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Exercise 6.19 discusses the headline "Domestic cats kill many more wild birds in the United States than scientists thought," and estimates the proportion of domestic cats that hunt outside. A separate study \(^{23}\) used KittyCams to record all activity of \(n=55\) domestic cats that hunt outdoors. The video footage showed that the mean number of kills per week for these cats was 2.4 with a standard deviation of \(1.51 .\) Find and interpret a \(99 \%\) confidence interval for the mean number of kills per week by US household cats that hunt outdoors.

Short Answer

Expert verified
The 99% confidence interval for the mean number of kills per week by US household cats that hunt outdoors would be calculated using the working in steps one through three and would turn out to be \(2.4 \pm 2.576 * \frac{1.51}{\sqrt{55}}\).

Step by step solution

01

Calculate the Standard Error

The standard error (SE) is the standard deviation divided by the square root of the sample size. So, the standard error can be calculated as follows: \[SE = \frac{SD}{\sqrt{n}}\] Substituting the given values, we have: \[SE = \frac{1.51}{\sqrt{55}}\]
02

Find the margin of error

The margin of error is the Z score times the standard error (SE). The Z score for a 99% confidence level is 2.576. So, the margin of error can be calculated as follows: \[MoE = Z * SE\] Substituting the values, we have: \[MoE = 2.576 * \frac{1.51}{\sqrt{55}}\]
03

Calculate the Confidence Interval

A confidence interval is calculated as the sample mean plus or minus the margin of error. Hence, the confidence interval can be calculated as follows: \[CI = \bar{X} \pm MoE\] Substituting the values, we have: \[CI = 2.4 \pm 2.576 * \frac{1.51}{\sqrt{55}}\] This yields the range for the 99% confidence interval for the mean number of kills per week by US household cats that hunt outdoors.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Standard Error Calculation
Understanding standard error is crucial for evaluating the precision of an estimated statistic. In statistics, it is particularly important when dealing with samples from a larger population.

The standard error (\text{SE}) is a measure of how much we would expect the sample mean to differ from the true population mean if we were to take many different samples. It's calculated by dividing the sample's standard deviation (\text{SD}) by the square root of the sample size (\text{n}). Mathematically, it is represented as: \[ SE = \frac{SD}{\sqrt{n}} \]
For example, if a study on outdoor hunting by domestic cats reports a standard deviation of 1.51 across a sample size of 55, the standard error would be: \[ SE = \frac{1.51}{\sqrt{55}} \]
A smaller standard error indicates a more precise estimate of the population mean. This calculation is the foundation for constructing confidence intervals and performing hypothesis tests.
Margin of Error
The margin of error helps in determining the range within which the true population parameter is likely to fall. It takes into account sample variability and is influenced by the standard error and the level of confidence desired.

The formula to compute the margin of error (\text{MoE}) involves the Z score (which reflects the confidence level) and the standard error: \[ MoE = Z \times SE \]
The Z score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. For a 99% confidence interval, the Z score is typically 2.576.

By applying the standard error from our previous example, the margin of error would be: \[ MoE = 2.576 \times \frac{1.51}{\sqrt{55}} \]
This means that the true mean of the population will be within this margin of error around the sample mean, with 99% confidence.
Z Score
A Z score is a statistical metric that tells us how many standard deviations an element is from the mean. It's a way of standardizing scores on different scales to easily compare them.

In the context of confidence intervals, the Z score is linked to the confidence level—the higher the confidence level (like 99% versus 95%), the higher the Z score. A Z score of 2.576 corresponds to a 99% confidence level, which means that we can be 99% certain that the population parameter lies within the confidence interval we calculate.

To find the Z score for different confidence levels, tables or statistical software are often used. It is critical to select the right Z score for the confidence level you need in your studies or research.
Sample Mean
The sample mean is the average value of a sample and it is used to estimate the population mean. If the sample is representative of the population, then the sample mean should be a good estimate of the population mean.

In statistics, the sample mean is represented by \( \bar{X} \) and it is calculated by summing all the numbers in the sample and then dividing by the number of observations in the sample. \[ \bar{X} = \frac{1}{n} \sum_{i=1}^{n} X_i \]
In the KittyCams study, the sample mean of 2.4 tells us that the average number of kills per week by the observed sample of domestic cats that hunt outdoors was 2.4. This statistic becomes the center of the confidence interval we want to estimate.
Statistics in Biology
When applying statistics in biology, accuracy is often as critical as the biological observations themselves. Statistics allow biologists to make sense of complex data and draw reliable conclusions from experiments and studies.

In the context of our KittyCams study, statistical methods enable researchers to estimate the mean number of kills per week for all US household cats that hunt outdoors, based on a sample. By using a confidence interval, they can account for the variability within the sample and quantify the uncertainty associated with their estimate. This branch of science requires a strong understanding of statistical concepts like standard error, margin of error, Z scores, and sample means to ensure that the biological conclusions drawn are supported by solid data analysis.

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Most popular questions from this chapter

A survey \(^{4}\) of 1060 randomly selected US teens ages 13 to 17 found that 605 of them say they have made a new friend online. (a) Find and interpret a \(90 \%\) confidence interval for the proportion, \(p\), of all US teens who have made a new friend online. (b) Give the best estimate for \(p\) and give the margin of error for the estimate. (c) Use the interval to determine whether we can be \(90 \%\) confident that more than half of US teens have made a new friend online.

Situations comparing two proportions are described. In each case, determine whether the situation involves comparing proportions for two groups or comparing two proportions from the same group. State whether the methods of this section apply to the difference in proportions. (a) Compare the proportion of students who use a Windows-based \(\mathrm{PC}\) to the proportion who use a Mac. (b) Compare the proportion of students who study abroad between those attending public universities and those at private universities. (c) Compare the proportion of in-state students at a university to the proportion from outside the state. (d) Compare the proportion of in-state students who get financial aid to the proportion of outof-state students who get financial aid.

If random samples of the given size are drawn from a population with the given mean and standard deviation, find the standard error of the distribution of sample means. Samples of size 1000 from a population with mean 28 and standard deviation 5

Do Ovulating Women Affect Men's Speech? Studies suggest that when young men interact with a woman who is in the fertile period of her menstrual cycle, they pick up subconsciously on subtle changes in her skin tone, voice, and scent. A study introduced in Exercise \(\mathrm{B} .23\) suggests that men may even change their speech patterns around ovulating women. The men were randomly divided into two groups with one group paired with a woman in the fertile phase of her cycle and the other group with a woman in a different stage of her cycle. The same women were used in the two different stages. For the men paired with a less fertile woman, 38 of the 61 men copied their partner's sentence construction in a task to describe an object. For the men paired with a woman at peak fertility, 30 of the 62 men copied their partner's sentence construction. The experimenters hypothesized that men might be less likely to copy their partner during peak fertility in a (subconscious) attempt to attract more attention to themselves. Use the normal distribution to test at a \(5 \%\) level whether the proportion of men copying sentence structure is less when the woman is at peak fertility.

Use a t-distribution to find a confidence interval for the difference in means \(\mu_{1}-\mu_{2}\) using the relevant sample results from paired data. Give the best estimate for \(\mu_{1}-\) \(\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed, and that differences are computed using \(d=x_{1}-x_{2}\). A \(95 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=3.7, s_{d}=\) 2.1, \(n_{d}=30\)

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