Answer:
a. Predicted Amount = $109.46
b. Confidence Interval = (94.84,124.08)
c. Interval = (110.6883,188.8517)
Step-by-step explanation:
Given
ŷ = 17.49 + 1.0334x.
SSE = 1541.4
a.
ŷ = 17.49 + 1.0334(89)
ŷ = 109.4626
ŷ = 109.46 --- Approximated
Predicted Amount = $109.46
b.
ŷ = 17.49 + 1.0334(89)
ŷ = 109.4626
ŷ = 109.46
First we calculate the standard deviation
variance = SSE/(n-2)
v = 1541.4/(9-2)
v = 1541.4/7
v = 220.2
s = √v
s = √220.2
s = 14.839
Then we calculate mean(x) and ∑(x - (mean(x))²
X --- Y -- Mean(x) --- ∑(x - (mean(x))²
148 -- 161 -- 43-- 1849
96 || 105|| -9 || 81
91 ||101 || -14 || 196
110 || 142 || 5 || 25
90 || 100 || -15 || 225
102 || ||120 ||-3|| 9
136 || 167 ||31 ||961
90 || 140 ||-15 ||225
82 || 98 ||-23 || 529
Sum 945 || 1134|| 0 ||4100
Mean (x) = 945/9 = 105
∑(x - (mean(x))² = 4100
α = 1 - 95% = 5%
α/2 = 2.5% = 0.025
tα,df = n − 2 = t0.025,7 =2.365
Confidence interval = 109.46 ± 2.365 * 14.839 √((1/9)+ (89-105)²/4100
Confidence Interval = (109.46 ± 14.62)
Confidence Interval = (94.84,124.08)
c.
ŷ = 17.49 + 1.0334(128)
ŷ = 149.7652
ŷ = 149.77
Interval = 149.77 ± 2.365 * 14.839 √((1/9)+ (128-105)²/4100
Interval = 149.77 ± 39.0817
Interval = (110.6883,188.8517)
