Question

In a study where the capability of a gauge was measured by the weight of paper, the following measurements were obtained: 3.481 3.448 3.485 3.475 3.472 3.477 3.472 3.464 3.508 3.170 4.123 3.470 2.893 3.473 4.201 3.474 4.301 3.021 3.231 3.405

Part I Using R do the following: 1. Draw an appropriately titled histogram. Interpret the graph. 2. Draw an appropriately titled boxplot. Interpret the graph. 3. Compute the five-number summary. 4. Find the interquartile range. 5. Find the mean 6. Find the standard deviation. 7. What would be the appropriate measures of center and dispersion?

Part 2 Assuming the above sample data is coming from a normal population, construct a 95% confidence interval for the population mean.

Answer 1

1. R-code for histogram plot :

> x<-c(3.481, 3.448, 3.485, 3.475, 3.472, 3.477, 3.472, 3.464, 3.508, 3.170, 4.123, 3.470, 2.893, 3.473, 4.201, 3.474, 4.301, 3.021, 3.231, 3.405)

> hist(x,main="Distribution of Weight of paper",xlab="Weight of paper")

Plot :

Interpretation : Most of the papers weigh in a vert tight short range around 3.5 ; where as there are few papers weighing excessively high and excessively low.

2. R-code for boxplot :

> boxplot(x,main="Paper weight distribution",xlab="Frequency",ylab="Weight of paper")

Plot :

Interpretation : Again, we see there is highest number of papers weighing in the tight range around 3.5 ; whereas there are few outliers on either side.

3. 5-point summary is shown :

> summary(x)
Min. 1st Qu. Median Mean 3rd Qu. Max.
2.893 3.437 3.472 3.502 3.482 4.301  

4. IQR = Q3 - Q1 = 3.482 - 3.437 = 0.045

5. Mean = 3.502

6. Sd = 0.3493518

> sd(x)
[1] 0.3493518

7. Central measure = Mean ; Dispersion = Std. deviation . This is so because we observed in the plots that the data is symmetrically distributed and not a skewed one.

8. 95% CI =