Math / Data & Statistics

QuestionHow Many Birds Do Domestic Cats Kill? "Domestic cats kill many more wild birds in the United States than scientists thought," states a recent article. ${ }^{1}$ Researchers used a sample of $n=140$ households in the US with cats to estimate that $35 \%$ of household cats in the US hunt outdoors. A separate study ${ }^{2}$ 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 a $99 \%$ confidence interval for the mean number of kills per week by US household cats that hunt outdoors.
Round your answers to three decimal places.
The 99 confidence interval is i $\square$ to i $\square$ ${ }^{1}$ Milius, S., "Cats kill more than one billion birds each year," Science News, 183(4), February 23, 2013, revised March 8, 2014. Data approximated from information give in the article. ${ }^{2}$ Loyd KAT, et al., "Quantifying free-roaming domestic cat predation using animal-borne video cameras," Biological Conservation, 160(2013), 183-189.

Studdy Solution

STEP 1

What is this asking? We need to find the 99% confidence interval for the average number of kills per week for outdoor-hunting US household cats, given data from a study with KittyCams. Watch out! Don't get distracted by the 35% of household cats that hunt outdoors – we're only looking at the ones that do hunt outdoors!

STEP 2

1. Find the critical value.
2. Calculate the margin of error.
3. Construct the confidence interval.

STEP 3

We're dealing with a 99% confidence interval, which means we want to capture the true mean within our interval 99% of the time.
This leaves 1% split equally in the two tails, so $\frac{1\%}{2} = 0.005$ in each tail.

STEP 4

We need to find the critical value, often denoted as $z_{\alpha/2}$ or $t_{\alpha/2}$ , that corresponds to this tail probability.
Since our sample size ( $n = 55$ ) is greater than 30, we can use the z-distribution.
We're looking for the z-score that leaves 0.005 in the right tail.
Using a z-table or calculator, we find our critical value is $z_{0.005} = 2.576$ .

STEP 5

The margin of error tells us how much "wiggle room" we have around our sample mean.
It's calculated as: $\text{Margin of Error} = z_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ where $s$ is the sample standard deviation and $n$ is the sample size.

STEP 6

Let's plug in our values: $z_{\alpha/2} = 2.576$ , $s = 1.51$ , and $n = 55$ . $\text{Margin of Error} = 2.576 \cdot \frac{1.51}{\sqrt{55}}$ $\text{Margin of Error} = 2.576 \cdot \frac{1.51}{7.416}$ $\text{Margin of Error} = 2.576 \cdot 0.204$ $\text{Margin of Error} \approx 0.525$ So, our margin of error is approximately 0.525.

STEP 7

The confidence interval is built around our sample mean ( $\bar{x}$ ), which is given as 2.4.
We add and subtract the margin of error to create the interval: $\text{Confidence Interval} = \bar{x} \pm \text{Margin of Error}$

STEP 8

Plugging in our values, we get: $\text{Confidence Interval} = 2.4 \pm 0.525$ This gives us a lower bound of $2.4 - 0.525 = 1.875$ and an upper bound of $2.4 + 0.525 = 2.925$ .

STEP 9

The 99% confidence interval for the mean number of kills per week by US household cats that hunt outdoors is 1.875 to 2.925.

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Studdy Solution

STEP 1

STEP 2

1. Find the critical value.
2. Calculate the margin of error.
3. Construct the confidence interval.

STEP 3

We're dealing with a 99% confidence interval, which means we want to capture the true mean within our interval 99% of the time.
This leaves 1% split equally in the two tails, so $\frac{1\%}{2} = 0.005$ in each tail.

STEP 4

STEP 5

The margin of error tells us how much "wiggle room" we have around our sample mean.
It's calculated as: $\text{Margin of Error} = z_{\alpha/2} \cdot \frac{s}{\sqrt{n}}$ where $s$ is the sample standard deviation and $n$ is the sample size.

STEP 6

STEP 7

The confidence interval is built around our sample mean ( $\bar{x}$ ), which is given as 2.4.
We add and subtract the margin of error to create the interval: $\text{Confidence Interval} = \bar{x} \pm \text{Margin of Error}$

STEP 8

Plugging in our values, we get: $\text{Confidence Interval} = 2.4 \pm 0.525$ This gives us a lower bound of $2.4 - 0.525 = 1.875$ and an upper bound of $2.4 + 0.525 = 2.925$ .

STEP 9

The 99% confidence interval for the mean number of kills per week by US household cats that hunt outdoors is 1.875 to 2.925.

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Studdy Solution

STEP 1

STEP 2

1. Find the critical value.2. Calculate the margin of error.3. Construct the confidence interval.

STEP 3

We're dealing with a **99% confidence interval**, which means we want to capture the true mean within our interval 99% of the time.This leaves 1% split equally in the two tails, so 1%2=0.005\frac{1\%}{2} = 0.00521%​=0.005 in each tail.

STEP 4

STEP 5

STEP 6

STEP 7

STEP 8

STEP 9

The 99% confidence interval for the mean number of kills per week by US household cats that hunt outdoors is 1.875 to 2.925.

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

1. Find the critical value.
2. Calculate the margin of error.
3. Construct the confidence interval.

We're dealing with a 99% confidence interval, which means we want to capture the true mean within our interval 99% of the time.
This leaves 1% split equally in the two tails, so $\frac{1\%}{2} = 0.005$ in each tail.