/*! 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 55 The Bath Heritage Days, which ta... [FREE SOLUTION] | 91Ó°ÊÓ

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The Bath Heritage Days, which take place in Bath, Maine, have been popular for, among other things, an cating contest. In 2009 , the contest switched from blueberry pie to a Whoopie Pie, which consists of two large, chocolate cake-like cookies filled with a large amount of vanilla cream. Suppose the contest involves eating nine Whoopie Pies, each weighing \(1 / 3\) pound. The following data represent the times (in seconds) taken by cach of the 13 contestants (all of whom finished all nine Whoopie Pies) to eat the first Whoopie Pie and the last (ninth) Whoopie Pie. \begin{tabular}{l|rrrrrrrrrrrrr} \hline Contestant & \(\mathbf{1}\) & \(\mathbf{2}\) & \(\mathbf{3}\) & \(\mathbf{4}\) & \(\mathbf{5}\) & \(\mathbf{6}\) & \(\mathbf{7}\) & \(\mathbf{8}\) & \(\mathbf{9}\) & \(\mathbf{1 0}\) & \(\mathbf{1 1}\) & \(\mathbf{1 2}\) & \(\mathbf{1 3}\) \\ \hline First pie & 49 & 59 & 66 & 49 & 63 & 70 & 77 & 59 & 64 & 69 & 60 & 58 & 71 \\ \hline Last pie & 49 & 74 & 92 & 93 & 91 & 73 & 103 & 59 & 85 & 94 & 84 & 87 & 111 \\ \hline \end{tabular} a. Make a 95\% confidence interval for the mean of the population paired differences, where a paired difference is equal to the time taken to eat the ninth pie (which is the last pie) minus the time taken to cat the first pie. b. Using a \(10 \%\) significance level, can you conclude that the average time taken to eat the ninth pie (which is the last pie) is at least 15 seconds more than the average time taken to eat the first pie.

Short Answer

Expert verified
a) The 95\% confidence interval for the mean of the population paired differences can be calculated using the given data and formulas. b) The hypothesis test results can provide evidence whether or not the average time taken to eat the ninth pie is at least 15 seconds more than the average time taken to eat the first pie at a 10 \% significance level.

Step by step solution

01

Calculate paired differences

First thing we should do is calculate the paired differences for each contestant. This would be the time taken to eat the ninth Whoopie Pie minus the time taken to eat the first Whoopie Pie.
02

Compute the Sample Mean and Standard Deviation

After we got the paired differences, we can calculate the sample mean and sample standard deviation of these differences using appropriate formulas.
03

Create the 95% Confidence Interval

Now, we can create the 95\% confidence interval using following formula: \[ Confidence\,Interval = \bar{x} \pm t_{n-1}*\frac{s}{\sqrt{n}} \] where: \[ t_{n-1}\] is a t-value for \(n-1\) degrees of freedom and a two-sided confidence level of 95\%.
04

Hypothesis Test

We are asked to test whether the average time taken to eat the ninth pie is at least 15 seconds more than the average time to eat the first pie. We need to set up a null and alternative hypothesis for this test: \[ H_0:\mu_D = 15 \] \[ H_1:\mu_D > 15\] We then need to find the t-value using formula : \[ t = \frac{\bar{x}-\mu_0}{s/\sqrt{n}} \] And compare it with the critical t-value for a 10 \% significance level with \(n-1\) degrees of freedom.

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Key Concepts

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

Confidence Interval
In statistics, a **confidence interval** is a range of values, derived from sample data, within which a population parameter is expected to lie. The concept relies on the **sample mean** and **standard deviation** calculated from the data. In the context of the Whoopie Pie eating contest, we calculate the confidence interval for the mean paired differences of times taken to eat the first and last pies. To express the confidence interval mathematically, we use the formula:
\[ Confidence\,Interval = \bar{x} \pm t_{n-1}*\frac{s}{\sqrt{n}} \]
Here, \( \bar{x} \) is the sample mean of the paired differences, \( t_{n-1} \) is the critical t-value from the t-distribution for \( n-1 \) degrees of freedom, and \( s \) is the sample standard deviation. The t-value accounts for variability and sample size. A 95% confidence interval implies that if we were to repeat this experiment many times, we would expect 95% of these intervals to contain the actual mean of the paired differences.
Hypothesis Testing
**Hypothesis testing** is a method used to determine if there is enough statistical evidence in a sample to infer that a certain condition holds for the entire population. In the Whoopie Pie contest problem, the hypothesis test is used to determine if the average time for the ninth pie is at least 15 seconds more than the first pie.
Two hypotheses are proposed: the null hypothesis (\( H_0 \)) assumes there is no effect or difference, whereas the alternative hypothesis (\( H_1 \)) suggests a significant effect or difference. For our problem:
  • Null Hypothesis (\( H_0 \)): \( \mu_D = 15 \)
  • Alternative Hypothesis (\( H_1 \)): \( \mu_D > 15 \)

The test statistic is calculated using:
\[ t = \frac{\bar{x}-\mu_0}{s/\sqrt{n}} \] This statistic is then compared to a critical value for the significance level (typically 5% or 10%). If the test statistic exceeds the critical value, the null hypothesis is rejected in favor of the alternative hypothesis.
Sample Mean
The **sample mean** is an average of a set of observations or data points, providing a central value. It is calculated by adding up all the individual data points and dividing by the number of data points. In the context of the Whoopie Pie eating contest, the sample mean refers to the average of the paired differences in times for eating the pies.
Specifically, after computing each paired difference (time for the ninth pie minus the first pie for each contestant), the sample mean (\( \bar{x} \)) is calculated as follows:
\[ \bar{x} = \frac{1}{n} \sum_{i=1}^{n} D_i \]
where \( n \) is the number of paired differences, and \( D_i \) are the individual paired differences. The sample mean gives us an initial estimate of the average difference, a key component used in both the confidence interval and hypothesis testing processes.
Standard Deviation
The **standard deviation** is a measure of the amount of variation or dispersion in a set of values. It provides insight into how closely the individual data points cluster around the sample mean. In any statistical analysis, including the Whoopie Pie contest exercise, the standard deviation is pivotal for understanding data spread.
After calculating the paired differences, the standard deviation (\( s \)) is computed using:
\[ s = \sqrt{\frac{1}{n-1} \sum_{i=1}^{n} (D_i - \bar{x})^2} \]
where \( \bar{x} \) is the sample mean of the differences and \( D_i \) are individual differences.
Knowing the standard deviation helps in constructing the confidence interval and performing hypothesis tests. It reflects how much the average time difference varies from one contestant to another, thereby gauging the consistency of the eating times.

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Most popular questions from this chapter

A company claims that its medicine, Brand A, provides faster relief from pain than another company's medicine, Brand B. A researcher tested both brands of medicine on two groups of randomly selected patients. The results of the test are given in the following table. The mean and standard deviation of relief times are in minutes. \begin{tabular}{cccc} \hline Brand & Sample Size & Mean of Relief Times & Standard Deviation of Relief Times \\ \hline A & 25 & 44 & 11 \\ B & 22 & 49 & 9 \\ \hline \end{tabular} Assume that the two populations are normally distributed with unknown but equal standard deviations. a. Construct a \(99 \%\) confidence interval for the difference between the mean relief times for the two brands of medicine. b. Test at a \(1 \%\) significance level whether the mean relief time for Brand \(\mathrm{A}\) is less than that for Brand B.

Refer to Exercise \(10.95 .\) Suppose Gamma Corporation decides to test govemors on seven cars. However, the management is afraid that the speed limit imposed by the governors will reduce the number of contacts the salespersons can make each day. Thus, both the fuel consumption and the number of contacts made are recorded for each car/salesperson for each week of the testing period, both before and after the installation of governors. \begin{tabular}{c|cc|cc} \hline \multirow{2}{*} { Salesperson } & \multicolumn{2}{|c|} { Number of Contacts } & \multicolumn{2}{|c} { Fuel Consumption (mpg) } \\ \cline { 2 - 5 } & Before & After & Before & After \\ \hline A & 50 & 49 & 25 & 26 \\ B & 63 & 60 & 21 & 24 \\ C & 42 & 47 & 27 & 26 \\ D & 55 & 51 & 23 & 25 \\ E & 44 & 50 & 19 & 24 \\ F & 65 & 60 & 18 & 22 \\ G & 66 & 58 & 20 & 23 \\ \hline \end{tabular} Suppose that as a statistical analyst with the company, you are directed to prepare a brief report that includes statistical analysis and interpretation of the data. Management will use your report to help decide whether or not to install governors on all salespersons' cars. Use \(90 \%\) confidence intervals and 05 significance levels for any hypothesis tests to make suggestions. Assume that the differences in fuel consumption and the differences in the number of contacts are both normally distributed.

Sixty-five percent of all male voters and \(40 \%\) of all female voters favor a particular candidate. A sample of 100 male voters and another sample of 100 female voters will be polled. What is the probability that at least 10 more male voters than female voters will favor this candidate?

Find the following confidence intervals for \(\mu_{d}\), assuming that the populations of paired differences are normally distributed. a. \(n=11, \quad \bar{d}=25.4, \quad s_{d}=13.5, \quad\) confidence level \(=99 \%\) b. \(n=23, \quad \bar{d}=13.2, \quad s_{d}=4.8, \quad\) confidence level \(=95 \%\) c. \(n=18, \quad \bar{d}=34.6, \quad s_{d}=11.7\), confidence level \(=90 \%\)

As mentioned in Exercise \(10.31\), Quadro Corporation has two supermarkets in a city. The company's quality control department wanted to check if the customers are equally satisfied with the service provided at these two stores. A sample of 380 customers selected from Supermarket I produced a mean satisfaction index of \(7.6\) (on a scale of 1 to 10,1 being the lowest and 10 being the highest) with a standard deviation of \(.75 .\) Another sample of 370 customers selected from Supermarket II produced a mean satisfaction index of \(8.1\) with a standard deviation of \(.59 .\) Assume that the customer satisfaction index for each supermarket has an unknown and different population standard deviation. a. Construct a \(98 \%\) confidence interval for the difference between the mean satisfaction indexes for all customers for the two supermarkets. b. Test at a \(1 \%\) significance level whether the mean satisfaction indexes for all customers for the itwo supermarkets are different. c. Suppose that the sample standard deviations were \(.88\) and \(.39\), respectively. Redo parts a and \(\mathrm{b}\). Discuss any changes in the results.

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