A Harris corporation / University of Florida study was undertaken to determine whether a manufacturing process performed at a remote location can be established locally. Test devices were set up at both the old and new locations and voltage readings on the process were obtained. A "good" process was considered to be one with voltage reading of at least 9.2 volts (with larger readings being better than smaller readings). The table contains voltage readings for 30 production runs at each location.

Old New

Location Location

9.98 10.12 9.84 9.19 10.01 8.82

10.26 10.05 10.15 9.63 8.82 8.65

10.05 9.80 10.02 10.10 9.43 8.51

10.29 10.15 9.80 9.70 10.03 9.14

10.03 10.00 9.73 10.09 9.85 9.75

8.05 9.87 10.01 9.60 9.27 8.78

10.55 9.55 9.98 10.05 8.83 9.35

10.26 9.95 8.72 10.12 9.39 9.54

9.97 9.70 8.80 9.49 9.48 9.36

9.87 8.72 9.84 9.37 9.64 8.68

1- Construct a frequency histogram for the voltage readings at each location. Compare the two graphs and provide appropriate interpretations.

2- Use the computer to generate descriptive statistics (mean, median, standard deviation, and interquartile range) for each location. Interpret the results.

3- Construct a box plot for the voltage readings at each location. Do you detect any outliers? Specify.

Please could you answer all the questions. Thank you.

Question

31-01-2020
Mathematics

contestada

A Harris corporation / University of Florida study was undertaken to determine whether a manufacturing process performed at a remote location can be established locally. Test devices were set up at both the old and new locations and voltage readings on the process were obtained. A "good" process was considered to be one with voltage reading of at least 9.2 volts (with larger readings being better than smaller readings). The table contains voltage readings for 30 production runs at each location.

Old New

Location Location

9.98 10.12 9.84 9.19 10.01 8.82

10.26 10.05 10.15 9.63 8.82 8.65

10.05 9.80 10.02 10.10 9.43 8.51

10.29 10.15 9.80 9.70 10.03 9.14

10.03 10.00 9.73 10.09 9.85 9.75

8.05 9.87 10.01 9.60 9.27 8.78

10.55 9.55 9.98 10.05 8.83 9.35

10.26 9.95 8.72 10.12 9.39 9.54

9.97 9.70 8.80 9.49 9.48 9.36

9.87 8.72 9.84 9.37 9.64 8.68

1- Construct a frequency histogram for the voltage readings at each location. Compare the two graphs and provide appropriate interpretations.

2- Use the computer to generate descriptive statistics (mean, median, standard deviation, and interquartile range) for each location. Interpret the results.

3- Construct a box plot for the voltage readings at each location. Do you detect any outliers? Specify.

Please could you answer all the questions. Thank you.

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dfbustos dfbustos · Answer 1 · 2020-02-02T13:16:19+01:00

Answer:

See the explanation below

Step-by-step explanation:

Data given:

Old Location : 9.98,10.12,9.84,10.26,10.05,10.15,10.05,9.8,10.02,10.29,10.15,9.8,10.03,10,9.73,8.05,9.87,10.01,10.55,9.55,9.98,10.26,9.95,8.72,9.97,9.7,8.8,9.87,8.72,9.84

New Location

9.19,10.01,8.82,9.63,8.82,8.65,10.1,9.43,8.51,9.7,10.03,9.14,10.09,9.85,9.75,9.6,9.27,8.78,10.05,8.83,9.35,10.12,9.39,9.54,9.49,9.48,9.36,9.37,9.64,8.68

1- Construct a frequency histogram for the voltage readings at each location. Compare the two graphs and provide appropriate interpretations.

We can construct the frequency histograms with the following code in R

> old<-c(9.98,10.12,9.84,10.26,10.05,10.15,10.05,9.8,10.02,10.29,10.15,9.8,10.03,10,9.73,8.05,9.87,10.01,10.55,9.55,9.98,10.26,9.95,8.72,9.97,9.7,8.8,9.87,8.72,9.84)

> length(old)

[1] 30

> new<-c(9.19,10.01,8.82,9.63,8.82,8.65,10.1,9.43,8.51,9.7,10.03,9.14,10.09,9.85,9.75,9.6,9.27,8.78,10.05,8.83,9.35,10.12,9.39,9.54,9.49,9.48,9.36,9.37,9.64,8.68)

> length(new)

[1] 30

> hist(old)

> hist(new)

The first histogram on the figure attached for the old case we see a distribution not symmetric and skewed to the left, with Mean<Median<mode. And most of the values are between 9.5 and 10.5

For the second histogram on the figure attached for the new case we have a more symmetric distribution and less skewed to the right compared with the old case.

2- Use the computer to generate descriptive statistics (mean, median, standard deviation, and interquartile range) for each location. Interpret the results.

We can use the following R code

> summary(old)

Min. 1st Qu. Median Mean 3rd Qu. Max.

8.050 9.800 9.975 9.804 10.050 10.550

The IQR = 10.50-9.8=0.25

> summary(new)

Min. 1st Qu. Median Mean 3rd Qu. Max.

8.510 9.152 9.455 9.422 9.738 10.120

IQR= 9.738-9.152=0.586

We see that the mean for the old case is higher than the mean for the new case. The interquartile range is higher for the new case than the old case, And with this info we can conclude that the variation in the new case is higher than the variation for the old case.

3- Construct a box plot for the voltage readings at each location. Do you detect any outliers? Specify.

For this case we can use the following R code

> par(mfrow=c(1,2))

> boxplot(old,main="Boxplot Old case")

> boxplot(new,main="Boxplot new case")

The result is on the 3 figure attached.

We can see that we have outliers on the old case and for the new case none. Approximately 4 outilers are represented by circles in the old case. We see lower variation for the old case compared to the new case.