Question

An agribusiness performed a regression of wheat yield (bushels per acre) using observations on 26 test plots with four predictors (rainfall, fertilizer, soil acidity, hours of sun). The standard error was 1.29 bushels.
(a) Find the approximate width of a 95% prediction interval for wheat yield. (Round your answer to 2 decimal places.)
yˆi ±
(b) Find the approximate width using the quick rule. (Round your answer to 2 decimal places.)
yˆi ±
(c) The quick rule gives a similar result.
a. Yes
b. No

Answer 1

Answer:

A)y ± $t_{21}$ ( 1.29 )

B) y ± 2.58

C ) yes

Step-by-step explanation:

standard error = 1.29 bushels

approximate width of 95% prediction interval for an individual can be represented as ;  $y$ ± $t_{n-k-1}$ ( SE )

$t_{n-k-1} = critical value of t at n-k-1$ degree of freedom

n = 26

k = 4

a) Approximate width of a 95% prediction interval for wheat yield

$y$ ± $t_{n-k-1}$ ( SE )

Insert the values into the equation

n - k -1 = 26 - 4 - 1 = 21

y ± $t_{21}$ ( 1.29 ) you can get the value of $t_{21}$ from the t table for 21 degrees at 0.05 significance level

B)  The approximate width using the quick rule

y ± 2SE

y ± 2(1.29) = y ± 2.58

C) The quick rule gives a similar result = yes