A study of external disk drives finds the following. The least squares line was

Question

A study of external disk drives finds the following.

The least squares line was found to be "Price = 26.72 + 0.104 Capacity with Se = 13.19 and SE(b1) = 0.0038. Complete parts a through c below.

a) Find the predicted Price of a 800 GB hard drive.

b) Find the 95% confidence interval for the mean Price of all 800 GB disk drives.

c) Find the 95% prediction interval for the Price of a 800 GB hard drive.

Capacity (in 60 | 120 | 250 | 320 | 800 | 1500 | 4000 |x-100714| SD(x)= 1412.59 GB) Price (in $)129.50 | 34.00| 50.95|69.95| 95.00|205.00|439.00| y=131.91 | SD(y)=148.03

Explanation / Answer

SolutonA:

capacity <- c(60,120,250,320,800,1500,4000)
price <- c(29.5,34,50.95,69.95,95,205,439)
mod1.lm =lm(price~capacity)
predict(mod1.lm, newdata, interval="confidence")

predcited price of a 800 GB is 110.2792 Price.

SolutionB:

code in R:

capacity <- c(60,120,250,320,800,1500,4000)
price <- c(29.5,34,50.95,69.95,95,205,439)

mod1.lm =lm(price~capacity)

newdata = data.frame(capacity=800)

predict(mod1.lm, newdata, interval="confidence")

output:

predict(mod1.lm, newdata, interval="confidence")
fit lwr upr
1 110.2792 97.30188 123.2565

95% confidence interval for the mean Price of all 800 GB disk drives lies in between 97.30188 and

123.2565

lower limit=97.30188

upper limit=123.2565

Solutionc:

capacity <- c(60,120,250,320,800,1500,4000)
price <- c(29.5,34,50.95,69.95,95,205,439)
mod1.lm =lm(price~capacity)
newdata = data.frame(capacity=800)
predict(mod1.lm, newdata, interval="predict")

output:

95%  prediction interval for the Price of a 800 GB hard drive lies in between

73.96894 and 146.5894

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