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91Ó°ÊÓ

In a Nielsen global online survey of about 27,000 people from 53 different countries, \(61 \%\) of consumers indicated that they purchased more store brands during the economic downturn and \(91 \%\) indicated that they will continue to purchase the same number of store brands when the economy improves. The results were remarkably consistent across all regions of the world. \({ }^{13}\) The survey was conducted during September 2010 and had quotas for age and gender for each country, and the results are weighted to be representative of consumers with Internet access. (a) What is the sample? What is an appropriate population? (b) Find and interpret a \(99 \%\) confidence interval for the proportion of consumers who purchased more store brands during the economic downturn. (c) Find and interpret a \(99 \%\) confidence interval for the proportion of consumers who will continue to purchase the same number of store brands.

Short Answer

Expert verified
The sample of the survey consists of approximately 27,000 internet users from 53 countries, and the population it represents is the global population of consumers with internet access. The 99% confidence interval for the proportion of consumers who purchased more store brands during the economic downturn and those who will continue to do so, is obtained through the respective calculations and interpretations.

Step by step solution

01

Definition of Sample and Population

The sample is a group of about 27,000 people who participated in the Nielsen global online survey. This sample is composed of individuals from 53 different countries. The appropriate population would be all consumers with internet access across the globe, as the sample results are weighted to be representative of this group.
02

Calculation of Confidence Interval for Proportion of Consumers Who Purchased More Brands

A 99% confidence interval for the proportion can be calculated using the formula \[ \hat{p} \pm Z*\sqrt{\frac{\hat{p}(1-\hat{p})}{n}} \] where \(\hat{p}\) is the sample proportion, \(Z\) is the critical value for the 99% confidence level, and \(n\) is the sample size. For this case, \(\hat{p} = 0.61\), \(Z = 2.58\) for a 99% confidence level, and \(n = 27000\). Then substitute these values into the formula to get the confidence interval.
03

Interpretation of Confidence Interval for Proportion of Consumers Who Purchased More Brands

The calculated confidence interval represents the range in which we are 99% confident that the true proportion of consumers who purchased more store brands during the economic downturn falls.
04

Calculation of Confidence Interval for Proportion of Consumers Who Will Continue to Purchase the Same Number of Store Brands

Similar to step 2, the 99% confidence interval can be calculated with \(\hat{p} = 0.91\), \(Z = 2.58\), and \(n = 27000\). Then substitute these values into the formula to get the confidence interval.
05

Interpretation of Confidence Interval for Proportion of Consumers Who Will Continue to Purchase the Same Number of Store Brands

The calculated confidence interval represents the range in which we are 99% confident that the true proportion of consumers who will continue to purchase the same number of store brands when the economy improves falls.

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Key Concepts

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

Sample and Population
Understanding the concepts of sample and population is fundamental in statistics. A sample refers to a subset of individuals selected from a larger group, known as the population. In the exercise, the sample consists of approximately 27,000 people from 53 different countries. These participants have been specifically chosen for the Nielsen global online survey.

When exploring surveys and studies, it's crucial to identify whether the sample is representative of the population. Here, the population is all consumers with internet access worldwide. This distinction is vital as results are usually intended to be generalized beyond the sample to apply to the broader population.
A key element in this context is ensuring the sample reflects the population's characteristics, achieved through strategies such as random sampling and weighting the results as done in this survey.
Proportion
Proportion in statistics indicates the part of the population that exhibits a particular characteristic. It's often expressed as a percentage or fraction. In the given survey, two proportions are of interest:
  • 61% of consumers purchased more store brands during the economic downturn.
  • 91% of consumers intend to continue purchasing the same amount of store brands after the economy improves.
The calculation of proportions within the context of the survey is handled using the formula for a confidence interval for proportions.
The confidence interval provides a range of plausible values for the population proportion based on the sample data. Hence, it offers a statistical estimate that considers sample variability and the desired confidence level, assisting in making generalizations about the true population.
Statistical Survey
A statistical survey involves collecting data from a subset of a population to glean insights about the whole group. Surveys like the Nielsen global online survey gather extensive data on consumer behavior and preferences.

The design and implementation of a survey are crucial to its success. Important factors include defining clear objectives, selecting an appropriate sample, applying methodological rigor, and analyzing the data reliably. For instance, in this survey, quotas for age and gender were established to ensure the sample accurately represents the population.
Furthermore, weighting techniques adjust for any disparities and enhance the representativeness of the survey results. The thoughtful execution of these strategies ensures the survey results provide a credible, realistic picture of the population's behaviors or opinions.

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Most popular questions from this chapter

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 300 from a population with proportion 0.08

In Exercises 6.166 and \(6.167,\) use StatKey or other technology to generate a bootstrap distribution of sample differences in proportions and find the standard error for that distribution. Compare the result to the value obtained using the formula for the standard error of a difference in proportions from this section (and using the sample proportions to estimate the population proportions). Sample A has a count of 30 successes with \(n=100\) and Sample \(\mathrm{B}\) has a count of 50 successes with \(n=250\).

In Exercises 6.1 to \(6.6,\) if random samples of the given size are drawn from a population with the given proportion: (a) Find the mean and standard error of the distribution of sample proportions. (b) If the sample size is large enough for the Central Limit Theorem to apply, draw a curve showing the shape of the sampling distribution. Include at least three values on the horizontal axis. Samples of size 1000 from a population with proportion 0.70

In the mid-1990s a Nabisco marketing campaign claimed that there were at least 1000 chips in every bag of Chips Ahoy! cookies. A group of Air Force cadets collected a sample of 42 bags of Chips Ahoy! cookies, bought from locations all across the country, to verify this claim. \({ }^{41}\) The cookies were dissolved in water and the number of chips (any piece of chocolate) in each bag were hand counted by the cadets. The average number of chips per bag was \(1261.6,\) with standard deviation 117.6 chips. (a) Why were the cookies bought from locations all over the country? (b) Test whether the average number of chips per bag is greater than 1000 . Show all details. (c) Does part (b) confirm Nabisco's claim that every bag has at least 1000 chips? Why or why not?

NBA Free Throws In Exercise 6.10, we learn that the percent of free throws made in the \(\mathrm{NBA}\) (National Basketball Association) has been about \(75 \%\) for the last 50 years. If we take random samples of free throws in the NBA and compute the proportion of free throws made, what percent of samples of size \(n=200\) will have a sample proportion greater than \(80 \%\) ? Use the fact that the sample proportions are normally distributed and compute the mean and standard deviation of the distribution.

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