The Starting Salary Case The following partial MINITAB regression output for the

Question

The Starting Salary Case The following partial MINITAB regression output for thestarting salary data relates to predicting the starting salary of amarketing graduate having a grade point average of 3.25. Predicted Values for New Observation: New Obs.: 1 Fit: 33.362 SE Fit: 0.213 95% CI: (32.813, 33.911) 95% PI: (31.878, 34.846) A) Report (as shown on the computer output) a point estimateof and a 95 percent confidence interval for the mean startingsalary of all marketing graduates having a grade point average of3.25. B)Report ( as shown on the computer output) a pointaverage prediction of and a 95 percent prediction interval for thestarting salray of an individual marketing graduate having a gradepoint average of 3.25. C) Rembering that s = .536321 and the distance value equals(sy/s)2 use sy from the computer output to handcalculate the distance value when x=3.25. D) Remembering that for the startain salary data n = 7,b0 = 14.816, and b1 = 5.7066, handcalculate (within rounding) the confidence interval of part (a) andthe prediction interval of part (b). The Starting Salary Case The following partial MINITAB regression output for thestarting salary data relates to predicting the starting salary of amarketing graduate having a grade point average of 3.25. Predicted Values for New Observation: New Obs.: 1 Fit: 33.362 SE Fit: 0.213 95% CI: (32.813, 33.911) 95% PI: (31.878, 34.846) A) Report (as shown on the computer output) a point estimateof and a 95 percent confidence interval for the mean startingsalary of all marketing graduates having a grade point average of3.25. B)Report ( as shown on the computer output) a pointaverage prediction of and a 95 percent prediction interval for thestarting salray of an individual marketing graduate having a gradepoint average of 3.25. C) Rembering that s = .536321 and the distance value equals(sy/s)2 use sy from the computer output to handcalculate the distance value when x=3.25. D) Remembering that for the startain salary data n = 7,b0 = 14.816, and b1 = 5.7066, handcalculate (within rounding) the confidence interval of part (a) andthe prediction interval of part (b).

Explanation / Answer

As mentioned in the part d) that b0 =14.816, and b1 = 5.7066. The regression equation is Y =14.816+5.7066X 95% CI: (32.813, 33.911) 95% PI: (31.878, 34.846) s = 0.536321 95% CI: (32.813, 33.911) 95% PI: (31.878, 34.846) s = 0.536321 95% PI: (31.878, 34.846) s = 0.536321 A) From the given minitab ouput the point estimate of and a 95percent confidence interval for the mean starting salary of allmarketing graduates having a grade point average of 3.25. is 32.813 +0.536321 = 33.3493 Which is the point estimate C)  x=3.25 then the value of the distance value Y =14.816+5.7066(3.25) = 33.3625 which is nothing but the fit value

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