/*! 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 224 Split the Bill? Exercise 2.153 o... [FREE SOLUTION] | 91Ó°ÊÓ

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Split 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 data in SplitBill includes the cost of what each person ordered (in Israeli shekels) and the payment method (Individual or Split). Some summary statistics are provided in Table 6.20 and both distributions are reasonably bell-shaped. Use this information to test (at a \(5 \%\) level ) if there is evidence that the mean cost is higher when people split the bill. You may have done this test using randomizations in Exercise 4.118 on page 302 .

Short Answer

Expert verified
The detailed conclusion will depend on the calculated p-value. If the p-value less than 0.05, we conclude there is sufficient evidence that the mean cost is higher when people split the bill.

Step by step solution

01

Define the Null and Alternative Hypotheses

The null hypothesis (\(H_0\)) is that the mean cost of meals is the same whether people pay individually or split the bill. So, \(H_0: \mu_{individual} = \mu_{split}\). The alternative hypothesis (\(H_a\)) is that the mean cost is higher when people split the bill. So, \(H_a: \mu_{split} > \mu_{individual}\).
02

Calculate the Test Statistic and P-value

We first calculate the difference between the means of the two groups. Then, we calculate the standard error of the difference in means. This involves using the given summary statistics and the formulas: \( SE = \sqrt{s^2_{split}/n_{split} + s^2_{individual}/n_{individual}}\). The test statistic (t) is then the difference in means divided by the SE. The p-value can be obtained using a t-distribution table look up or a statistical software.
03

Compare P-value with Significance Level and Draw Conclusions

The p-value is compared with the given significance level (0.05). If the p-value is less than 0.05, we reject the null hypothesis in favor of the alternative. This would mean that there is sufficient evidence at the 5% level to conclude that the mean cost is higher when people split the bill.

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Key Concepts

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

Null and Alternative Hypotheses
Understanding the foundation of hypothesis testing starts with the null and alternative hypotheses. In the context of our restaurant bill study, the null hypothesis ({H_{0}}) maintains that there is no difference in average meal costs between groups who pay individually and those who split the bill. It is expressed as {H_{0}: mu_{individual} = mu_{split}}. Put simply, it's the assumption of 'no effect or no difference'.
On the flip side, the alternative hypothesis ({H_{a}}) proposes that there is an effect, or in this case, that the mean cost of meals is higher when the bill is split. We articulate this as {H_{a}: mu_{split} > mu_{individual}}. This hypothesis reflects the research question we aim to answer. The outcomes of our test will either provide evidence against the null hypothesis or fail to do so, which indirectly supports the null hypothesis.
Test Statistic
Moving to the core of hypothesis testing, we compute the test statistic, a pivotal value derived from our sample data. This number measures how far our sample statistic lies from the null hypothesis. We calculate it by first determining the difference in average costs between the two groups. With that figure in hand, we compute the standard error (SE) which accounts for variability in our data. With the formula {SE = sqrt{s^2_{split}/n_{split} + s^2_{individual}/n_{individual}}}, we attain the value encapsulating the uncertainty around the mean difference.
Our test statistic is then this difference in means divided by the SE. It's a way to standardize our results, enabling us to compare our findings against a distribution that represents our null hypothesis. With this comparison, we can determine the likelihood of observing a test statistic at least as extreme as ours if the null hypothesis were true.
P-value
Subsequently, our attention turns to the p-value. This is the probability of observing a test statistic at least as extreme as the one computed, assuming that the null hypothesis is true. It is not the probability that the null hypothesis is true; rather, it shows the compatibility of the observed data with the null hypothesis.
Procuring the p-value can be done through a statistical software or manual look-up in a t-distribution table, since our test statistic follows a t-distribution. If this p-value falls below our predetermined significance level (typically 0.05), it signals that the observed data is quite unlikely under the null hypothesis, leading us to consider the alternative hypothesis as more plausible.
Statistical Significance
Lastly, we assess the statistical significance based on the p-value and our preselected significance level, which is often 5% or 0.05. This threshold helps us make a decision: if the p-value is less than 0.05, the results are statistically significant. It means the evidence against the null hypothesis is strong enough to reject it in favor of the alternative hypothesis.
In our restaurant bill example, if the p-value is less than 0.05, we have enough grounds to assert, with 95% confidence, that the mean cost is indeed higher when people split the bill. This bridges our statistical findings with a real-world conclusion, thereby enhancing our understanding of the study's implications.

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Most popular questions from this chapter

In Exercises 6.32 and 6.33, find a \(95 \%\) confidence interval for the proportion two ways: using StatKey or other technology and percentiles from a bootstrap distribution, and using the normal distribution and the formula for standard error. Compare the results. Proportion of home team wins in soccer, using \(\hat{p}=0.583\) with \(n=120\)

Test \(H_{0}: \mu=15\) vs \(H_{a}: \mu>15\) using the sample results \(\bar{x}=17.2, s=6.4,\) with \(n=40\)

In Exercises 6.156 to 6.161: (a) Find the relevant sample proportions in each group and the pooled proportion. (b) Complete the hypothesis test using the normal distribution and show all details. Test whether there is a difference between two groups in the proportion who voted, if 45 out of a random sample of 70 in Group 1 voted and 56 out of a random sample of 100 in Group 2 voted.

In Exercises 6.9 and 6.10 , indicate whether the Central Limit Theorem applies so that the sample proportions follow a normal distribution. In each case below, is the sample size large enough so that the sample proportions follow a normal distribution? (a) \(n=500\) and \(p=0.1\) (b) \(n=25\) and \(p=0.5\) (c) \(n=30\) and \(p=0.2\) (d) \(n=100\) and \(p=0.92\)

Who Watches More TV: Males or Females? The dataset StudentSurvey has information from males and females on the number of hours spent watching television in a typical week. Computer output of descriptive statistics for the number of hours spent watching TV, broken down by gender, is given: \(\begin{array}{l}\text { Descriptive Statistics: TV } \\ \text { Variable } & \text { Gender } & \mathrm{N} & \text { Mean } & \text { StDev } \\ \text { TV } & \mathrm{F} & 169 & 5.237 & 4.100 \\ & \mathrm{M} & 192 & 7.620 & 6.427 \\\ \text { Minimum } & \mathrm{Q} 1 & \text { Median } & \mathrm{Q} 3 & \text { Maximum } \\ & 0.000 & 2.500 & 4.000 & 6.000 & 20.000 \\ & 0.000 & 3.000 & 5.000 & 10.000 & 40.000\end{array}\) (a) In the sample, which group watches more TV, on average? By how much? (b) Use the summary statistics to compute a \(99 \%\) confidence interval for the difference in mean number of hours spent watching TV. Be sure to define any parameters you are estimating. (c) Compare the answer from part (c) to the confidence interval given in the following computer output for the same data: \(\begin{array}{l}\text { Two-sample T for TV } \\ \text { Gender } & \text { N } & \text { Mean } & \text { StDev } & \text { SE Mean } \\ \mathrm{F} & 169 & 5.24 & 4.10 & 0.32 \\ \mathrm{M} & 192 & 7.62 & 6.43 & 0.46 \\ \text { Difference } & =\mathrm{mu}(\mathrm{F})-\mathrm{mu}(\mathrm{M}) & & \end{array}\) Estimate for for difference: -2.383 r difference: (-3.836,-0.930) \(99 \% \mathrm{Cl}\) for (d) Interpret the confidence interval in context.

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