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Resistance training is a popular form of conditioning aimed at enhancing sports performance and is widely used among high school, college, and professional athletes, although its use for younger athletes is controversial. A random sample of 4111 patients aged 8-30 admitted to U.S. emergency rooms with the Consumer Product Safety Commission code "weightlifting" were obtained. These injuries were classified as "accidental" if caused by dropped weight or improper equipment use. Of the 4111 weightlifting injuries, 1552 were classified as accidental. \({ }^{10}\) Give a \(90 \%\) confidence interval for the proportion of weightlifting injuries in this age group that were accidental. Follow the four-step process as illustrated in Example 22.4.

Short Answer

Expert verified
90% CI for accidental injuries proportion: (0.3657, 0.3889).

Step by step solution

01

Define the Parameter

We need to estimate the proportion of weightlifting injuries that are accidental. Let the true proportion be denoted as \( p \). In our sample, we have 4111 total injuries, and 1552 of these are classified as accidental.
02

Calculate Point Estimate

The sample proportion \( \hat{p} \) is calculated by dividing the number of accidental injuries by the total number of injuries. Thus, \( \hat{p} = \frac{1552}{4111} \approx 0.3773 \).
03

Compute Standard Error

For a proportion, the standard error (SE) is computed as \( \text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( n = 4111 \) and \( \hat{p} = 0.3773 \). So, \( \text{SE} \approx \sqrt{\frac{0.3773(1-0.3773)}{4111}} \approx 0.0076 \).
04

Determine Z-Score for Confidence Level

For a 90% confidence interval, the Z-score is approximately 1.645, corresponding to the 5th percentile since 5% of the distribution is in the tails (2.5% in each tail).
05

Calculate Confidence Interval

The confidence interval is found using the formula: \( \hat{p} \pm Z \cdot \text{SE} \). Plugging the values we get: \( 0.3773 \pm 1.645 \cdot 0.0076 \). This results in the interval \( (0.3657, 0.3889) \).
06

Interpret the Confidence Interval

We are 90% confident that the true proportion of accidental weightlifting injuries in this age group is between 36.57% and 38.89%.

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Key Concepts

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

Proportion Estimation
Proportion estimation is a technique used to determine the fraction or percentage of a particular subgroup within a larger group. In the context of our exercise, we are interested in estimating the proportion of accidental weightlifting injuries.

To find this proportion, we start by calculating the sample proportion, symbolized as \( \hat{p} \). This is done by dividing the number of accidental injuries by the total number of injuries in the sample. For instance, if there are 1552 accidental injuries out of 4111 total injuries, our sample proportion would be \( \hat{p} = \frac{1552}{4111} \approx 0.3773 \). This gives us an initial estimate of the proportion of accidental injuries.

Understanding how to compute \( \hat{p} \) is crucial because it forms the foundation for further statistical analysis, such as constructing confidence intervals.
Standard Error
Standard error is an essential concept in statistics that gives us an idea of how much our sample estimate (like \( \hat{p} \)) might vary if we took different samples from the same population.

The formula for the standard error of a proportion is \( \text{SE} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( \hat{p} \) is the sample proportion and \( n \) is the total number of observations in the sample.

For example, in our exercise with an estimated proportion of 0.3773 and a sample size of 4111, the standard error is \( \text{SE} \approx \sqrt{\frac{0.3773 \times (1-0.3773)}{4111}} \approx 0.0076 \).

This tells us how much the sample proportion \( \hat{p} \) is expected to vary from the true population proportion we aim to estimate. A smaller standard error indicates a more precise estimate of the population proportion.
Normal Distribution
The normal distribution is a crucial concept in statistics, often used to approximate the distribution of various sample statistics, including proportions.

When dealing with large samples, the sample proportion \( \hat{p} \) will tend to follow a normal distribution. This assumption simplifies the process of constructing confidence intervals, as the bell-shaped curve of the normal distribution provides a predictable way of calculating the range within which we expect the true proportion to lie with a certain level of confidence.

In our exercise, the normal distribution helps us utilize the Z-score for a 90% confidence interval, which ensures that we can accurately estimate the range that likely includes the true proportion of accidental injuries.
Statistical Inference
Statistical inference involves the process of using data from a sample to make estimates or test hypotheses about a population.

In the case of estimating a confidence interval, statistical inference gives us the framework to make a prediction about the proportion of accidental weightlifting injuries in the entire population based on our sample data.

This involves calculating the sample proportion, determining the standard error, finding the appropriate Z-score, and then constructing the confidence interval. By calculating these steps, as we did with the interval \( (0.3657, 0.3889) \), we are essentially making an educated prediction about the true population proportion.

The beauty of statistical inference is that it allows researchers to draw meaningful conclusions even when examining only a portion of the overall population.

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Most popular questions from this chapter

Explain whether we can use the \(z\) test for a proportion in these situations. (a) You toss a coin 10 times in order to test the hypothesis \(H_{0}: p=0.5\) that the coin is balanced. (b) A local candidate contacts an SRS of 900 of the registered voters in his district to see if there is evidence that more than half support the bill he is sponsoring. (c) A college president says, " \(99 \%\) of the alumni support my firing of Coach Boggs." You contact an SRS of 200 of the college's 15,000 living alumni to test the hypothesis \(H_{0}: p=0.99\).

The Monterey Bay Aquarium, founded in 1984 , is situated on the beautiful coast of Monterey Bay in the historic Cannery Row district. In 1985 , the aquarium began a survey program that involved randomly sampling visitors as they exit for the day. The survey includes visitor demographic information, use of social media, and opinions on their aquarium visit. In 2015, the survey included 356 visitors over age 65 , of which 52 used a mobile device such as an Android phone or iPad during their visit. \({ }^{11}\) (a) What is the margin of error of the large-sample \(95 \%\) confidence interval for the proportion of visitors over 65 who used a mobile device during their visit? (b) How large a sample is needed to get the common \(\pm 3\) percentage point margin of error? Use the 2015 sample as a pilot study to get \(p^{*}\).

In the National AIDS Behavioral Surveys sample of 2673 adult heterosexuals, \(0.2 \%\) (that's \(0.002\) as a decimal fraction) had both received a blood transfusion and had a sexual partner from a group at high risk of AIDS. Explain why we can't use the large-sample confidence interval to estimate the proportion \(p\) in the population who share these two risk factors.

The value of the \(z\) statistic for the Exercise \(22.22\) is \(2.53\). This test is (a) not significant at either \(\alpha=0.05\) or \(\alpha=0.01\). (b) significant at \(\alpha=0.05\) but not at \(\alpha=0.01\). (c) significant at both \(\alpha=0.05\) and \(\alpha=0.01\).

Canada has much stronger gun control laws than the United States, and Canadians support gun control more strongly than do Americans. A sample survey asked a random sample of 1505 adult Canadians, "Do you agree or disagree that all firearms should be registered?" Of the 1505 people in the sample, 1288 answered either "Agree strongly" or "Agree somewhat."9 (a) The survey dialed residential telephone numbers at random in all 10 Canadian provinces (omitting the sparsely populated northern territories). Based on what you know about sample surveys, what is likely to be the biggest weakness in this survey? (b) Nonetheless, act as if we have an SRS from adults in the Canadian provinces. Give a \(95 \%\) confidence interval for the proportion who support registration of all firearms.

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