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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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Most popular questions from this chapter

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 \(90 \%\) confidence interval for \(\mu_{1}-\mu_{2}\) using the sample results \(\bar{x}_{1}=10.1, s_{1}=2.3, n_{1}=50\) and \(\bar{x}_{2}=12.4, s_{2}=5.7, n_{2}=50 .\)

Examine the results of a study \(^{45}\) investigating whether fast food consumption increases one's concentration of phthalates, an ingredient in plastics that has been linked to multiple health problems including hormone disruption. The study included 8,877 people who recorded all the food they ate over a 24 -hour period and then provided a urine sample. Two specific phthalate byproducts were measured (in \(\mathrm{ng} / \mathrm{mL}\) ) in the urine: DEHP and DiNP. Find and interpret a \(95 \%\) confidence interval for the difference, \(\mu_{F}-\mu_{N},\) in mean concentration between people who have eaten fast food in the last 24 hours and those who haven't. The mean concentration of DiNP in the 3095 participants who had eaten fast food was \(\bar{x}_{F}=10.1\) with \(s_{F}=38.9\) while the mean for the 5782 participants who had not eaten fast food was \(\bar{x}_{N}=7.0\) with \(s_{N}=22.8\)

Is Gender Bias Influenced by Faculty Gender? Exercise 6.215 describes a study in which science faculty members are asked to recommend a salary for a lab manager applicant. All the faculty members received the same application, with half randomly given a male name and half randomly given a female name. In Exercise \(6.215,\) we see that the applications with female names received a significantly lower recommended salary. Does gender of the evaluator make a difference? In particular, considering only the 64 applications with female names, is the mean recommended salary different depending on the gender of the evaluating faculty member? The 32 male faculty gave a mean starting salary of \(\$ 27,111\) with a standard deviation of \(\$ 6948\) while the 32 female faculty gave a mean starting salary of \(\$ 25,000\) with a standard deviation of \(\$ 7966 .\) Show all details of the test.

Who Exercises More: Males or Females? The dataset StudentSurvey has information from males and females on the number of hours spent exercising in a typical week. Computer output of descriptive statistics for the number of hours spent exercising, broken down by gender, is given: \(\begin{array}{l}\text { Descriptive Statistics: Exercise } \\ \text { Variable } & \text { Gender } & \mathrm{N} & \text { Mean } & \text { StDev } \\\ \text { Exercise } & \mathrm{F} & 168 & 8.110 & 5.199 \\ & \mathrm{M} & 193 & 9.876 & 6.069\end{array}\) \(\begin{array}{rrrrr}\text { Minimum } & \text { Q1 } & \text { Median } & \text { Q3 } & \text { Maximum } \\ 0.000 & 4.000 & 7.000 & 12.000 & 27.000 \\\ 0.000 & 5.000 & 10.000 & 14.000 & 40.000\end{array}\) (a) How many females are in the dataset? How many males? (b) In the sample, which group exercises more, on average? By how much? (c) Use the summary statistics to compute a \(95 \%\) confidence interval for the difference in mean number of hours spent exercising. Be sure to define any parameters you are estimating. (d) Compare the answer from part (c) to the confidence interval given in the following computer output for the same data: Two-sample \(\mathrm{T}\) for Exercise Gender N Mean StDev SE Mean \(\begin{array}{lllll}\mathrm{F} & 168 & 8.11 & 5.20 & 0.40 \\ \mathrm{M} & 193 & 9.88 & 6.07 & 0.44\end{array}\) Difference \(=\operatorname{mu}(F)-\operatorname{mu}(M)\) Estimate for difference: -1.766 \(95 \%\) Cl for difference: (-2.932,-0.599)

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 women's tears on men, levels of testosterone are measured in 50 men after they sniff women's tears and after they sniff a salt solution. The order of the two treatments was randomized and the study was double-blind.

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