Question

question 6 (essay worth 4 points) (comparing data hc) the distribution of test scores for two random samples of students is shown in the stem - and - leaf plots.
group a:
5 | 8
6 | 3 8
7 | 2 3 5 5 6
8 | 0 2 4 5 5 5 7 7
9 | 1 3 4 5 6 7 7 8 8 8 9
10 | 0 0 0 0
key: 5|8 represents 58
group b:
5 | 4 7
6 | 2 3 7
7 | 1 1 2 3 5 6 7 8 8 8 8
8 | 2 2 3 3 3 3 4 4 4 5 7 9
9 | 1 2 8
10 | 0 0
key: 5|4 represents 54
part a: calculate the mean, median, range, and interquartile range for each data set (2 points)
part b: using the appropriate measures, compare the center and variability of the two groups of students. explain your reasoning based on the shapes of the distributions (2 points)

Answer:

Part A: Group A

Step 1: List all data points from stem - leaf plot

• For Group A:
• Stem 5: Leaf 8 → 58
• Stem 6: Leaf 3, 8 → 63, 68
• Stem 7: Leaf 2, 3, 5, 5, 6 → 72, 73, 75, 75, 76
• Stem 8: Leaf 0, 2, 4, 5, 5, 5, 7, 7 → 80, 82, 84, 85, 85, 85, 87, 87
• Stem 9: Leaf 1, 3, 4, 5, 6, 7, 7, 8, 8, 8, 9 → 91, 93, 94, 95, 96, 97, 97, 98, 98, 98, 99
• Stem 10: Leaf 0, 0, 0 → 100, 100, 100
• Now, count the number of data points: \(n_A=1 + 2+5 + 8+11 + 3=30\)

Step 2: Calculate the mean (\(\bar{x}_A\))

The formula for the mean is \(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\)

• \(\sum x_A=58+(63 + 68)+(72+73 + 75+75 + 76)+(80+82 + 84+85+85+85+87+87)+(91+93 + 94+95+96+97+97+98+98+98+99)+(100+100+100)\)
• \(58 = 58\)
• \(63 + 68=131\)
• \(72+73 + 75+75 + 76=72 + 73+75\times2+76=72+73 + 150+76 = 371\)
• \(80+82 + 84+85+85+85+87+87=80+82+84+(85\times3)+(87\times2)=80 + 82+84+255+174=675\)
• \(91+93 + 94+95+96+97+97+98+98+98+99=91+93+94+95+96+97\times2+98\times3+99=91+93+94+95+96 + 194+294+99=1056\)
• \(100\times3 = 300\)
• \(\sum x_A=58+131+371+675+1056+300=2591\)
• \(\bar{x}_A=\frac{2591}{30}\approx86.37\)

Step 3: Calculate the median

Since \(n = 30\) (even), the median is the average of the \(\frac{n}{2}=15^{th}\) and \((\frac{n}{2}+ 1)=16^{th}\) values.

• Order the data: After listing, the 15th and 16th values are in the stem 8 group.
• The 15th value: Let's count the cumulative number of data points.
• Stem 5: 1, Stem 6: 2 (cumulative 3), Stem 7: 5 (cumulative 8), Stem 8: 8 (cumulative \(8 + 8=16\)). So the 15th and 16th values are in stem 8.
• The data in stem 8: 80, 82, 84, 85, 85, 85, 87, 87. The 15th value (since cumulative up to stem 7 is 8, 15 - 8 = 7th in stem 8) is 85, 16th value is 87.
• Median \(M_A=\frac{85 + 87}{2}=86\)

Step 4: Calculate the range

Range = Maximum - Minimum

• Maximum value in Group A: 100, Minimum value: 58
• Range \(R_A=100 - 58 = 42\)

Step 5: Calculate the interquartile range (IQR)

• First, find \(Q_1\) (25th percentile) and \(Q_3\) (75th percentile).
• \(n = 30\), \(Q_1\) is the value at the \(\frac{n}{4}=7.5^{th}\) term (round up to 8th term), \(Q_3\) is the value at the \(\frac{3n}{4}=22.5^{th}\) term (round up to 23rd term).
• Cumulative count:
• Stem 5: 1, Stem 6: 2 (cumulative 3), Stem 7: 5 (cumulative 8). So \(Q_1\) is the 8th term, which is the last term in stem 7: 76.
• To find \(Q_3\): cumulative up to stem 8: \(8 + 8=16\), cumulative up to stem 9: \(16+11 = 27\). The 23rd term is in stem 9. The data in stem 9: 91, 93, 94, 95, 96, 97, 97, 98, 98, 98, 99. The 23rd term (23 - 16 = 7th in stem 9) is 97.
• \(IQR_A=Q_3 - Q_1=97 - 76 = 21\)

Group B

Step 1: List all data points from stem - leaf plot

• For Group B:
• Stem 5: Leaf 4, 7 → 54, 57
• Stem 6: Leaf 2, 3, 7 → 62, 63, 67
• Stem 7: Leaf 1, 1, 2, 3, 5, 6, 7, 8, 8, 8, 8 → 71, 71, 72, 73, 75, 76, 77, 78, 78, 78, 78
• Stem 8: Leaf 2, 2, 3, 3, 3, 3, 4, 4, 5, 7, 9 → 82, 82, 83, 83, 83, 83, 84, 84, 85, 87, 89
• Stem 9: Leaf 1, 2, 8 → 91, 92, 98
• Stem 10: Leaf 0, 0 → 100, 100
• Number of data points: \(n_B=2 + 3+11 + 11+3 + 2=32\)

Step 2: Calculate the mean (\(\bar{x}_B\))

\(\bar{x}=\frac{\sum_{i = 1}^{n}x_i}{n}\)

• \(\sum x_B=(54 + 57)+(62+63 + 67)+(71+71 + 72+73 + 75+76+77+78+78+78+78)+(82+82 + 83+83+83+83+84+84+85+87+89)+(91+92 + 98)+(100+100)\)
• \(54 + 57 = 111\)
• \(62+63 + 67=192\)
• \(71+71 + 72+73 + 75+76+77+78+78+78+78=71\times2+72+73+75+76+77+(78\times4)=142+72+73+75+76+77+312 = 757\)
• \(82+82 + 83+83+83+83+84+84+85+87+89=82\times2+(83\times4)+(84\times2)+85+87+89=164+332+168+85+87+89 = 925\)
• \(91+92 + 98=281\)
• \(100\times2 = 200\)
• \(\sum x_B=111+192+757+925+281+200=2466\)
• \(\bar{x}_B=\frac{2466}{32}\approx77.06\)

Step 3: Calculate the median

Since \(n = 32\) (even), the median is the average of the \(\frac{n}{2}=16^{th}\) and \((\frac{n}{2}+ 1)=17^{th}\) values.

• Cumulative count:
• Stem 5: 2, Stem 6: 3 (cumulative 5), Stem 7: 11 (cumulative \(5 + 11 = 16\)), Stem 8: 11 (cumulative \(16+11 = 27\))
• The 16th value is the last term in stem 7: 78, the 17th value is the first term in stem 8: 82
• Median \(M_B=\frac{78 + 82}{2}=80\)

Step 4: Calculate the range

Range = Maximum - Minimum

• Maximum value in Group B: 100, Minimum value: 54
• Range \(R_B=100 - 54 = 46\)

Step 5: Calculate the interquartile range (IQR)

• \(n = 32\), \(Q_1\) is the value at the \(\frac{n}{4}=8^{th}\) term, \(Q_3\) is the value at the \(\frac{3n}{4}=24^{th}\) term.
• Cumulative count:
• Stem 5: 2, Stem 6: 3 (cumulative 5), Stem 7: 11 (cumulative \(5 + 11 = 16\)). The 8th term is in stem 7. The data in stem 7: 71, 71, 72, 73, 75, 76, 77, 78, 78, 78, 78. The 8th term is 78.
• To find \(Q_3\): cumulative up to stem 8: \(16 + 11=27\). The 24th term is in stem 8. The data in stem 8: 82, 82, 83, 83, 83, 83, 84, 84, 85, 87, 89. The 24th term (24 - 16 = 8th in stem 8) is 84.
• \(IQR_B=Q_3 - Q_1=84 - 78 = 6\)

Part B: Comparison of Center and Variability

Center (Mean and Median)

• For Group A, the mean is approximately \(86.37\) and the median is \(86\). For Group B, the mean is approximately \(77.06\) and the median is \(80\). The mean and median of Group A are higher than those of Group B, indicating that on average, the test scores of Group A are higher than those of Group B.

Variability (Range and IQR)

• The range of Group A is \(42\) and the range of Group B is \(46\). The IQR of Group A is \(21\) and the IQR of Group B is \(6\). A larger IQR for Group A and a slightly larger range for Group B. The IQR of Group A is larger, which means that the middle 50% of the data in Group A is more spread out than in Group B. The range of Group B is larger, which is mainly due to the minimum value being lower in Group B.

Final Answers (Part A)

• Group A:
• Mean: \(\approx86.37\)
• Median: \(86\)
• Range: \(42\)
• IQR: \(21\)
• Group B:
• Mean: \(\approx77.06\)
• Median: \(80\)
• Range: \(46\)
• IQR: \(6\)

(For Part B, the comparison is as explained above, focusing on center (mean/median) and variability (range/IQR) with respect to the distribution shapes. Group A has a right - skewed? No, actually, looking at the stem - leaf, Group A has more values in the higher stems, maybe slightly left - skewed? Wait, no, the lower stems have fewer values. Group B has values more spread in the middle stems. But the key is to use mean/median for center and range/IQR for variability. The center of Group A is higher, and the variability (IQR) of Group A is higher, while range of Group B is higher.)