Question

section ii: free response show all your work.

1. you are interested in the number of contacts older adults have on their smartphones. here are the data on the number of contacts for a random sample of 30 elderly adults with smartphones in a large city.

7 20 24 25 25 28 28 30 32 35
42 43 44 45 46 47 48 48 50 51
72 75 77 78 79 83 87 88 135 151
(a) construct a histogram of these data with a bin width of 20.
(b) are there any outliers? justify your answer.
(c) would it be better to use mean and standard deviation or median and iqr to describe the center and variability of this distribution? explain.

Explanation:

Part (a) - Histogram Construction

Step1: Define Bins

Bin width = 20. Bins: \(0 - 19\), \(20 - 39\), \(40 - 59\), \(60 - 79\), \(80 - 99\), \(100 - 119\), \(120 - 139\), \(140 - 159\).

Step2: Count Frequencies

• \(0 - 19\): Data = 7 → Frequency = 1
• \(20 - 39\): Data = 20,24,25,25,28,28,30,32,35 → Frequency = 9
• \(40 - 59\): Data = 42,43,44,45,46,47,48,48,50,51 → Frequency = 10
• \(60 - 79\): Data = 72,75,77,78,79 → Frequency = 5
• \(80 - 99\): Data = 83,87,88 → Frequency = 3
• \(100 - 119\): No data → Frequency = 0
• \(120 - 139\): Data = 135 → Frequency = 1
• \(140 - 159\): Data = 151 → Frequency = 1

Step3: Draw Histogram

Plot bins on x - axis, frequency on y - axis. Bars with heights equal to frequencies.

Part (b) - Outlier Detection

Step1: Find Quartiles

First, order data: 7,20,24,25,25,28,28,30,32,35,42,43,44,45,46,47,48,48,50,51,72,75,77,78,79,83,87,88,135,151.
\(n = 30\). \(Q_1\) (25th percentile): Position \(=\frac{30 + 1}{4}=7.75\)th term. \(Q_1=28+\ 0.75\times(30 - 28)=29.5\)
\(Q_3\) (75th percentile): Position \(=\frac{3(30 + 1)}{4}=23.25\)th term. \(Q_3 = 78+\ 0.25\times(79 - 78)=78.25\)

Step2: Calculate IQR

\(IQR=Q_3 - Q_1=78.25 - 29.5 = 48.75\)

Step3: Find Fences

Lower fence \(=Q_1-1.5\times IQR=29.5-1.5\times48.75=29.5 - 73.125=- 43.625\)
Upper fence \(=Q_3 + 1.5\times IQR=78.25+1.5\times48.75=78.25 + 73.125 = 151.375\)
Data points: 7 (above lower fence? 7 > - 43.625, no), 135 (135 < 151.375), 151 (151 < 151.375). So no outliers (151 is less than upper fence 151.375, 7 is above lower fence). Wait, recalculate \(Q_1\) and \(Q_3\) with \(n = 30\) (even). \(Q_1\): average of 7th and 8th terms (7th = 28, 8th = 30) → \(Q_1=\frac{28 + 30}{2}=29\)
\(Q_3\): average of 23rd and 24th terms (23rd = 78, 24th = 79) → \(Q_3=\frac{78+79}{2}=78.5\)
\(IQR = 78.5 - 29=49.5\)
Lower fence \(=29-1.5\times49.5=29 - 74.25=- 45.25\)
Upper fence \(=78.5 + 1.5\times49.5=78.5+74.25 = 152.75\)
Data: 151 < 152.75, 135 < 152.75, 7 > - 45.25. So no outliers? Wait, 151 is close. Wait, maybe miscalculation. Wait, correct method for even \(n\): \(Q_1\) is median of first 15 terms: first 15 terms: 7,20,24,25,25,28,28,30,32,35,42,43,44,45,46. Median of 15 terms is 8th term = 30. \(Q_3\) is median of last 15 terms: 47,48,48,50,51,72,75,77,78,79,83,87,88,135,151. Median of 15 terms is 8th term = 78. So \(Q_1 = 30\), \(Q_3=78\), \(IQR = 78 - 30 = 48\)
Upper fence \(=78+1.5\times48=78 + 72 = 150\). 151 > 150 → 151 is an outlier. 135 < 150, 7 > \(30 - 1.5\times48=30 - 72=- 42\). So 151 is an outlier.

Step2: Justify

Using \(IQR\) method, upper fence \(=Q_3 + 1.5IQR=78+1.5\times48 = 150\). 151 > 150, so 151 is an outlier.

Part (c) - Measure Selection

Step1: Analyze Distribution

The distribution has an outlier (151) and is skewed (long tail on right).

Step2: Compare Measures

Mean and standard deviation are sensitive to outliers and skewness. Median and IQR are resistant to outliers and skewness. Since the distribution is skewed with an outlier, median and IQR are better to describe center and variability.

Final Answers (Summarized)

(a) Histogram with frequencies as calculated.
(b) Yes, 151 is an outlier (justification via IQR).
(c) Median and IQR (explanation: skewed with outlier, resistant measures better).

For part (b) detailed answer:

Answer:

Yes, 151 is an outlier. Justification: Using the IQR method, \(Q_1 = 30\), \(Q_3 = 78\), \(IQR=48\). Upper fence \(=78 + 1.5\times48=150\). Since \(151>150\), 151 is an outlier.

For part (c) detailed answer: