/*! 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 69 In a 2017 Harris poll conducted ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In a 2017 Harris poll conducted for Uber Eats, 438 of 1019 U.S. adults polled said they were "picky eaters." a. What proportion of the respondents said they were picky eaters? b. Find a \(95 \%\) confidence interval for the population proportion of U.S. adults who say they are picky eaters. c. Would a \(90 \%\) confidence interval based on this sample be wider or narrower than the \(95 \%\) interval? Give a reason for your answer. d. Construct the \(90 \%\) confidence interval. Was your conclusion in part c correct?

Short Answer

Expert verified
a) The proportion of 'picky eaters' in the sample is 0.430. b) The 95% confidence interval for the population proportion of 'picky eaters' is \([0.400, 0.460]\). c) A 90% confidence interval would be narrower than a 95% confidence interval. d) The 90% confidence interval is \([0.405, 0.455]\), confirming that it is narrower than the 95% interval.

Step by step solution

01

Proportion Calculation

To answer part a, we need to find out the proportion of respondents who identified as picky eaters. This is achieved by dividing the number of 'picky eaters' (438) by the total number of respondents (1019). So, the proportion \(p\) would be \( \frac{438}{1019} = 0.430 \).
02

Confidence Interval Calculation

For part b, we are asked to compute a 95% confidence interval for the population proportion. The formula for a confidence interval is given by \( p \pm Z \cdot \sqrt{\frac{p(1 - p)}{n}} \) , where \( Z \) is the Z value from the standard normal distribution for the desired confidence level (1.96 for 95% confidence), \( p \) is the sample proportion and \( n \) is the size of the sample. Plugging in our values, we get \( 0.430 \pm 1.96 \cdot \sqrt{\frac{0.430 \cdot (1 - 0.430)}{1019}} = 0.430 \pm 0.030 \). Thus the 95% confidence interval is \([0.400, 0.460]\).
03

Comparing Confidence Intervals

In part c, we are asked to predict whether a 90% confidence interval would be wider or narrower than the 95% interval. A lower confidence level means a narrower confidence interval because we are less certain about where the true population parameter lies. Therefore, a 90% confidence interval will be narrower than a 95% confidence interval.
04

Construction of 90% Confidence Interval

To answer part d, we are asked to construct the 90% confidence interval. The Z value for a 90% confidence interval is 1.645. Using the same formula as in Step 2, we find the 90% confidence interval to be \( 0.430 \pm 1.645 \cdot \sqrt{\frac{0.430 \cdot (1 - 0.430)}{1019}} = 0.430 \pm 0.025 \). Thus the 90% confidence interval is \([0.405, 0.455]\). This confirms our earlier prediction that a 90% confidence interval would be narrower than a 95% confidence interval.

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.

Statistical Inference
Statistical inference is a cornerstone of data analysis, allowing us to draw conclusions about a population from a sample. It involves using sample data to make estimates or test hypotheses about a population parameter. A key aspect of statistical inference is the ability to quantify the uncertainty of an estimate, which is often done through the construction of confidence intervals.

When we speak of the 'population', we're referring to the entire group that we're interested in studying, such as all U.S. adults in the context of eating habits. Our sample, which is a subset, ideally represents this larger group, though it naturally comes with some degree of error since we can't poll every single person.

Once we take a sample and compute statistics like the mean or proportion from it, we're faced with a new problem: how can we be sure that our sample statistics reflect the true population parameters? This is where statistical inference provides tools - like confidence intervals - to suggest with a certain confidence level, such as 95% or 90%, that the population parameter lies within a certain range.
Population Proportion
Population proportion reflects the fraction of individuals in a population who possess a particular attribute. In the Harris poll conducted for Uber Eats, if we want to know the proportion of all U.S. adults that are picky eaters, we're seeking the population proportion.

It is generally not feasible to assess every individual in a population. Thus, we use sample proportions—such as the 438 out of 1019 U.S. adults who identified as picky eaters—as estimates. However, because this is a sample of the larger population, there is uncertainty tied to the estimate. To address this, we express our level of confidence in this proportion estimate by using confidence intervals, providing a range that, statistically speaking, has a high probability of encompassing the true population proportion.
Sampling Distribution
The sampling distribution is a theoretical distribution that describes the probabilities of various possible outcomes for a statistic if we were to take many samples from a population. This distribution helps us understand how our sample statistic, like the proportion of picky eaters in a sample, would behave if we were to repeat our sampling process numerous times.

The Central Limit Theorem tells us that for a large enough sample size, the sampling distribution of the sample proportion will be approximately normal. This is pivotal because it allows us to utilize the properties of the normal distribution to make inferences about the population. Thus, when we talk about the standard error and calculate the margin of error in constructing a confidence interval, we are essentially relying on the nature of the sampling distribution to gauge our estimates' reliability.
Z Value
The Z value plays an integral role in statistical inference, particularly when building confidence intervals and conducting hypothesis tests. It represents the number of standard deviations a given value is away from the mean in a standard normal distribution. In context, it reveals how 'extreme' a sample statistic is compared to what we would expect if the null hypothesis were true.

For example, to construct a 95% confidence interval, we need a Z value that corresponds to the middle 95% of the bell curve—excluding 2.5% from each tail. This Z value is commonly 1.96. Choosing different confidence levels means selecting different Z values; a 90% interval has a narrower range (and thus a smaller Z value of 1.645) because it excludes only 5% from the two tails combined instead of 5%. These Z values are used to calculate the margin of error and, consequently, the confidence interval.

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

Chapman University conducts an annual Survey of American Fears. One of the objectives of this survey is to collect annual data on the fears, worries, and concerns of Americans. In 2017 the survey sampled 1207 participants. One of the survey findings was that \(16 \%\) believe that Bigfoot is a real creature. Identify the sample and population. Is the value \(16 \%\) a parameter or a statistic? What symbol would be use for this value?

A large collection of one-digit random numbers should have about \(50 \%\) odd and \(50 \%\) even digits, because five of the ten digits are odd \((1,3,5,7\), and 9\()\) and five are even \((0,2,4,6\), and 8\()\). a. Find the proportion of odd-numbered digits in the following lines from a random number table. Count carefully. $$ \begin{array}{lll} 57.283 \mathrm{pt} & 74834 & 81172 \\ \hline 89281 & 48134 & 71185 \end{array} $$ b. Does the proportion found in part a represent \(\hat{p}\) (the sample proportion) or \(p\) (the population proportion)? c. Find the error in this estimate, the difference between \(\hat{p}\) and \(p\) (or \(\hat{p}-p)\).

In 2017 , the journal Obesity reported on trends in sugar-sweetened beverage (SSB) consumption. A random sample of youths aged 12 to 19 years old were asked to monitor all food and beverages consumed in a 24 -hour period. The study was done in 2003 and repeated in 2014 . The numbers who consumed a sugary beverage such as soda or fruit juice in a day are shown in the table. (Bleich et al., "Trends in Beverage Consumption among Children and Adults, 2003-2014," Obesity, vol. 26 [2018]: 432-441. doi:10.1002/oby.22056) $$ \begin{array}{|l|l|} \hline \text { Consumed SSB } & \mathbf{2 0 0 3} & \mathbf{2 0 1 4} \\ \hline \text { Yes } & 3416 & 2682 \\ \hline \text { No } & 685 & 1419 \\ \hline \end{array} $$ a. Calculate and compare the percentages of youths in this age group who consumed an SSB during the recording period. b. Check that the conditions for using a two-population confidence interval hold. c. Find the \(95 \%\) confidence interval for the difference in the proportion of youth consuming an SSB in 2003 and 2014. Based on your confidence interval, do you think there has been a change in sugar-sweetened beverage consumption among this age group? Explain.

a. If a rifleman's gunsight is adjusted correctly, but he has shaky arms. the bullets might be scattered widely around the bull's-eye target. Draw a sketch of the target with the bullet holes. Does this show variation (lack of precision) or bias? b. Draw a second sketch of the target if the shots are unbiased and have precision (little variation). The rifleman's aim is not perfect, so your sketches should show more than one bullet hole.

In a simple random sample of 1200 Americans age 20 and over, the proportion with diabetes was found to be \(0.115\) (or \(11.5 \%)\). a. What is the standard error for the estimate of the proportion of all Americans age 20 and over with diabetes? b. Find the margin of error, using a \(95 \%\) confidence level, for estimating this proportion. c. Report the \(95 \%\) confidence interval for the proportion of all Americans age 20 and over with diabetes. d. According to the Centers for Disease Control and Prevention, nationally, \(10.7 \%\) of all Americans age 20 or over have diabetes. Does the confidence interval you found in part c support or refute this claim? Explain.

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.