/*! 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 197 Effect of Splitting the Bill Exe... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 197

Effect of Splitting the Bill Exercise 2.153 on page 105 describes a study to compare the cost of restaurant meals when people pay individually versus splitting the bill as a group. In the experiment half of the people were told they would each be responsible for individual meal costs and the other half were told the cost would be split equally among the six people at the table. The 24 people paying individually had a mean cost of 37.29 Israeli shekels with a standard deviation of 12.54 , while the 24 people splitting the bill had a higher mean cost of 50.92 Israeli shekels with a standard deviation of 14.33. The raw data can be found in SplitBill and both distributions are reasonably bell-shaped. Use this information to find and interpret a \(95 \%\) confidence interval for the difference in mean meal cost between these two situations.

Short Answer

Expert verified
Without performing the actual calculations, the steps detailed above should lead to the 95% confidence interval for the difference in mean meal cost. The presence or absence of zero in this interval will denote whether a significant difference between the two situations exists.

Step by step solution

01

Formulate the Confidence Interval Formula

The formula for a confidence interval around a difference in means is given by \((\overline{X}_1 - \overline{X}_2) \pm Z \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\), where \(\overline{X}_i\) is the mean, \(s_i\) the standard deviation, \(n_i\) the sample size for group i, and Z the z-score associated with our confidence level. For a 95% confidence level, Z is approximately 1.96.
02

Compute the Confidence Interval

Plugging in the given values in the formula: \(\overline{X}_1 = 37.29\), \(s_1 = 12.54\), \(n_1 = 24\), \(\overline{X}_2 = 50.92\), \(s_2 = 14.33\), \(n_2 = 24\), and \(Z = 1.96\), we can calculate the confidence interval: \(CI = (37.29 - 50.92) \pm 1.96 \cdot \sqrt{\frac{12.54^2}{24} + \frac{14.33^2}{24}}\).
03

Interpret the Confidence Interval

Once calculated, the confidence interval represents the range of plausible values for the difference in mean meal cost between the two situations. If the interval contains zero, it would indicate that there is no significant difference between the two means at the 95% confidence level. If the interval does not contain zero, it indicates that a significant difference exists.

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 91影视!

Key Concepts

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

Difference in Means
When conducting a study to compare the costs of restaurant meals based on how bills are paid, examining the difference in means is crucial. The mean is essentially the average cost in each group, and the difference in means helps us to assess whether there is a statistically significant difference between two setups: paying individually or splitting the bill equally.
In this exercise, the people who paid individually had a mean cost of 37.29 shekels, while the group splitting the bill had a mean of 50.92 shekels. The difference between these two means is one of the primary values of interest in this experiment.
By subtracting the individual payment mean from the group payment mean, we find the difference in means: \[\overline{X}_1 - \overline{X}_2 = 37.29 - 50.92 = -13.63\]This negative difference indicates that, on average, those who paid individually spent less compared to those who split the bill, highlighting an important trend in how payment methods can influence spending behavior.
Standard Deviation
Understanding standard deviation is key to interpreting the variability of data in each group. Standard deviation tells us how spread out the meal costs are around their respective means in both scenarios.
In our exercise: - For those paying individually, the standard deviation is 12.54 shekels. - For those splitting the bill, the standard deviation is 14.33 shekels.
These values indicate the degree of variation in meal costs. A lower standard deviation implies that meal costs are closer to the mean, whereas a higher standard deviation suggests a wider range of meal costs.
These standard deviations are incorporated into further calculations, such as the confidence interval, as they affect the precision of our estimates regarding the mean differences. A larger standard deviation can lead to a wider confidence interval, reflecting greater uncertainty about the precise difference in means.
95% Confidence Level
A 95% confidence level is a popular statistical threshold used to estimate the range of values that is likely to contain the true difference in means. When we use this confidence level, we are saying that if we were to repeat this study many times, 95% of the time the calculated confidence interval would encompass the true mean difference.
The formula for our confidence interval around the difference in means includes a crucial term: \[CI = (\overline{X}_1 - \overline{X}_2) \pm Z \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]Where \(Z\) is the z-score that corresponds to the chosen confidence level, here approximately 1.96 for 95%.
Once we compute the confidence interval, it provides a range of plausible values for the difference in means. Importantly, if zero is not within this interval, we can say with 95% confidence that a significant difference exists between the two bill scenarios. If zero lies within the interval, it indicates that there might not be a significant difference in spending behavior between the two payment methods.

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

In Exercises 6.7 and 6.8 , compute the standard error for sample proportions from a population with the given proportion using three different sample sizes. What effect does increasing the sample size have on the standard error? Using this information about the effect on the standard error, discuss the effect of increasing the sample size on the accuracy of using a sample proportion to estimate a population proportion. A population with proportion \(p=0.75\) for sample sizes of \(n=40, n=300,\) and \(n=1000 .\)

Use the t-distribution to find a confidence interval for a mean \(\mu\) given the relevant sample results. Give the best point estimate for \(\mu,\) the margin of error, and the confidence interval. Assume the results come from a random sample from a population that is approximately normally distributed. A \(99 \%\) confidence interval for \(\mu\) using the sample results \(\bar{x}=88.3, s=32.1,\) and \(n=15\)

6.122 Be Nice to Pigeons, As They Remember Your Face In a study \(^{30}\) conducted in Paris, France, equal amounts of pigeon feed were spread on the ground in two adjacent locations. A person was present in both sites, with one acting hostile and running at the birds to scare them away and the other acting neutral and just observing. The two people were randomly exchanged between the two sites throughout and the birds quickly learned to avoid the hostile person's site and to eat at the site of the neutral person. At the end of the training session, both people behaved neutrally but the birds continued to remember which one was hostile. In the most interesting part of the experiment, when the two people exchanged coats (orange worn by the hostile one and yellow by the neutral one throughout training), the pigeons were not fooled and continued to recognize and avoid the hostile person. The quantity measured is difference in number of pigeons at the neutral site minus the hostile site. With \(n=32\) measurements, the mean difference in number of pigeons is 3.9 with a standard deviation of 6.8 . Test to see if this provides evidence that the mean difference is greater than zero, meaning the pigeons can recognize faces (and hold a grudge!)

A survey of 1000 adults in the US conducted in March 2011 asked "Do you favor or oppose 'sin taxes' on soda and junk food?" The proportion in favor of taxing these foods was \(32 \% .10\) (a) Find a \(95 \%\) confidence interval for the proportion of US adults favoring taxes on soda and junk food. (b) What is the margin of error? (c) If we want a margin of error of only \(1 \%\) (with \(95 \%\) confidence \()\), what sample size is needed?

Use a t-distribution to answer the question. Assume the sample is a random sample from a distribution that is reason ably normally distributed and we are doing inference for a sample mean. Find endpoints of a t-distribution with 0.025 beyond them in each tail if the sample has size \(n=25\)
See all solutions

Recommended explanations on Math Textbooks

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation

Geometry

Read Explanation

Decision Maths

Read Explanation

Calculus

Read Explanation

Logic and Functions

Read Explanation
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.