AE311 LRA 5

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Problem 1: a. Calculate the mean, median, first quartile and 3rd quartile.

5.501

The mean weight of tea bags is 5.501 grams. The computed mean weight from the sample minimally differs from the labeled weight of a tea bag which is 5 grams.

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5.515

The median weight of tea bags is 5.515 grams. The median weight of tea bags is greater than the mean weight computed as well as the labeled weight of 5 grams.

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25%
5.44

75%
5.57

The first quartile is 5.44 grams while the third quartile is 5.57 grams.

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2. Calculate the variance, standard deviation, and coefficient of variation.

0.0112

The variance is 0.0112 squared.

0.1058

The standard deviation is 0.1058 grams.

## [1] 1.92%

The Coefficient of variation is 1.92%.

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3. Why should the company be concerned about the central tendency and variation measures?

The company should be concerned with the measures of central tendency and variation in order to monitor the production process and make necessary adjustments should there be problems as indicated by these measures.

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4. Boxplot of data

The boxplot shows that there is some degree of skewness of the data. Particularly, it can be seen that the data is skewed to the right where a greater number of lower values have higher frequencies.

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5. Based on the sample data, it can be seen that the company somehow meets the requirement set forth regarding the weight of tea bags as presented on the label. The labeled weight is 5.5 grams while the computed mean weight is 5.501 grams. The computed median, on the other hand, is 5.515 grams.

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Problem 2: Rosner’s Test on Data

## Warning in pander.default(ros.out): No pander.method for "gofOutlier", reverting
## to default.

• distribution: Normal

• statistic:

  R.1	R.2
  2.538	2.472

• sample.size: 50

• parameters:

  k
  2

• alpha: 0.05

• crit.value:

  lambda.1	lambda.2
  3.128	3.12

• n.outliers: 0

• alternative: Up to 2 observations are not from the same Distribution.

• method: Rosner’s Test for Outliers

• data: 5.65, 5.44, 5.42, 5.4, 5.53, 5.34, 5.54, 5.45, 5.52, 5.41, 5.57, 5.4, 5.53, 5.54, 5.55, 5.62, 5.56, 5.46, 5.44, 5.51, 5.47, 5.4, 5.47, 5.61, 5.53, 5.32, 5.67, 5.29, 5.49, 5.55, 5.77, 5.57, 5.42, 5.58, 5.5, 5.32, 5.5, 5.53, 5.58, 5.58, 5.61, 5.45, 5.44, 5.25, 5.56, 5.63, 5.5, 5.57, 5.67 and 5.36

• data.name: df

• bad.obs: 0

• all.stats:

  i	Mean.i	SD.i	Value	Obs.Num	R.i+1	lambda.i+1	Outlier
  0	5.501	0.1058	5.77	31	2.538	3.128	FALSE
  1	5.496	0.0995	5.25	44	2.472	3.12	FALSE

The Rosner’s Test shows that there are no significant outliers in the data.

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Problem 3:

1. Compute for the mean, standard deviation, and coefficient of variation for each group. Interpret these measures.

_43.7__30.32__16.93__7.128_

  [1] 38.74%

  [1] 23.51%

The mean time it takes for smokers to fall asleep is 43.7 minutes, while it takes the nonsmokers only 30.32 minutes.

The standard deviation for smokers is greater than that of nonsmokers where we have 16.93 and 7.128 minutes for smokers and nonsmokers respectively.

The data for nonsmokers is less variable as indicated by the coefficient of variation of 23.51% as compared to a CV value of 38.74% for smokers.

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2. Comment on what kind of impact smoking appears to have on the time required to fall sleep.

Smoking appears to prolong the time required to fall asleep.