Wildlife biologists inspect 165 deer taken by hunters and find 34 or 20.6% of them carry ticks that test positive for Lyme disease. Calculate the 99% margin of error and confidence interval for the proportion of deer that may carry ticks. How many deer are needed to be inspected if the margin of error were to be cut in half? Comment on the statistical concerns about this study if it were reported in the news.

To calculate the margin of error and confidence interval for the proportion of deer that may carry ticks, we can use the formula:

Margin of Error = Critical Value * Standard Error

where the critical value is determined based on the desired level of confidence, and the standard error is calculated using the formula:

Standard Error = sqrt((p * (1 - p)) / n)

where p is the proportion of deer carrying ticks and n is the sample size.

Given that the proportion of deer carrying ticks is 20.6% or 0.206, and the sample size is 165, we can calculate the margin of error and confidence interval as follows:

Standard Error = sqrt((0.206 * (1 - 0.206)) / 165)
≈ sqrt(0.162464 / 165)
≈ 0.0356

To determine the critical value for a 99% confidence level, we need to use a Z-table or calculator. The critical value is approximately 2.576.

Margin of Error = 2.576 * 0.0356
≈ 0.0919

Now we can calculate the confidence interval:

Lower Limit of Confidence Interval = Proportion - Margin of Error
= 0.206 - 0.0919
≈ 0.114

Upper Limit of Confidence Interval = Proportion + Margin of Error
= 0.206 + 0.0919
≈ 0.298

Therefore, the 99% confidence interval for the proportion of deer that may carry ticks is approximately 0.114 to 0.298.
The margin of error is approximately 0.0919.

To cut the margin of error in half, we need to find the new sample size required. The margin of error is inversely proportional to the square root of the sample size (n). Thus, to cut the margin of error in half, the sample size needs to be quadrupled.

Statistical concerns about this study if reported in the news:

1. Sample Size: The study only examines 165 deer, which may not be representative of the entire population. A larger sample size is typically preferred for more reliable results.

2. Generalization: The study focuses on deer taken by hunters, which may not accurately represent the entire deer population or their tick-carrying behavior.

3. Selection Bias: The study relies on deer brought in by hunters, which could introduce bias if certain types of hunters or geographic areas are overrepresented.

4. Causation vs. Correlation: The study identifies ticks carrying Lyme disease, but it does not establish a causal relationship between the deer and the transmission of Lyme disease to humans.

5. External Factors: The study does not consider other possible factors that may contribute to the presence of ticks or Lyme disease, such as geography, climate, or the prevalence of tick-carrying hosts.

6. Reporting Inaccuracy: News reports often simplify complex statistical concepts, potentially leading to misinterpretation or misunderstandings by the audience. It is important to present study limitations and context accurately.

It is crucial to consider these statistical concerns and communicate them accurately when reporting scientific studies to ensure an accurate understanding of the findings and their implications.

To calculate the margin of error and confidence interval for the proportion of deer carrying ticks, we can use the formula:

Margin of Error = Z * sqrt((p * (1-p)) / n)

Where:
- Z is the z-score corresponding to the desired confidence level. For a 99% confidence level, the z-score is approximately 2.576.
- p is the observed proportion (in decimal form) of deer carrying ticks. In this case, p = 0.206.
- n is the sample size, which is 165 in this case.

Let's calculate the margin of error:

Margin of Error = 2.576 * sqrt((0.206 * (1-0.206)) / 165)
≈ 0.067

To find the confidence interval, we can use the formula:

Confidence Interval = p ± Margin of Error

Confidence Interval ≈ 0.206 ± 0.067
≈ (0.139, 0.273)

Now, to determine how many deer would need to be inspected if the margin of error were to be cut in half, we can use the margin of error formula with the new desired margin of error:

0.033 = 2.576 * sqrt((0.206 * (1-0.206)) / n)

By rearranging the formula and solving for n, we can find the required sample size:

n = (2.576^2 * (0.206 * (1-0.206))) / 0.033^2
≈ 673.71

Therefore, approximately 674 deer would need to be inspected if the margin of error were to be cut in half.

Finally, in terms of the statistical concerns about this study if reported in the news, there are a few points to consider. First, the sample size of 165 is relatively small, which may limit the generalizability of the results to the entire deer population. Additionally, the study only considers deer taken by hunters, which may not represent the entire population of deer. Moreover, the accuracy of tick testing and the representativeness of tick-infected deer among the tested sample could also affect the validity of the study. These concerns should be acknowledged when interpreting the results.