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(SHOW R CODE): 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

Part 1

1. The histogram shows a bell shaped distribution  for the data.

2.The boxplot shows a very lower amount of variation in the data. Again the central line (i.e. median) is close to the upper boundary (i.e. third quartile). Thus the distribution seems negatively skewed. In addition, a number of outliers are observed from the boxplot.

3. Five number summary

Min. 1st Qu. Median    3rd Qu. Max.
2.893 3.437 3.473    3.482 4.301

4. IQR=3 rd Quartile-1 st Quartile=3.482-3.437=.045

5. Mean=3.502

6. SD=0.3493518

7. Since the distribution is skewed and has a number of outliers as observed from the boxplot, appropriate measure of center is median=3.473 and that of dispersion as QD=IQR/2=0.0225

Part 2

95% CI for population mean (based on the t distribution as population variance is not known and sample size is small) is

(3.338698, 3.665702)

R Code

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,col="red",xlab="paper weight",ylab="frequency",main="Histogram of the weight of paper")
boxplot(x,col="blue",xlab="",ylab="paper weight",main="Boxplot of the weight of paper")
summary(x)
sd(x)
t.test(x)