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.

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.