An agribusiness performed a regression of wheat yield (bushels per acre) using o

Question

An agribusiness performed a regression of wheat yield (bushels per acre) using observations on 30 test plots with four predictors (rainfall, fertilizer, soil acidity, hours of sun). The standard error was 1.08 bushels.

Find the approximate width of a 95% prediction interval for wheat yield. (Round your answer to 2 decimal places.)

An agribusiness performed a regression of wheat yield (bushels per acre) using observations on 30 test plots with four predictors (rainfall, fertilizer, soil acidity, hours of sun). The standard error was 1.08 bushels.

Explanation / Answer

Solution:

a) From the given information: SE =1.08, n=30 and k = 4
The degrees of freedom(df) is,
df = n-k-1
= 30-4-1
= 30-5
= 25
Looking for the critical value of t from t table for 25 degrees at 0.05 level of significance for a two-tailed test,
The t-critical value at the 0.05 level of significance is: 2.1098
Thus, the 95 percent prediction interval for an individual's salary is:
yi ± t25(1.08)
yi ± (2.1098)(1.08)
yi ± 2.28
b) The quick 95 percent prediction for interval for individual y value is given by:
yi ±2SE
yi ±2(1.08)
yi ±2.16
It can be observed that the quick rule will give almost similar results as the 95 percent prediction interval.

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