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A researcher claims that at least \(10 \%\) of all football helmets have manufacturing flaws that could potentially cause injury to the wearer. A sample of 200 helmets revealed that 16 helmets contained such defects. (a) Does this finding support the researcher's claim? Use \(\alpha=\) 0.01. Find the \(P\) -value (b) Explain how the question in part (a) could be answered with a confidence interval.

Short Answer

Expert verified
The P-value is greater than \(0.01\). The confidence interval includes \(0.1\), supporting the claim.

Step by step solution

01

Define Hypotheses

Set up the null and alternative hypotheses. The null hypothesis \( H_0: p \geq 0.1 \), implies that at least \( 10\% \) of all helmets have defects. The alternative hypothesis \( H_1: p < 0.1 \), states that less than \( 10\% \) have defects.
02

Calculate the Sample Proportion

Find the sample proportion \( \hat{p} \). In this case, \( \hat{p} = \frac{16}{200} = 0.08 \).
03

Compute the Test Statistic

Use the formula for the test statistic for proportions: \( z = \frac{\hat{p} - p_0}{\sqrt{\frac{p_0(1-p_0)}{n}}} \). Substitute \( \hat{p} = 0.08 \), \( p_0 = 0.1 \), and \( n = 200 \). Calculate \( z = \frac{0.08 - 0.1}{\sqrt{\frac{0.1 \times 0.9}{200}}} \).
04

Find the P-value

Use the computed \( z \)-value to find the \( P \)-value from a standard normal table. The \( P \)-value represents the probability of observing a test statistic as extreme as, or more extreme than, the observed value under \( H_0 \).
05

Decision Making

Compare the \( P \)-value with the significance level \( \alpha = 0.01 \). If \( P \)-value \(< \alpha \), reject \( H_0 \). Otherwise, do not reject \( H_0 \).
06

Confidence Interval Approach

Find a \(99\%\) confidence interval for the true proportion \( p \). That interval is \( \hat{p} \pm z \times \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( z \) is the critical value for \( \alpha = 0.01 \). If \( 0.1 \) is within this interval, it supports the claim.

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Key Concepts

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

Proportion Testing
Proportion testing is a method used to determine if a particular proportion in a population matches a claimed value. In our football helmet example, we're examining whether 10% or more helmets show defects, as claimed by the researcher. We start with two hypotheses:
  • The null hypothesis ( H_0 ) assumes the researcher's claim is true, suggesting that at least 10% of helmets have defects.
  • The alternative hypothesis ( H_1 ) indicates that fewer than 10% have defects, challenging the researcher's claim.
To test these hypotheses, we calculate the sample proportion, which is simply the number of defective helmets divided by the total number. Here, it is 16 out of 200 or 0.08. Once we have our sample proportion, we calculate the test statistic. This value helps us understand how far away our observed sample is from the claimed 10%. If it's too far, we might reject H_0, suggesting the claim doesn't hold. Understanding this proportion testing helps us make evidence-based decisions about population characteristics.
Confidence Interval
A confidence interval provides a range where we expect the true population proportion to lie. In the context of our example, it's a range constructed around the sample proportion of helmets with defects. Calculating a confidence interval allows us to check if the claim (10% helmets are defective) falls within a plausible range based on our sample data.To construct a 99% confidence interval, we begin with the sample proportion (0.08) and add and subtract the margin of error. The margin of error is calculated using: \[ \text{Margin of error} = z \times \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \] where \( z \) is the critical value from a standard normal distribution, reflecting our desired confidence level (alpha = 0.01). The confidence interval gives us a picture of uncertainty around our sample proportion. If the claimed proportion (0.1 or 10%) is within this interval, it suggests that the true proportion might indeed be at least 10%, supporting the researcher's claim. It's a useful method for confirming test results without conducting further hypothesis testing.
Significance Level
The significance level, denoted by \( \alpha \), is a threshold used in hypothesis testing to decide whether to reject the null hypothesis (H_0). Practically, it's the probability of rejecting H_0 when it is actually true (a type I error). In our problem, we use a significance level of 0.01, which means we're okay with a 1% chance of mistakenly rejecting H_0.Choosing a significance level is crucial as it influences the strength of our decision. The smaller the \( \alpha \), the more stringent the test. For instance, using 0.01 instead of a more common 0.05 implies we require more substantial evidence against H_0 before rejecting it. This is especially important in contexts like manufacturing, where the implications of defects can be significant. By comparing our \( P \)-value to \( \alpha \), we have a clear rule for decision-making: if \( P \)-value < \( \alpha \), we reject H_0. Otherwise, we don't reject H_0. This helps in consistently determining the validity of claims about population proportions.
P-value
The \( P \)-value plays a crucial role in hypothesis testing by quantifying the extremeness of our sample data, assuming the null hypothesis (H_0) is true. In a simple sense, it tells us how likely it is to observe a result as extreme as the one obtained through the sample, given that H_0 holds true. For our helmet defect example, once we compute the test statistic, we use it to find the \( P \)-value from a standard normal distribution table. This \( P \)-value reflects the probability of observing 16 or fewer defective helmets (given 200 sampled) if indeed 10% truly are defective.The decision rule involves comparing the \( P \)-value to our significance level (\alpha = 0.01). If this value is smaller than 0.01, we have strong evidence against H_0 and would reject it, suggesting that less than 10% of helmets likely have defects. This is key in substantiating or refuting the researcher's claim, thereby illustrating the importance of \( P \)-value in statistical hypothesis testing.

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Most popular questions from this chapter

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