Question

Chapter 3 – Numerically Summarizing Data

1. Section 3.1 Measures of Central Tendency

pH in Water:

The acidity of alkalinity of a solution is measured using pH. A pH less than 7 is acidic; a pH greater than 7 is alkaline. The following data represent the pH in samples of bottled water and tap water.

Tap	7.64	7.45	7.47	7.50	7.68	7.69
	7.45	7.10	7.56	7.47	7.52	7.47
Bottled	5.15	5.09	5.26	5.20	5.02	5.23
	5.28	5.26	5.13	5.26	5.21	5.24

1. Determine the mean, median and mode pH for each type of water.
2. Comment on the differences between the two water types.
3. Suppose the pH of 7.10 in tap water was incorrectly recorded as 1.70. How does this affect the mean? The median?
4. What property of the median does this illustrate?

2. Section 3.2 Measures of Dispersion

pH in Water:

Use the same pH data table in the above question to answer the following.

1. Which type of water has more dispersion in pH using the range as the measure of dispersion?
2. Which type of water has more dispersion in pH using the standard deviation as the measure of dispersion?

3. Section 3.2 Measures of Dispersion

The Empirical Rule:

SAT Math scores have a bell-shaped distribution with a mean of 515 and a standard deviation of 114.

1. What percentage of SAT scores is between 401 and 629?
2. What percentage of SAT scores is less than 401 or greater than 629?

4. Section 3.4 Measures of Position and Outliers

You Explain it – Percentiles & Quartiles:

1. The 5^thpercentile of the weight of males 36 months of age is 12.0 kg.
2. The 95^thpercentile of the length of newborn females is 53.8 cm.

One variable that is measured by online homework systems is the amount of time a student spends on homework for each section of the text. The following is a summary of the number of minutes a student spends for each section of the text for the fall 2014 semester in a college statistics class at UHWO.

Q1 = 42         Q2 = 51.5         Q3 = 72.5

3. Provide an interpretation of these results.
4. Determine and interpret the interquartile range.
5. Suppose a student spends 2 hours doing homework for a section. Is this an outlier?
6. Do you believe that the distribution of time spend doing homework is skewed or symmetric? Explain your answer.

5. Section 3.5 The Five Number Summary & Boxplots

Got a Headache?

The following data represent the weight in grams of a random sample of 25 Tylenol tablets.

0.608	0.601	0.606	0.602	0.611
0.608	0.610	0.610	0.607	0.600
0.608	0.608	0.605	0.609	0.605
0.610	0.607	0.611	0.608	0.610
0.612	0.598	0.600	0.605	0.603

1. Construct a boxplot.
2. Use the boxplot and quartiles to describe the shape of the distribution.

Answer 1

a.

For tap water,

Mean = (7.64 + 7.45 + 7.47 + 7.50 + 7.68 + 7.69 + 7.45 + 7.10 + 7.56 + 7.47 + 7.52 + 7.47) / 12 = 7.5

The data in ascending order is,

7.10 7.45 7.45 7.47 7.47 7.47 7.50 7.52 7.56 7.64 7.68 7.69

For n = 12 (even), the median is the mean of (n/2)th and (n/2)+1th element or the mean of 6th and 7th element.

Median = (7.47 + 7.50)/2 =  7.485

The maximum frequency in the data is for 7.47. Thus, the mode is 7.47

For bottled water,

Mean = (5.15 + 5.09 + 5.26 + 5.20 + 5.02 + 5.23 + 5.28 + 5.26 + 5.13 + 5.26 + 5.21 + 5.24) / 12 = 5.19

The data in ascending order is,

5.02 5.09 5.13 5.15 5.20 5.21 5.23 5.24 5.26 5.26 5.26 5.28

For n = 12 (even), the median is the mean of (n/2)th and (n/2)+1th element or the mean of 6th and 7th element.

Median = (5.21 + 5.23)/2 =  5.22

The maximum frequency in the data is for 5.26. Thus, the mode is 5.26

b)

All measures of center (mean, median and mode) for tap water is greater than that of bottled water. Thus, the bottled water is more acidic than the tap water.

c)

Changing the value of 7.10 to 1.70 in tap water, the mean is,

Mean = (7.64 + 7.45 + 7.47 + 7.50 + 7.68 + 7.69 + 7.45 + 1.70 + 7.56 + 7.47 + 7.52 + 7.47) / 12 = 7.05

The data in ascending order is,

1.7 7.45 7.45 7.47 7.47 7.47 7.50 7.52 7.56 7.64 7.68 7.69

For n = 12 (even), the median is the mean of (n/2)th and (n/2)+1th element or the mean of 6th and 7th element.

Median = (7.47 + 7.50)/2 =  7.485

The mean is reduced from 7.5 to 7.05. The median is unchanged, since the middlemost elements (5th and 6th) are unchanged.

d.

This illustrates that the median is unaffected by the extreme values in the data.