I used 2.55 and 2.6 z critical value for calculating Interval Estimate but I still got them wrong. Please help.

Question

The following table contains data on the number of pages printed per minute and the retail price for a random sample of printers.

PPM(first column),Price (Second Column)->
4	236.65
3	294.68
3.6	341.49
1.8	169.41
2.7	288.65
4	436.54
2	155.91
3.5	336.83
3.1	298.57
3.6	334.64
3.1	335.45
4.1	326.45

Part E

Provide a 96% prediction interval for the retail price of a printer if this printer prints ____________.

Round the interval estimates to 8 decimal places if possible.

 (a) 2.1 pages per minutes =   and 

 (b) 3.1 pages per minutes =   and 

 (c) 4.1 pages per minutes =   and 

Answer 1

First the basics. You are examining Bivariate data which implies a regression analysis. The test statistic for the parameters will be a t distribution, NOT a z distribution. Also you need to do a regression. Your Z values would not be needed . Also, they seem to be Z values for a 99 % confidence interval.

Since you do NOT know the population parameters you need to estimate them from the sample. Here are the parameters for your bivariate data, and the estimates (96% confidence interval) of the prices.

mean ppm =. 3.208333 mean price. 296.2725

PPM lwr upr

1 2.1 142.5401 277.4666

2 3.1 250.6984 324.9819

3 4.1 307.1824 424.1716

Using R the following code will produce the answer.

# create a data frame

PPM = c(4,3,3.6,1.8,2.7,4,2,3.5,3.1,3.6,3.1,4.1)

Price = c(236.65, 294.68, 341.49, 169.41, 288.65, 436.54, 155.91, 336.83, 298.57,

334.64, 335.45, 326.45)

df = data.frame(PPM=PPM, Price = Price)

# Fit a linear regression model and print a summary.

model = lm(Price ~ PPM)

summary(model)

# To find the confidence interval in R, create a new data.frame

# with the desired value to predict.

# The prediction is made with the predict() function.

# The interval argument is set to ‘confidence’ to output the mean interval.

# Confidence interval is set using level command

new.dat <- data.frame(PPM = c(2.1,3.1,4.1))

predict(model, newdata = new.dat, interval = 'confidence',level=.96)

NOTE: if you are expected to do this by hand you can calculate it with the correct formula. See https://rpubs.com/aaronsc32/regression-confidence-prediction-intervals#:~:text=Confidence%20(Mean)%20Intervals&text=To%20find%20the%20confidence%20interval,to%20output%20the%20mean%20interval. for a good example.

here is the r code which will calculate the above