91Ó°ÊÓ

As was mentioned in Exercise \(9.38\), The Bath Heritage Days, which take place in Bath, Maine, switched to a Whoopie Pie eating contest in 2009 . Suppose the contest involves eating nine Whoopie Pies, each weighing \(1 / 3\) pound. The following data represent the times (in seconds) taken by each of the 13 contestants (all of whom finished all nine Whoopie Pies) to eat the first Whoopie Pie and the last (ninth) Whoopie Pie. $$ \begin{array}{l|rrrrrrrrrrrr} \hline \text { Contestant } & \mathbf{1} & \mathbf{2} & \mathbf{3} & \mathbf{4} & \mathbf{5} & \mathbf{6} & \mathbf{7} & \mathbf{8} & \mathbf{9} & \mathbf{1 0} & \mathbf{1 1} & \mathbf{1 2} & \mathbf{1 3} \\ \hline \text { First pie } & 49 & 59 & 66 & 49 & 63 & 70 & 77 & 59 & 64 & 69 & 60 & 58 & 71 \\ \hline \text { Last pie } & 49 & 74 & 92 & 93 & 91 & 73 & 103 & 59 & 85 & 94 & 84 & 87 & 111 \\ \hline \end{array} $$ 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 eat the first pie. b. Using the \(10 \%\) 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.

Step by step solution

01

Calculate the paired differences

To start, subtract the time taken to eat the first pie from the time taken to eat the last pie. Do this for each contestant. The resulting differences are your 'paired differences.'

02

Find the mean difference

Calculate the mean of these paired differences. The mean is the sum of the differences divided by the number of differences.

03

Calculate the standard deviation

Use the paired differences obtained in Step 1 to calculate the standard deviation. The standard deviation measures the average distance between each data point and the mean.

04

Create the 95% confidence interval

Using the mean difference and standard deviation, construct the 95% confidence interval. The formula for this is: \(\overline{x}\) ± \(t_{n-1, \frac{α}{2}}\) * \(\frac{s}{\sqrt{n}}\) where \(\overline{x}\) is the sample mean, \(t_{n-1, \frac{α}{2}}\) is the t-value from the t-distribution table (with degree of freedom n-1 and \(\frac{α}{2}\)), and s is the standard deviation, and n is the number of pairs.

05

Hypothesis Testing

Next, perform a hypothesis test to determine if the average time taken to eat the ninth pie is at least 15 seconds more than that of the first pie. Have a null hypothesis (H_0: \( \mu_D = 15 \)) stating the mean difference is 15 seconds and an alternate hypothesis (H_1: \( \mu_D > 15 \)) stating the mean difference is greater than 15 seconds. If the t-value is greater than the critical value from the t-distribution table, reject the null hypothesis.

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Key Concepts

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

Confidence Interval

A confidence interval is a range of values that is used to estimate an unknown population parameter. In the context of this problem, it helps us understand the variation in time differences when contestants eat the ninth Whoopie Pie compared to the first. This interval gives us a range in which we expect the true mean paired difference to lie with a certain level of confidence, typically 95%.

To calculate this interval, you'll first need the mean difference and the standard deviation of the paired differences. The formula for constructing a confidence interval is:

• \( \overline{x} \pm t_{n-1, \frac{\alpha}{2}} \cdot \frac{s}{\sqrt{n}} \)

Where:

• \( \overline{x} \) is the sample mean of the differences
• \( t_{n-1, \frac{\alpha}{2}} \) is the t-value from the t-distribution table
• \( s \) is the standard deviation
• \( n \) is the number of paired observations

This approach helps us to determine not just an average but a reliable range reflecting real-world variability.

Hypothesis Testing

Hypothesis testing is a statistical method used to make decisions about the data population. In this problem, we want to test if eating the ninth pie takes significantly longer than the first by at least 15 seconds. This involves setting up two hypotheses:

• Null Hypothesis \( (H_0): \mu_D = 15 \)
• Alternative Hypothesis \( (H_1): \mu_D > 15 \)

The null hypothesis suggests that there is no significant difference, while the alternative suggests there is.

We perform the test by calculating a test statistic and comparing it to a critical value from the t-distribution. If the calculated t-value exceeds the critical value, we reject the null hypothesis. This rejection means that our data supports the claim that the ninth pie takes significantly longer to eat by more than 15 seconds.

This method helps us quantify our levels of confidence and uncertainty about our conclusions.

Mean Difference

The mean difference is the average of all paired differences between participants' times. It tells us the central tendency of the differences in time taken for eating the ninth pie compared to the first.

To calculate this, sum up all the paired differences and divide them by the number of pairs:

• Mean Difference \( \overline{D} = \frac{\sum D_i}{n} \)

Where \( D_i \) represents each individual difference and \( n \) is the number of participants.

In this exercise, the mean difference helps to answer whether, on average, contestants take longer to eat the ninth pie. This value is critical for both forming the confidence interval and performing hypothesis testing. By understanding the mean difference, we get a clearer view of the overall shift in participants' eating performance.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. For this exercise, it shows how spread out the contestants' time differences are from the mean difference.

To calculate standard deviation for the paired differences, use:

• \( s = \sqrt{\frac{\sum (D_i - \overline{D})^2}{n-1}} \)

Where \( D_i \) are the individual paired differences, \( \overline{D} \) is the mean of those differences, and \( n \) is the number of pairs.

A small standard deviation indicates that the time differences are close to the mean, while a larger standard deviation means they are more spread out. Understanding this helps us in interpreting the confidence interval and hypothesis test results, as high variability can influence these outcomes by making it harder to detect true differences.