Use the data from the Copier Service Example to answer the following. a. What is

Question

Use the data from the Copier Service Example to answer the following.

a. What is a 95% confidence interval for yhat when xh= 4?

b. What is a 95% confidence interval for yhat when xh = 8?

c. What is a 95% confidence interval for the average of the next 25 observations when xh = 4?

Copiersx) Time(y) (xi-x)2 (yi-y)2 (xi-x)(yi-y) 2 20 10 3,166 175 4 60 1 265 (65) 3 46 4 916 (87) 2 41 10 1,244 (71) 1 12 17 4,130 (64) 10 137 24 3,689 607 5 68 0 68 (41) 5 89 0 162 64 1 4 17 5,222 (72) 2 32 10 1,960 (89) 9 144 15 4,588 610 10 156 24 6,357 797 6 93 1 280 100 3 36 4 1,621 (121) 4 72 1 18 (17) 8 100 8 563 190 7 105 4 826 201 8 131 8 2,996 438 10 127 24 2,574 507 4 57 1 371 (77) 5 66 0 105 (51) 7 101 4 612 173 7 109 4 1,071 229 5 74 0 5 (11) 9 134 15 3,333 520 7 112 4 1,277 250 2 18 10 3,395 (117) 5 73 0 11 (16) 7 111 4 1,206 243 6 96 1 389 118 8 123 8 2,184 374 5 90 0 189 69 2 20 10 3,166 (113) 2 28 10 2,330 (97) 1 3 17 5,368 (73) 4 57 1 371 (77) 5 86 0 95 49 9 132 15 3,106 502 7 112 4 1,277 250 1 27 17 2,427 (49) 9 131 15 2,996 493 2 34 10 1,786 (85) 2 27 10 2,427 (99) 4 61 1 233 (61) 5 77 0 1 4 sum 230 3,432 340 80,377 5,410 mean 5.111

Explanation / Answer

## Input first 2 columns as a data frame. First save the spreadsheet as a .csv file in your working directory
newd = read.csv("Untitled.csv", header = TRUE)
newd = newd[1:2]

## Fit a linear regression model for y on x
fit = lm(y~x, data = newd)
summary(fit)
## Looking at the results (p-value), x is a significant factor in determining y

### PART (a)
newd4 = data.frame(x = c(4))
pred4 = predict(fit, newdata = newd4, interval = "confidence", level = 0.95)

## fit lwr upr
## 59.56084 56.67078 62.45089 <-- [Answer (a)]

### PART (b)
newd8 = data.frame(x = c(8))
pred8 = predict(fit, newdata = newd8, interval = "confidence", level = 0.95)

## fit lwr upr
## 119.7018 115.8157 123.5879 <-- [Answer (b)]

### PART (c)

## Number of observations = nrow(newd) = 45
## Average of next 25 observations can only be done for first 21 rows

avg_25 = list()
for(i in 1:nrow(newd))
{
if(nrow(newd)-i>=24)
{
avg_25[i] = mean(newd$y[i:(i+24)])
}
else
{
avg_25[i] = NA
}
}

newd$z = as.double(avg_25)
rm(avg_25)
newd1 = subset(newd, subset = (newd$z > 0))

plot(y = newd1$z, x = newd1$x)

## Model (linear regression) this new variable z (average of next 25 observations) on x
fit_z = lm(z~x, data = newd1)
summary(fit_z)
## Looking at the results (p-value), x is NOT a significant factor in determining z

newd4 = data.frame(x = c(4))
pred4_z = predict(fit_z, newdata = newd4, interval = "confidence", level = 0.95)

## fit lwr upr
## 83.96217 81.84072 86.08362 <-- [Answer (c)]

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