Question

1. Listed below are altitudes​ (thousands of​ feet) and outside air temperatures​ (°F) recorded during a flight. Find the​ (a) explained​ variation, (b) unexplained​ variation, and​ (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear​ correlation, so it is reasonable to use the regression equation when making predictions. For the prediction​ interval, use a​ 95% confidence level with the altitude of 6327 ft​ (or 6.327 thousand​ feet).

Altitude	4	10	13	22	29	31	32
Temperature	60	38	24	−5	−31	−41	−60

a. Find the explained variation. ​(Round to two decimal places as​ needed.)

b. find the unexplained variation ( two decimal places)

c. the indicated prediction interval (two decimal places)

2. The table below lists weights​ (carats) and prices​ (dollars) of randomly selected diamonds. Find the​ (a) explained​ variation, (b) unexplained​ variation, and​ (c) indicated prediction interval. There is sufficient evidence to support a claim of a linear​ correlation, so it is reasonable to use the regression equation when making predictions. For the prediction​ interval, use a​ 95% confidence level with a diamond that weighs 0.8 carats.

Weight	0.3	0.4	0.5	0.5	1.0	0.7
Price	​$510	​$1154	​$1338	​$1405	​$5654	​$2264

a. Find the explained variation. ​(Round to the nearest whole number as​ needed.)

b. Find the unexplained variation (Round to the nearest whole number as​ needed.)

c. Find the prediction interval (Round to the nearest whole number as​ needed.)

Answer 1

## R command

## Question 1)
Altitude=c(4,10,13,22,29,31,32)
Temperature=c(60, 38, 24, -5, -31, -41, -60)

model.aov=aov(Temperature~Altitude)
summary(model.aov)
# a. The explained variation is 11653.
# b. The unexplained variation is 202.

model=lm(Temperature~Altitude)
summary(model)

predict(model, data.frame(Altitude=6.327), level=1-0.05, interval="prediction")

# c. The 95% prediction interval with the altitude of 6327 is (32.8414, 71.43875)

## Question 2

##
Weight=c(0.3,0.4,0.5,0.5,1.0,0.7)
Price=c(510,1154,1338,1405,5654,2264)
model.aov1=aov(Price~Weight)
summary(model.aov1)

# a. The explained variation 16024104.

# b. The unexplained variation 1107789.

model.1=lm(Price~Weight)
summary(model.1)
predict(model.1, data.frame(Weight=0.8), level=1-0.05, interval="prediction")

# c. The 95% prediction interval of price when weighs=0.8 carats is (2031.154, 5414.442)

## End