/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 10 Consider two independent binomia... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider two independent binomial experiments. In the first one, 40 trials had 15 successes. In the second one, 60 trials had 6 successes. (a) Is it appropriate to use a normal distribution to approximate the \(\hat{p}_{1}-\hat{p}_{2}\) distribution? Explain. (b) Find a \(95 \%\) confidence interval for \(p_{1}-p_{2}\) (c) IBased on the confidence interval you computed, can you be \(95 \%\) confident that \(p_{1}\) is more than \(p_{2} ?\) Explain.

Short Answer

Expert verified
(a) Yes, normal approximation is valid. (b) CI is (0.107, 0.443). (c) Yes, \( p_1 > p_2 \) with 95% confidence.

Step by step solution

01

Check Normal Approximation Conditions

To decide if it's appropriate to use a normal distribution for \( \hat{p}_1 - \hat{p}_2 \), we need to check if both sample sizes allow for this approximation. For each group, \( n\hat{p} \) and \( n(1-\hat{p}) \) should be at least 5. For the first experiment: \( \hat{p}_1 = \frac{15}{40} = 0.375 \), so \( 40 \times 0.375 = 15 \) and \( 40 \times 0.625 = 25 \). Both are greater than 5. For the second experiment: \( \hat{p}_2 = \frac{6}{60} = 0.1 \), so \( 60 \times 0.1 = 6 \) and \( 60 \times 0.9 = 54 \). Both are greater than 5. Thus, a normal approximation is appropriate.
02

Calculate Standard Error

The standard error (SE) for \( \hat{p}_1 - \hat{p}_2 \) is given by \[ SE = \sqrt{\frac{\hat{p}_1(1-\hat{p}_1)}{n_1} + \frac{\hat{p}_2(1-\hat{p}_2)}{n_2}} \]. Using the estimates: \( \hat{p}_1 = 0.375 \), \( 1 - \hat{p}_1 = 0.625 \), \( n_1 = 40 \), \( \hat{p}_2 = 0.1 \), \( 1 - \hat{p}_2 = 0.9 \), \( n_2 = 60 \). Substituting these into the formula yields: \[ SE = \sqrt{\frac{0.375 \times 0.625}{40} + \frac{0.1 \times 0.9}{60}} \] \[ SE = \sqrt{\frac{0.234375}{40} + \frac{0.09}{60}} = \sqrt{0.005859375 + 0.0015} = \sqrt{0.007359375} \approx 0.0858 \].
03

Compute Confidence Interval

We will use the standard normal distribution for a 95% confidence interval, which corresponds to \( Z_{0.025} = 1.96 \). The confidence interval is given by \( (\hat{p}_1 - \hat{p}_2) \pm Z \times SE \). So, \[ CI = (0.375 - 0.1) \pm 1.96 \times 0.0858 \] \[ CI = 0.275 \pm 0.168 \] \[ CI = (0.107, 0.443) \].
04

Interpret the Confidence Interval

The 95% confidence interval for \( p_1 - p_2 \) is \( (0.107, 0.443) \). Since this interval is entirely greater than 0, we can confidently say, with 95% certainty, that \( p_1 \) is greater than \( p_2 \).

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Key Concepts

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

Binomial Distribution
A binomial distribution models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success. In simple terms, imagine flipping a coin a certain number of times and counting how many heads (successes) you get. Each flip is independent of the others, and each has the same probability of resulting in heads. Here, the formula for the binomial distribution is expressed as:
  • Probability of exactly k successes in n trials: \(P(X=k) = \binom{n}{k} p^k (1-p)^{n-k}\)
Where \(n\) is the total number of trials, \(p\) is the probability of success on any given trial, and \(\binom{n}{k}\) is the binomial coefficient.
In our example exercises, the first experiment had 40 trials with a probability of success of 0.375, and the second had 60 trials with a success probability of 0.1. This setup shows the use of binomial distribution in real-life data collection, like experiments or surveys.
Confidence Interval
A confidence interval provides a range of values that likely contain the difference between two population proportions with a given level of confidence. It's a critical concept in statistical inference, as it offers insight into how much difference there truly is between populations. For a confidence interval to be useful, we often express it as a percentage, such as 95%. This means we are 95% confident that the true difference lies within our calculated interval.
In practice, to calculate a 95% confidence interval for the difference in proportions \(p_1 - p_2\), we use the formula:
  • \( CI = (\hat{p}_1 - \hat{p}_2) \pm Z \times SE \)
Here \(Z\) is the Z-score corresponding to the desired level of confidence, often 1.96 for a 95% confidence level, and \(SE\) is the standard error of the difference between sample proportions. In our example, the confidence interval calculated was \((0.107, 0.443)\), suggesting with 95% confidence that the true proportion of successes differs by this range.
Normal Approximation
Normal approximation is a technique used when certain conditions are met to approximate a binomial distribution using a normal distribution. This is essential when dealing with larger sample sizes, as it simplifies computations and helps in constructing confidence intervals and hypothesis tests more easily.
To be eligible for using normal approximation, the rule of thumb is:
  • Both \(n\hat{p}\) and \(n(1-\hat{p})\) should be 5 or greater.
This ensures the shape of the binomial distribution becomes bell-shaped or normal. In the given exercise, both experiments meet these conditions, making the normal approximation valid for calculating the confidence interval for \(\hat{p}_1 - \hat{p}_2\). This validity simplifies the calculation of statistical measures and helps derive meaningful insights from the data.
Hypothesis Testing
Hypothesis testing is a fundamental method in statistics used to determine if there is enough evidence in a sample to support a particular claim about a population. It provides a framework to test assumptions or hypotheses using sample data.
The process involves:
  • Stating a null hypothesis (\(H_0\)), which is a default assumption, and an alternative hypothesis (\(H_a\)), which indicates the effect or difference we suspect.
  • Calculating a test statistic and comparing it to a critical value or using a p-value to see if it falls within a pre-set significance level, such as 0.05.
In the context of our example, based on the calculated 95% confidence interval, the hypothesis test helps determine if we can reasonably state that the proportion of successes in the first experiment is indeed greater than in the second. Since the entire confidence interval lies above zero, it supports the claim that \(p_1 > p_2\) at a 95% confidence level, leading to the rejection of the null hypothesis that \(p_1 \leq p_2\).

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Most popular questions from this chapter

What price do farmers get for their watermelon crops? In the third week of July, a random sample of 40 farming regions gave a sample mean of \(\bar{x}=\6.88\) per 100 pounds of watermelon. Assume that \(\sigma\) is known to be \(1.92\) per 100 pounds (Reference: Agricultural Statistics\(,\) U.S. Department of Agriculture). (a) Find a \(90 \%\) confidence interval for the population mean price (per 100 pounds) that farmers in this region get for their watermelon crop. What is the margin of error? (b) Sample Size Find the sample size necessary for a \(90 \%\) confidence level with maximal margin of error \(E=0.3\) for the mean price per 100 pounds of watermelon. (c) A farm brings 15 tons of watermelon to market. Find a \(90 \%\) confidence interval for the population mean cash value of this crop. What is the margin of error? Hint: 1 ton is 2000 pounds.

Overproduction of uric acid in the body can be an indication of cell breakdown. This may be an advance indication of illness such as gout, leukemia, or lymphoma (Reference: Manual of Laboratory and Diagnostic Tests by F. Fischbach). Over a period of months, an adult male patient has taken eight blood tests for uric acid. The mean concentration was \(\bar{x}=5.35 \mathrm{mg} / \mathrm{dl} .\) The distribution of uric acid in healthy adult males can be assumed to be normal, with \(\sigma=1.85 \mathrm{mg} / \mathrm{dl}\) (a) Find a \(95 \%\) confidence interval for the population mean concentration of uric acid in this patient's blood. What is the margin of error? (b) What conditions are necessary for your calculations? (c) Interpret your results in the context of this problem. (d) Sample Size Find the sample size necessary for a \(95 \%\) confidence level with maximal margin of error \(E=1.10\) for the mean concentration of uric acid in this patient's blood.

In a combined study of northern pike, cutthroat trout, rainbow trout, and lake trout, it was found that 26 out of 855 fish died when caught and released using barbless hooks on flies or lures. All hooks were removed from the fish (Source: \(A\) National Symposium on Catch and Release Fishing, Humboldt State University Press). (a) Let \(p\) represent the proportion of all pike and trout that die (i.e., \(p\) is the mortality rate) when caught and released using barbless hooks. Find a point estimate for \(p\). (b) Find a \(99 \%\) confidence interval for \(p,\) and give a brief explanation of the meaning of the interval. (c) Is the normal approximation to the binomial justified in this problem? Explain.

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 P/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). (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.

Diagnostic Tests: Total Calcium Over the past several months, an adult patient has been treated for tetany (severe muscle spasms). This condition is associated with an average total calcium level below 6 mg/dl (Reference: Manual of Laboratory and Diagnostic Tests by F. Fischbach). Recently, the patient's total calcium tests gave the following readings (in mg/dl). $$ \begin{array}{ccccccc} 9.3 & 8.8 & 10.1 & 8.9 & 9.4 & 9.8 & 10.0 \\ 9.9 & 11.2 & 12.1 & & & & \end{array} $$ (a) Use a calculator to verify that \(\bar{x}=9.95\) and \(s \approx 1.02\) (b) Find a \(99.9 \%\) confidence interval for the population mean of total calcium in this patient's blood. (c) Interpretation Based on your results in part (b), does it seem that this patient still has a calcium deficiency? Explain.

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