/*! 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 28 Suppose the number of successes ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose the number of successes observed in \(n=500\) trials of a binomial experiment is 27 . Find a \(95 \%\) confidence interval for \(p\). Why is the confidence interval narrower than the confidence interval in Exercise \(8.27 ?\)

Short Answer

Expert verified
Question: Determine a 95% confidence interval for the probability of success, p, in a binomial experiment given n = 500 trials and 27 observed successes. Compare the result to a confidence interval from Exercise 8.27 and explain why the interval found here is narrower. Answer: The 95% confidence interval for the probability of success, p, is (0.0399, 0.0681). The interval found in this problem is narrower than the interval found in Exercise 8.27 because the sample size is larger (500 trials). With a larger sample size, the standard error decreases, which means the width of the confidence interval also decreases, giving a more precise estimate of the population parameter.

Step by step solution

01

Calculate the sample proportion

Let \(\hat{p}\) represent the sample proportion. To find \(\hat{p}\), divide the number of successes (27) by the total number of trials (500). \(\hat{p} = \frac{27}{500} = 0.054\)
02

Find the associated normal distribution standard deviation

The standard deviation of the normal approximation to the binomial distribution is given by: \(\sigma_{\hat{p}}= \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\) Substitute \(\hat{p} = 0.054\) and \(n = 500\) to calculate the standard deviation. \(\sigma_{\hat{p}}= \sqrt{\frac{0.054(1-0.054)}{500}} \approx 0.0072\)
03

Determine the critical value

For a 95% confidence interval, the corresponding critical value, \(Z_{\frac{\alpha}{2}}\), is 1.96. (This value reflects the necessary standard deviations away from the mean to capture 95% of the data).
04

Calculate the confidence interval

Confidence interval is given by the formula: \(\hat{p} \pm Z_{\frac{\alpha}{2}} \times \sigma_{\hat{p}}\) Substitute \(\hat{p}\), \(Z_{\frac{\alpha}{2}}\), and \(\sigma_{\hat{p}}\) to compute the confidence interval. \(0.054 \pm 1.96 \times 0.0072\) Calculate the lower and upper limits of the confidence interval: Lower limit: \(0.054 - 1.96 \times 0.0072 \approx 0.0399\) Upper limit: \(0.054 + 1.96 \times 0.0072 \approx 0.0681\) Thus, the 95% confidence interval for \(p\) is \((0.0399, 0.0681)\).
05

Explain the difference in width between the two confidence intervals

The confidence interval found here is narrower than the confidence interval in Exercise 8.27. The key reason for this difference is the sample size. In this problem, there are 500 trials as opposed to a smaller number of trials in Exercise 8.27. As the sample size increases, the standard error (and thus, the width of the confidence interval) decreases, giving a more precise estimate of the population parameter.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Binomial Distribution
The binomial distribution is a probability distribution that models the number of successes in a fixed number of trials. Each trial is independent, and the probability of success remains constant across all trials. In simpler terms, it's like flipping a coin several times and counting how many times it lands on heads. It's crucial to use when each trial has only two possible outcomes, like success and failure.
For instance, in a binomial experiment with 500 trials, such as measuring the number of successful attempts to convince a customer to buy a product, the outcomes in each attempt are either a sale (success) or not a sale (failure).
Understanding this distribution helps assess the chances of different numbers of successes occurring over many attempts, which is ideal for estimating probabilities in real-world scenarios.
Sample Proportion
The sample proportion is a way to express the number of successes as a fraction of the total number of trials. It provides an estimate of the probability of success in a binomial experiment. To find the sample proportion, divide the number of observed successes by the total number of trials:
  • Let \( \hat{p} \) represent the sample proportion.
  • Calculate \( \hat{p} = \frac{\text{Number of Successes}}{\text{Total Number of Trials}} \).
In the context of our exercise, with 27 successes out of 500 trials, the sample proportion is \( \hat{p} = 0.054 \).
This value reflects the estimated success rate of the experiment, giving a snapshot of results relative to the total set, making it essential for further statistical calculations like confidence intervals.
Standard Deviation
Standard deviation, especially in the context of binomial distribution, measures how much the sample proportion might vary from the actual population proportion. It's crucial because it provides an indication of the uncertainty or variability in our sample.
In our situation, the standard deviation of the sample proportion is given by:\[\sigma_{\hat{p}} = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\] Using \( \hat{p} = 0.054 \) and \( n = 500 \), we find the standard deviation is approximately \( 0.0072 \). This small value suggests low variability, indicating that our sample proportion is relatively stable given the sample size. This calculation is vital when building the confidence interval, as it determines the margin of error in our sample results.
Critical Value
The critical value in statistics helps determine the range of a confidence interval, indicating how far we are willing to stretch from the sample mean to infer about the population mean. For a 95% confidence interval, the critical value comes from the standard normal distribution, often noted as \( Z_{\frac{\alpha}{2}} \).
With a 95% confidence interval, the critical value \( Z_{\frac{\alpha}{2}} = 1.96 \) suggests that 95% of the data will fall within 1.96 standard deviations from the mean.
  • This ensures that we have a high degree of confidence that our interval contains the true population proportion.
  • The critical value influences the size of the confidence interval by being multiplied with the standard deviation.
Knowing this value helps us construct the interval accurately, providing a reliable estimate of the actual proportion in the general population.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

An experimenter fed different rations, \(A\) and \(B\), to two groups of 100 chicks each. Assume that all factors other than rations are the same for both groups. Of the chicks fed ration \(\mathrm{A}, 13\) died, and of the chicks fed ration \(\mathrm{B}, 6\) died. a. Construct a \(98 \%\) confidence interval for the true difference in mortality rates for the two rations. b. Can you conclude that there is a difference in the mortality rates for the two rations?

Suppose you wish to estimate the mean \(\mathrm{pH}\) of rainfalls in an area that suffers heavy pollution due to the discharge of smoke from a power plant. You know that \(\sigma\) is approximately \(.5 \mathrm{pH},\) and you wish your estimate to lie within .1 of \(\mu\), with a probability near \(.95 .\) Approximately how many rainfalls must be included in your sample (one pH reading per rainfall)? Would it be valid to select all of your water specimens from a single rainfall? Explain.

Find a \(99 \%\) lower confidence bound for the binomial proportion \(p\) when a random sample of \(n=400\) trials produced \(x=196\) successes.

In an Advertising Age white paper concerning the changing role of women as "breadwinners" in the American family, it was reported that according to their survey with JWT, working men reported doing 54 minutes of household chores a day, while working women reported tackling 72 minutes daily. But when examined more closely, Millennial men reported doing just as many household chores as the average working women, 72 minutes, compared to an average of 54 minutes among both Boomer men and Xer men. \({ }^{6}\) The information that follows is adapted from these data and is based on random samples of 1136 men and 795 women. a. Construct a \(95 \%\) confidence interval for the average time all men spend doing household chores. b. Construct a \(95 \%\) confidence interval for the average time women spend doing household chores.

In addition to teachers and administrative staff, schools also have many other employees, including bus drivers, custodians, and cafeteria workers. In Auburn, WA, the average hourly wage is \(\$ 16.92\) for bus drivers, \(\$ 17.65\) for custodians, and \(\$ 12.86\) for cafeteria workers. \({ }^{23}\) Suppose that a second school district employs \(n=36\) bus drivers who earn an average of \(\$ 13.45\) per hour with a standard deviation of \(s=\$ 2.84 .\) Find a \(95 \%\) confidence interval for the average hourly wage of bus drivers in school districts similar to this one. Does your confidence interval enclose the Auburn, WA average of \(\$ 16.92 ?\) What can you conclude about the hourly wages for bus drivers in this second school district?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.