/*! 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 11 In a sample of 1000 randomly sel... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a sample of 1000 randomly selected consumers who had opportunities to send in a rebate claim form after purchasing a product, 250 of these people said they never did so ("Rebates: Get What You Deserve", Consumer Reports, May 2009: 7). Reasons cited for their behavior included too many steps in the process, amount too small, missed deadline, fear of being placed on a mailing list, lost receipt, and doubts about receiving the money. Calculate an upper confidence bound at the \(95 \%\) confidence level for the true proportion of such consumers who never apply for a rebate. Based on this bound, is there compelling evidence that the true proportion of such consumers is smaller than \(1 / 3\) ? Explain your reasoning.

Short Answer

Expert verified
The upper confidence bound is 0.2726, which is less than 1/3, indicating the true proportion is likely smaller than 1/3.

Step by step solution

01

Identify the Sample Proportion

First, identify the sample proportion of consumers who did not send in the rebate claim form. This is calculated by dividing the number of consumers who did not send in the claim form by the total number of consumers in the sample. In this case, the sample proportion \( \hat{p} \) is \( \frac{250}{1000} = 0.25 \).
02

Determine the Confidence Level

We are given a confidence level of 95%. This means we will use a Z-score that corresponds to the upper 95% of the standard normal distribution. For a 95% confidence interval, the Z-score is 1.645.
03

Calculate the Standard Error

The standard error (SE) of the sample proportion \( \hat{p} \) is calculated using the formula: \( SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \), where \( n \) is the sample size. Here, \( \hat{p} = 0.25 \) and \( n = 1000 \), so \( SE = \sqrt{\frac{0.25 \times 0.75}{1000}} = \sqrt{\frac{0.1875}{1000}} = 0.0137 \approx 0.0137 \).
04

Find the Upper Confidence Bound

The formula for the upper confidence bound for the proportion is given by: \( \hat{p} + Z \times SE \). Substituting the values we have: \( 0.25 + 1.645 \times 0.0137 = 0.2726 \).
05

Compare to the Threshold of 1/3

We now compare the upper confidence bound of 0.2726 to the hypothesized proportion \( \frac{1}{3} = 0.3333 \). Since 0.2726 is less than 0.3333, the upper confidence bound supports the hypothesis that the true proportion of consumers who don't apply for a rebate is smaller than \( \frac{1}{3} \).

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.

Sample Proportion
The sample proportion is a key concept when analyzing data collected from a survey or experiment. In this context, it represents the fraction of individuals within a sample that exhibit a particular trait or behavior, such as not sending in a rebate claim form. To calculate the sample proportion, divide the number of individuals with the trait by the total number of individuals in the sample.

For example, if 250 out of 1000 consumers did not return their rebate form, the sample proportion is calculated as \( \hat{p} = \frac{250}{1000} = 0.25 \).

Understanding the sample proportion enables researchers to make predictions and draw conclusions about a larger population. It serves as a foundation for further statistical calculations such as confidence intervals or hypothesis tests.
Standard Error
The standard error (SE) measures the potential variability of a sample statistic from the true population parameter due to random sampling. It quantifies how much a sample proportion, like our \(\hat{p} = 0.25\), could differ from the actual population proportion due to chance.

In this exercise, the formula for the standard error of the sample proportion \( SE \) is given by: \[ SE = \sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \] where \( \hat{p} \) is the sample proportion and \( n \) is the sample size.

For the consumer rebate scenario, with \( \hat{p} = 0.25 \) and \( n = 1000 \), the standard error is calculated as: \[ SE = \sqrt{\frac{0.25 \times 0.75}{1000}} \approx 0.0137 \]

A smaller standard error suggests a more accurate sample estimate of the population parameter, while a larger standard error suggests less precision.
Z-score
The Z-score represents how many standard deviations a data point is from the mean of a distribution. In hypothesis testing, Z-scores are used to determine the likelihood of observing a result given a standard normal distribution.

For confidence intervals, the Z-score represents the critical value for a given confidence level. For example, a common confidence level is 95%, and the Z-score corresponding to 95% confidence is approximately 1.645.

This Z-score is used in conjunction with the standard error to compute confidence bounds around a sample estimate. The Z-score allows us to understand the probability of finding the observed sample proportion (or more extreme values) assuming the null hypothesis is true.
Hypothesis Testing
Hypothesis testing is a method to make decisions or inferences about population parameters based on sample data. It involves setting up two opposing hypotheses: the null hypothesis \(H_0\) and the alternative hypothesis \(H_a\).

In this exercise, you are assessing whether the true proportion of consumers who skip sending in rebate forms is less than \(\frac{1}{3}\). Your null hypothesis might state that the true proportion is \(\geq \frac{1}{3}\), while your alternative hypothesis suggests it's \(< \frac{1}{3}\).
  • Null hypothesis: \( H_0: p \geq \frac{1}{3} \)
  • Alternative hypothesis: \( H_a: p < \frac{1}{3} \)


By calculating the upper confidence bound of the sample proportion and comparing it to \(\frac{1}{3}\), you can determine if there is compelling evidence to reject the null hypothesis in favor of the alternative. If the upper bound is less than \(\frac{1}{3}\), as in this case, you have sufficient evidence to support the claim that fewer consumers send in rebate forms than \(\frac{1}{3}\).

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

Nine Australian soldiers were subjected to extreme conditions, which involved a 100 -min walk with a 25 -lb pack when the temperature was \(40^{\circ} \mathrm{C}\left(104^{\circ} \mathrm{F}\right)\). One of them overheated (above \(39^{\circ} \mathrm{C}\) ) and was removed from the study. Here are the rectal Celsius temperatures of the other eight at the end of the walk ("Neural Network Training on Human Body Core Temperature Data," Combatant Protection and Nutrition Branch, Aeronautical and Maritime Research Laboratory of Australia, DSTO TN-0241, 1999): \(\begin{array}{llllllll}38.4 & 38.7 & 39.0 & 38.5 & 38.5 & 39.0 & 38.5 & 38.6\end{array}\) We would like to get a \(95 \%\) confidence interval for the population mean. a. Compute the \(t\)-based confidence interval of Section 8.3. b. Check for the validity of part (a). c. Generate a bootstrap sample of 999 means. d. Use the standard deviation for part (c) to get a \(95 \%\) confidence interval for the population mean. e. Investigate the distribution of the bootstrap means to see if part (d) is valid. f. Use part (c) to form the \(95 \%\) confidence interval using the percentile method. g. Compare the intervals and explain your preference. h. Based on your knowledge of normal body temperature, would you say that body temperature can be influenced by environment?

The article "Evaluating Tunnel Kiln Performance" (Amer. Ceramic Soc. Bull., Aug. 1997: 59-63) gave the following summary information for fracture strengths (MPa) of \(n=169\) ceramic bars fired in a particular kiln: \(x=89.10\), \(s=3.73\). a. Calculate a (two-sided) confidence interval for true average fracture strength using a confidence level of \(95 \%\). Does it appear that true average fracture strength has been precisely estimated? b. Suppose the investigators had believed a priori that the population standard deviation was about \(4 \mathrm{MPa}\). Based on this supposition, how large a sample would have been required to estimate \(\mu\) to within \(.5 \mathrm{MPa}\) with \(95 \%\) confidence?

The Associated Press (October 9, 2002) reported that in a survey of 4722 American youngsters aged \(6-19,15 \%\) were seriously overweight (a body mass index of at least 30 , this index is a measure of weight relative to height). Calculate and interpret a confidence interval using a \(99 \%\) confidence level for the proportion of all American youngsters who are seriously overweight.

The one-sample \(t\) CI for \(\mu\) is also a confidence interval for the population median \(\tilde{\mu}\) when the population distribution is normal. We now develop a CI for \(\tilde{\mu}\) that is valid whatever the shape of the population distribution as long as it is continuous. Let \(X_{1}, \ldots, X_{n}\) be a random sample from the distribution and \(Y_{1}, \ldots, Y_{n}\) denote the corresponding order statistics (smallest observation, second smallest, and so on). a. What is \(P\left(X_{1}<\tilde{\mu}\right) ?\) What is \(P\left(\left\\{X_{1}<\tilde{\mu}\right\\} \cap\right.\) \(\left.\left\\{X_{2}<\tilde{\mu}\right\\}\right) ?\) b. What is \(P\left(Y_{n}<\tilde{\mu}\right)\) ? What is \(P\left(Y_{1}>\tilde{\mu}\right)\) ? [Hint: What condition involving all of the \(X_{i}\) 's is equivalent to the largest being smaller than the population median?] c. What is \(P\left(Y_{1}<\tilde{\mu}

The superintendent of a large school district, having once had a course in probability and statistics, believes that the number of teachers absent on any given day has a Poisson distribution with parameter \(\lambda\). Use the accompanying data on absences for 50 days to derive a large-sample CI for \(\lambda\). [Hint: The mean and variance of a Poisson variable both cqual \(\lambda, s 0\) $$ Z=\frac{X-\lambda}{\sqrt{\lambda / n}} $$ has approximately a standard normal distribution. Now proceed as in the derivation of the interval for \(p\) by making a probability statement (with probability \(1-\alpha\) ) and solving the resulting inequalities for \(\lambda\) (see the argument just after \((8.10))\) ]. \begin{tabular}{l|lll} Number of absences & 0 & 1 & 2 \\ \hline Frequency & 1 & 4 & 8 \end{tabular}

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.