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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.

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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.

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Most popular questions from this chapter

Football Air Pressure During the NFL's 2014 AFC championship game, officials measured the air pressure on game balls following a tip that one team's balls were under-inflated. In exercise 6.124 we found that the 11 balls measured for the New England Patriots had a mean psi of 11.10 (well below the legal limit) and a standard deviation of 0.40. Patriot supporters could argue that the under-inflated balls were due to the elements and other outside effects. To test this the officials also measured 4 balls from the opposing team (Indianapolis Colts) to be used in comparison and found a mean psi of \(12.63,\) with a standard deviation of 0.12. There is no significant skewness or outliers in the data. Use the t-distribution to determine if the average air pressure in the New England Patriot's balls was significantly less than the average air pressure in the Indianapolis Colt's balls.

In Exercises 6.203 and \(6.204,\) use Stat Key or other technology to generate a bootstrap distribution of sample differences in means and find the standard error for that distribution. Compare the result to the standard error given by the Central Limit Theorem, using the sample standard deviations as estimates of the population standard deviations. Difference in mean commuting time (in minutes) between commuters in Atlanta and commuters in St. Louis, using \(n_{1}=500, \bar{x}_{1}=29.11,\) and \(s_{1}=20.72\) for Atlanta and \(n_{2}=500, \bar{x}_{2}=21.97,\) and \(s_{2}=14.23\) for St. Louis

In Exercises 6.103 and 6.104 , find a \(95 \%\) confidence interval for the mean two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the t-distribution and the formula for standard error. Compare the results. Mean distance of a commute for a worker in Atlanta, using data in Commute Atlanta with \(\bar{x}=\) 18.156 miles, \(s=13.798,\) and \(n=500\)

Use the t-distribution to find a confidence interval for a difference in means \(\mu_{1}-\mu_{2}\) given the relevant sample results. Give the best estimate for \(\mu_{1}-\mu_{2},\) the margin of error, and the confidence interval. Assume the results come from random samples from populations that are approximately normally distributed. A \(99 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=501, s_{1}=115, n_{1}=400\) and \(\bar{x}_{2}=469, s_{2}=96, n_{2}=200 .\)

A data collection method is described to investigate a difference in means. In each case, determine which data analysis method is more appropriate: paired data difference in means or difference in means with two separate groups. To study the effect of sitting with a laptop computer on one's lap on scrotal temperature, 29 men have their scrotal temperature tested before and then after sitting with a laptop for one hour.

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