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Chapter 6: Problem 19

"Domestic cats kill many more wild birds in the United States than scientists thought," states a recent article. \({ }^{3}\) 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) Find and interpret a \(95 \%\) confidence interval for the proportion of household cats in the US that hunt outdoors. (b) Is it plausible that the proportion of household cats in the US hunting outdoors is \(0.45 ?\) Is it plausible that it is \(0.30 ?\)

Short Answer

Expert verified
The 95% confidence interval estimate for the proportion of household cats in the U.S. that hunt outdoors can be calculated using above steps. The plausibility of a given proportion value, like 0.45 or 0.30, can be determined by checking if it falls within this confidence interval.

Step by step solution

01

Calculate Standard Error

First, calculate the standard error for the proportion. The formula is \(\sqrt{ \frac{p(1-p)}{n} }\), where \(p = 0.35\) represents the sample proportion of household cats that hunt outdoors, and \(n = 140\) is the number of households in the sample. The standard error gives the average distance that the observed values fall from the population mean.
02

Calculate Confidence Interval

Next, calculate the 95% confidence interval. For this, we need the standard normal value for 95% confidence, which is approximately 1.96. The confidence interval is then \(p \pm z \times SE = 0.35 \pm 1.96 \times SE\). This interval estimates the range in which the true population proportion falls with a 95% confidence level.
03

Interpret Confidence Interval

If the interval includes a value, it means that this value is plausible for the population proportion given the data we have. If it does not include a value, then that value is considered implausible at the 95% confidence level.
04

Evaluate Plausibility of Proportion Values

Here, we need to check if the values 0.45 and 0.30 are within the estimated confidence interval. If they fall within the calculated interval, then it is plausible; if not, it is implausible.

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Key Concepts

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

Understanding Standard Error
When working with sample data, it's important to assess how much the sample proportion might differ from the actual population proportion. This is where the **standard error (SE)** becomes crucial. The standard error provides an estimate of the variation or the typical distance between the sample proportion and the population proportion.
The formula used to calculate the standard error for a proportion is:
  • \( SE = \sqrt{ \frac{p(1-p)}{n} } \)
Here, \(p\) represents the sample proportion鈥攊n this case, the proportion of household cats that hunt outdoors鈥攁nd \(n\) is the number of observations in the sample, specifically, the households surveyed. Calculating the SE helps us understand the reliability of our sample data in representing the entire population. The smaller the SE, the more accurate our sample proportion is considered to be. This estimation is crucial in determining the range of our confidence interval.
The Concept of Population Proportion
The **population proportion** refers to the proportion of subjects with a particular characteristic within an entire population. In this example, it's the proportion of domestic cats in the US that hunt outdoors. As it might be expensive or impractical to survey every household in the country, researchers use samples.
By considering a smaller, manageable number of households and assuming the sample is representative, researchers approximate this proportion with the sample proportion \(p\). With a sample proportion of 0.35, researchers estimate that about 35% of all domestic cats hunt outdoors.
However, to confirm how closely this sample proportion reflects the population proportion, researchers calculate a confidence interval, providing a range that likely contains the true proportion. This combination of sampled data and statistical analysis helps give more accurate insights into the behavior of cats across the entire US.
Exploring Plausibility Analysis
**Plausibility analysis** involves checking whether certain values are reasonable given the data and the statistical estimates we've made, such as the confidence interval. Once you've established a confidence interval around a sample proportion, you can conduct plausibility checks to determine if certain hypothesized values of population proportions could be correct.
When you want to know if 0.45 or 0.30, for instance, is a plausible value for the proportion of household cats hunting outdoors, you compare these values to your confidence interval. If the confidence interval includes these values, it鈥檚 plausible that such a proportion could exist in the overall population based on the sample data.
If a value lies outside this interval, it means, at the given confidence level (usually set at 95% for many analyses), there isn鈥檛 enough statistical evidence to consider that value plausible. This type of analysis is beneficial for decision-making based on statistical data, impacting ecological studies or policy-making concerning domestic cat behavior and its environmental effects.

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Most popular questions from this chapter

What Gives a Small P-value? In each case below, two sets of data are given for a two-tail difference in means test. In each case, which version gives a smaller \(\mathrm{p}\) -value relative to the other? (a) Both options have the same standard deviations and same sample sizes but: Option 1 has: \(\quad \bar{x}_{1}=25 \quad \bar{x}_{2}=23\) $$ \text { Option } 2 \text { has: } \quad \bar{x}_{1}=25 \quad \bar{x}_{2}=11 $$ (b) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\left.\bar{x}_{2}=17\right)\) and same sample sizes but: Option 1 has: \(\quad s_{1}=15 \quad s_{2}=14\) $$ \text { Option } 2 \text { has: } \quad s_{1}=3 \quad s_{2}=4 $$ (c) Both options have the same means \(\left(\bar{x}_{1}=22,\right.\) \(\left.\bar{x}_{2}=17\right)\) and same standard deviations but: Option 1 has: \(\quad n_{1}=800 \quad n_{2}=1000\) $$ \text { Option } 2 \text { has: } \quad n_{1}=25 \quad n_{2}=30 $$

Football Air Pressure During the NFL's 2014 AFC championship game, officials measured the air pressure on game balls following a tip that one team's balls were under-inflated. In exercise 6.124 we found that the 11 balls measured for the New England Patriots had a mean psi of 11.10 (well below the legal limit) and a standard deviation of 0.40. Patriot supporters could argue that the under-inflated balls were due to the elements and other outside effects. To test this the officials also measured 4 balls from the opposing team (Indianapolis Colts) to be used in comparison and found a mean psi of \(12.63,\) with a standard deviation of 0.12. There is no significant skewness or outliers in the data. Use the t-distribution to determine if the average air pressure in the New England Patriot's balls was significantly less than the average air pressure in the Indianapolis Colt's balls.

Assume the samples are random samples from distributions that are reasonably normally distributed, and that a t-statistic will be used for inference about the difference in sample means. State the degrees of freedom used. Find the proportion in a t-distribution less than -1.4 if the samples have sizes \(n_{1}=30\) and \(n_{2}=40\)

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 \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the paired data in the following table:. $$ \begin{array}{lccccc} \hline \text { Case } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} \\ \hline \text { Treatment 1 } & 22 & 28 & 31 & 25 & 28 \\ \text { Treatment 2 } & 18 & 30 & 25 & 21 & 21 \\ \hline \end{array} $$

Use a t-distribution and the given matched pair sample results to complete the test of the given hypotheses. Assume the results come from random samples, and if the sample sizes are small, assume the underlying distribution of the differences is relatively normal. Assume that differences are computed using \(d=x_{1}-x_{2}\). Test \(H_{0}: \mu_{1}=\mu_{2}\) vs \(H_{a}: \mu_{1} \neq \mu_{2}\) using the paired difference sample results \(\bar{x}_{d}=15.7, s_{d}=12.2\) \(n_{d}=25 .\)
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