83.0%RSq(\mathrm { S } = 5.651 \quad \mathrm { R } - \mathrm { Sq } = 83.0 \% \quad \mathrm { R } - \mathrm { Sq } ( Adj ) =82.7%) = 82.7 \%

Predicted values

 Fit  StDev Fit 95.0% CI 95.0% PI 53.353.168(42.72,63.98) (31.61,75.09) \begin{array} { l c c c } \text { Fit } & \text { StDev Fit } & 95.0 \% \text { CI } & 95.0 \% \text { PI } \\53.35 & 3.168 & ( 42.72,63.98 ) & ( 31.61,75.09 ) \end{array}
For a person who studies for 2 years, obtain the 95% prediction interval and write a statement interpreting the
Interval.


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