Question

Suppose the following tables present the number of specimens that tested positive for Type A and Type B influenza in a country during a flu season.
Type A
35
96
179
206
260
396
586
962
1597
2662
3756
4837
5568
5466
5686

Type B:
54
94
120
138
150
234
261
366
484
704
843
1083
1151
1073
1152

(a) Find the mean and median number of Type A cases. Round the answers to at least one decimal place.

(b) Find the mean and median number of Type B cases. Round the answers to at least one decimal place.

(c) A public health official says that there are more than twice as many cases of Type A influenza than Type B. Do these data support this claim?

Answer 1

To analyze the provided data for Type A and Type B influenza cases, we will calculate the mean and median for both types and assess the claim made by the public health official.

### (a) Type A Cases

#### Data:
35, 96, 179, 206, 260, 396, 586, 962, 1597, 2662, 3756, 4837, 5568, 5466, 5686

#### Mean Calculation:
1. **Sum the cases**:
\[
\text{Total} = 35 + 96 + 179 + 206 + 260 + 396 + 586 + 962 + 1597 + 2662 + 3756 + 4837 + 5568 + 5466 + 5686 = 47442
\]
2. **Count the number of entries**: There are 15 values.
3. **Calculate the mean**:
\[
\text{Mean} = \frac{\text{Total}}{\text{Number of values}} = \frac{47442}{15} \approx 3162.8
\]

#### Median Calculation:
1. **Arrange the data** (already provided in ascending order).
2. **Identify the median**:
- Since there are 15 values (an odd number), the median is the middle value. The median is the 8th value.
- The 8th value is **962**.

**Mean and Median for Type A**:
- Mean: **3162.8**
- Median: **962**

### (b) Type B Cases

#### Data:
54, 94, 120, 138, 150, 234, 261, 366, 484, 704, 843, 1083, 1151, 1073, 1152

#### Mean Calculation:
1. **Sum the cases**:
\[
\text{Total} = 54 + 94 + 120 + 138 + 150 + 234 + 261 + 366 + 484 + 704 + 843 + 1083 + 1151 + 1073 + 1152 = 7150
\]
2. **Count the number of entries**: There are 15 values.
3. **Calculate the mean**:
\[
\text{Mean} = \frac{\text{Total}}{\text{Number of values}} = \frac{7150}{15} \approx 476.7
\]

#### Median Calculation:
1. **Arrange the data** (already provided in ascending order).
2. **Identify the median**:
- Since there are 15 values (an odd number), the median is the middle value. The median is the 8th value.
- The 8th value is **366**.

**Mean and Median for Type B**:
- Mean: **476.7**
- Median: **366**

### (c) Public Health Official's Claim

The public health official claims that there are more than twice as many cases of Type A influenza than Type B.

To support the claim, we can check:
\[
\text{Twice the mean of Type B} = 2 \times \text{Mean of Type B} \approx 2 \times 476.7 \approx 953.4
\]

**Now compare with the mean of Type A**:
\[
\text{Mean of Type A} \approx 3162.8
\]

Since:
\[
3162.8 > 953.4
\]

This supports the public health official's claim that there are indeed more than twice as many cases of Type A influenza than Type B.

### Summary:
- **Type A**: Mean = **3162.8**, Median = **962**
- **Type B**: Mean = **476.7**, Median = **366**
- The claim that there are more than twice as many cases of Type A influenza than Type B is supported by these data.