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Basic Computation: Confidence Interval for p Consider \(n=200\) binomial trials with \(r=80\) successes. (a) Cbeck Requirements Is it appropriate to use a normal distribution to approximate the \(\hat{p}\) distribution? (b) Find a \(95 \%\) confidence interval for the population proportion of successes \(p\). (c) Interpretation Explain the meaning of the confidence interval you computed.

Short Answer

Expert verified
(a) Yes, normal approximation is appropriate. (b) CI: (0.3322, 0.4678). (c) We are 95% confident the true proportion is between 33.22% and 46.78%.

Step by step solution

01

Check Requirements

To determine if it's appropriate to use a normal distribution to approximate the distribution of \( \hat{p} \), we need to ensure that both \( n \cdot \hat{p} \) and \( n \cdot (1 - \hat{p}) \) are greater than 5. Here, \( \hat{p} = \frac{r}{n} = \frac{80}{200} = 0.4 \). Therefore, \( n \cdot \hat{p} = 200 \times 0.4 = 80 \) and \( n \cdot (1 - \hat{p}) = 200 \times 0.6 = 120 \). Both are greater than 5, so the normal approximation is appropriate.
02

Calculate the Confidence Interval

To find a 95% confidence interval, we use the formula for the confidence interval of a population proportion: \( \hat{p} \pm Z \cdot \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \). Here, \( Z \approx 1.96 \) for a 95% confidence level. So, the confidence interval is:\[0.4 \pm 1.96 \cdot \sqrt{\frac{0.4 \cdot 0.6}{200}}\]Let's calculate it:\( \sqrt{\frac{0.4 \cdot 0.6}{200}} \approx 0.03464 \). Now, calculate the margin of error: \( 1.96 \times 0.03464 \approx 0.0678 \).Therefore, the confidence interval is \( 0.4 \pm 0.0678 \), or \( (0.3322, 0.4678) \).
03

Interpretation of the Confidence Interval

The 95% confidence interval \( (0.3322, 0.4678) \) means that we are 95% confident that the true population proportion of successes \( p \) is between 33.22% and 46.78%. This implies that if we conducted many samples of 200 trials each, about 95% of the intervals calculated from those samples would contain the true population proportion.

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Key Concepts

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

normal approximation
The normal approximation is a useful method for simplifying the calculation of probabilities in binomial distributions. When dealing with a large number of trials, the binomial distribution can be approximated by a normal distribution. This makes calculations much more manageable. The key requirement for using the normal approximation is the rule of thumb that both \( n \cdot \hat{p} \) and \( n \cdot (1 - \hat{p}) \) must be greater than 5. This ensures that the binomial distribution is sufficiently symmetrical to be approximated by the normal curve.
  • For our example, \( \hat{p} = 0.4 \), \( n = 200 \).
  • Calculating \( n \cdot \hat{p} = 200 \times 0.4 = 80 \), which is greater than 5.
  • Calculating \( n \cdot (1 - \hat{p}) = 200 \times 0.6 = 120 \), also greater than 5.
In this context, both conditions are satisfied, making it appropriate to use the normal approximation for calculating our confidence interval. This approach simplifies the process by using the properties of the normal distribution to estimate probabilities.
population proportion
The process of estimating the population proportion is at the heart of understanding how a sample can reflect the wider population. The population proportion, denoted as \( p \), represents the fraction of the total population that exhibits a particular characteristic. In practice, we often use the sample proportion (\( \hat{p} \)) as an estimate of this true population proportion.
  • Sample proportion (\( \hat{p} \)) is calculated by dividing the number of successes (\( r \)) by the total trials (\( n \)).
  • In our exercise, \( \hat{p} = \frac{r}{n} = \frac{80}{200} = 0.4 \).
The concept of confidence intervals becomes important here. These intervals give us a range that likely includes the true population proportion. For a 95% confidence interval, you can say that 95% of the time, the interval constructed from samples will contain the true population proportion. This is crucial for making informed decisions based on sample data.
binomial distribution
The binomial distribution is a fundamental concept in statistics, often used to model the number of successes in a fixed number of independent trials of a binary outcome. It is characterized by two parameters: the number of trials \( n \) and the probability of success in each trial \( p \).
  • A binomial distribution is described by the probability mass function: \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \), where \( \binom{n}{k} \) is the binomial coefficient.
  • In our context, each trial results in either a success or a failure, such as flipping a coin or defect/non-defect in a manufacturing process.
To use a binomial distribution in practical applications, it helps to understand its properties, like mean and variance. The mean of a binomial distribution is \( np \), and the variance is \( np(1-p) \). In large sample sizes, the binomial distribution starts resembling the normal distribution, which is why the normal approximation can be beneficial for calculations, especially when dealing with large \( n \).

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Most popular questions from this chapter

Finance: \(\mathrm{P} / \mathrm{E}\) Ratio The price of a share of stock divided by a company's estimated future earnings per share is called the P/E ratio. High P/E ratios usually indicate "growth" stocks, or maybe stocks that are simply overpriced. Low \(\mathrm{P} / \mathrm{E}\) ratios indicate "value" stocks or bargain stocks. A random sample of 51 of the largest companies in the United States gave the following P/E ratios (Reference: Forbes). $$ \begin{array}{rrrrrrrrrrrr} 11 & 35 & 19 & 13 & 15 & 21 & 40 & 18 & 60 & 72 & 9 & 20 \\ 29 & 53 & 16 & 26 & 21 & 14 & 21 & 27 & 10 & 12 & 47 & 14 \\ 33 & 14 & 18 & 17 & 20 & 19 & 13 & 25 & 23 & 27 & 5 & 16 \\ 8 & 49 & 44 & 20 & 27 & 8 & 19 & 12 & 31 & 67 & 51 & 26 \\ 19 & 18 & 32 & & & & & & & & & \end{array} $$ (a) Use a calculator with mean and sample standard deviation keys to verify that \(\bar{x} \approx 25.2\) and \(s \approx 15.5\). (b) Find a \(90 \%\) confidence interval for the \(\mathrm{P} / \mathrm{E}\) population mean \(\mu\) of all large U.S. companies. (c) Find a \(99 \%\) confidence interval for the \(\mathrm{P} / \mathrm{E}\) population mean \(\mu\) of all large U.S. companies. (d) Interpretation Bank One (now merged with J.P. Morgan) had a P/E of 12 , AT\&T Wireless had a \(\mathrm{P} / \mathrm{E}\) of 72 , and Disney had a \(\mathrm{P} / \mathrm{E}\) of 24 . Examine the confidence intervals in parts (b) and (c). How would you describe these stocks at the time the sample was taken? (e) Check Requirements In previous problems, we assumed the \(x\) distribution was normal or approximately normal. Do we need to make such an assumption in this problem? Why or why not? Hint: See the central limit theorem in Section \(6.5\).

Answer true or false. Explain your answer. A larger sample size produces a longer confidence interval for \(\mu\).

Basic Computation: Confidence Interval A random sample of size 81 has sample mean 20 and sample standard deviation \(3 .\) (a) Check Requirements Is it appropriate to use a Student's \(t\) distribution to compute a confidence interval for the population mean \(\mu\) ? Explain. (b) Find a \(95 \%\) confidence interval for \(\mu\). (c) Interpretation Explain the meaning of the confidence interval you computed.

Expand Your Knowledge: Alternate Method for Confidence Intervals When \(\sigma\) is unknown and the sample is of size \(n \geq 30\), there are two methods for computing confidence intervals for \(\mu\). Method 1: Use the Student's \(t\) distribution with d.f. \(=n-1\). This is the method used in the text. It is widely employed in statistical studies. Also, most statistical software packages use this method. Method 2: When \(n \geq 30\), use the sample standard deviation \(s\) as an estimate for \(\sigma\), and then use the standard normal distribution. This method is based on the fact that for large samples, \(s\) is a fairly good approximation for \(\sigma\). Also, for large \(n\), the critical values for the Student's \(t\) distribution approach those of the standard normal distribution. Consider a random sample of size \(n=31\), with sample mean \(\bar{x}=45.2\) and sample standard deviation \(s=5.3\). (a) Compute \(90 \%, 95 \%\), and \(99 \%\) confidence intervals for \(\mu\) using Method 1 with a Student's \(t\) distribution. Round endpoints to two digits after the decimal. (b) Compute \(90 \%, 95 \%\), and \(99 \%\) confidence intervals for \(\mu\) using Method 2 with the standard normal distribution. Use \(s\) as an estimate for \(\sigma\). Round endpoints to two digits after the decimal. (c) Compare intervals for the two methods. Would you say that confidence intervals using a Student's \(t\) distribution are more conservative in the sense that they tend to be longer than intervals based on the standard normal distribution? (d) Repeat parts (a) through (c) for a sample of size \(n=81\). With increased sample size, do the two methods give respective confidence intervals that are more similar?

Answer true or false. Explain your answer. For the same random sample, when the confidence level \(c\) is reduced, the confidence interval for \(\mu\) becomes shorter.

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