/*! 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 29 Which foams more when you pour i... [FREE SOLUTION] | 91Ó°ÊÓ

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Which foams more when you pour it, Coke or Pepsi? Here are measurements by Diane Warfield on the foam volume \((\mathrm{mL})\) after pouring a 12-oz can of Coke, based on a sample of 12 cans: \(\begin{array}{llllll}312.2 & 292.6 & 331.7 & 355.1 & 362.9 & 331.7 \\ 292.6 & 245.8 & 280.9 & 320.0 & 273.1 & 288.7\end{array}\) and here are measurements for Pepsi, based on a sample of 12 cans: \(\begin{array}{rrrrrr}148.3 & 210.7 & 152.2 & 117.1 & 89.7 & 140.5 \\ 128.8 & 167.8 & 156.1 & 136.6 & 124.9 & 136.6\end{array}\) a. Verify graphically that normality is an appropriate assumption. b. Calculate a \(99 \%\) confidence interval for the population difference in mean volumes. c. Does the upper limit of your interval in (b) give a \(99 \%\) lower confidence bound for the difference between the two \(\mu\) 's? If not, calculate such a bound and interpret it in terms of the relationship between the foam volumes of Coke and Pepsi. d. Summarize in a sentence what you have leamed about the foam volumes of Coke and Pepsi.

Short Answer

Expert verified
Coke foams significantly more than Pepsi, with the 99% confidence interval showing a positive difference.

Step by step solution

01

Graphical Verification of Normality

To verify normality, we can use a quantile-quantile (Q-Q) plot for each dataset. Plot the sorted values of each dataset against the quantiles of a standard normal distribution. If the points roughly form a straight line, the data can be considered to be normally distributed.
02

Calculate Mean and Standard Deviation

Calculate the sample mean and standard deviation for both Coke and Pepsi datasets. For Coke:\[\bar{x}_{\text{Coke}} = \frac{\sum x_i}{n} = \frac{312.2 + 292.6 + \ldots + 288.7}{12}\]Calculate the standard deviation similarly. Repeat for Pepsi.
03

Compute the 99% Confidence Interval for the Mean Difference

Use the formula for the confidence interval of the difference between two means: \[\bar{x}_1 - \bar{x}_2 \pm t_{\alpha/2, df} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}\]where \(t_{\alpha/2, df}\) is the t-value for a 99% confidence interval and DF is calculated using the formula for unequal variances. Plug in the calculated means, standard deviations, and sample sizes.
04

Interpretation of the Confidence Interval

Check if 0 is within the computed confidence interval. If not, it indicates a significant difference in mean foam volumes. Interpret the interval for practical significance.
05

Determine a 99% Lower Confidence Bound

The upper limit of the confidence interval in Step 3 is not necessarily a lower bound. If it's not, use: \[\bar{x}_1 - \bar{x}_2 - t_{\alpha, df} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]This bound gives a better gauge of how smaller the mean Coke foaming volume is compared to Pepsi.
06

Summarize the Findings

Summarize the results: Coke foams more than Pepsi, evidenced by higher mean foam volumes and a significant difference reflected in the confidence interval.

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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 provides a range of values that is likely to contain the population parameter, such as the mean difference between two groups, with a certain level of confidence, typically 95% or 99%. In this exercise, we are working with foam volumes of Coke and Pepsi, and our goal is to determine a 99% confidence interval for the difference in their mean foam volumes.

To do this, we first calculate the sample mean and standard deviation for both Coke and Pepsi. These will give us an idea of the typical volume of foam for each drink. Next, we use the formula for the confidence interval of the difference in means. The formula, \[\bar{x}_1 - \bar{x}_2 \pm t_{\alpha/2, df} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]indicates that we need the t-value for a 99% confidence interval, along with the degrees of freedom calculated for unequal variances.

If the confidence interval does not include zero, it suggests there is a significant difference between the two means. This information helps us to conclude whether Coke generally foams more than Pepsi with 99% confidence.
Normality Verification
Normality verification is an important step in hypothesis testing because many statistical tests, including the t-test used here, assume normal distribution of data. To verify normality, we use a tool called the Quantile-Quantile (Q-Q) plot.

This plot involves graphing the sorted data values of the foam volumes against the quantiles of a standard normal distribution. If the data points on the Q-Q plot fall along a straight line, it suggests that the data is approximately normally distributed.

In our analysis of Coke and Pepsi foam volumes, checking for normality allows us to justify the use of the t-test for the confidence interval calculation. If either dataset shows significant deviation from normality, we might need to consider alternative statistical techniques that do not assume normal distribution.
Mean and Standard Deviation
The mean and standard deviation are two fundamental statistical measures used to describe the characteristics of a dataset.

The mean, or average, is calculated by adding all the data points together and then dividing by the number of points. It provides an estimate of the central tendency of the data. For Coke and Pepsi, the individual mean foam volumes give us an idea of how much each drink foams on average.

The standard deviation measures the spread or variability within the data. Calculating this involves determining how much each data point deviates from the mean. It tells us how consistent the foam measurements are across different cans. A smaller standard deviation means the values are closely grouped around the mean, while a larger one indicates more variability.

These two values are crucial for calculating confidence intervals and for understanding the underlying distribution and consistency of the foam volumes produced by Coke and Pepsi.
Quantile-Quantile Plot
A Quantile-Quantile (Q-Q) plot is a graphical tool used to assess if a dataset follows a specific distribution, most commonly the normal distribution.

To create a Q-Q plot, you first sort the data and then plot the ordered values against theoretical quantiles from a normal distribution. If the data follows a normal distribution, then the plotted points should lie approximately on a straight diagonal line.

In our exercise examining foam volumes of Coke and Pepsi, Q-Q plots help visually verify that the assumption of data normality holds. This step is essential because many statistical methods rely on this assumption. If the Q-Q plot shows significant deviation from the line, alternative analytical techniques may be necessary.

Q-Q plots are a powerful tool for quickly assessing distributional characteristics, helping ensure the selected statistical analyses are appropriate for the data at hand.

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Most popular questions from this chapter

The article "Supervised Exercise Versus NonSupervised Exercise for Reducing Weight in Obese Adults" ( . Sport. Med. Phys. Fit., 2009: 85-90) reported on an investigation in which participants were randomly assigned either to a supervised exercise program or a control group. Those in the control group were told only that they should take measures to lose weight. After 4 months, the sample mean decrease in body fat for the 17 individuals in the experimental group was \(6.2 \mathrm{~kg}\) with a sample standard deviation of \(4.5 \mathrm{~kg}\), whereas the sample mean and sample standard deviation for the 17 people in the control group were \(1.7 \mathrm{~kg}\) and \(3.1 \mathrm{~kg}\), respectively. Assume normality of the two body fat loss distributions (as did the investigators). a. Calculate a \(99 \%\) lower prediction bound for the body fat loss of a single randomly selected individual subjected to the supervised exercise program. Can you be highly confident that such an individual will actually lose body fat? b. Does it appear that true average decrease in body fat is more than \(2 \mathrm{~kg}\) larger for the experimental condition than for the control condition? Carry out a test of appropriate hypotheses using a significance level of \(.01\)

How does energy intake compare to energy expenditure? One aspect of this issue was considered in the article "Measurement of Total Energy Expenditure by the Doubly Labelled Water Method in Professional Soccer Players" (J.Sports Sci., 2002: 391-397), which contained the accompanying data (MJ/day). $$ \begin{array}{lccccccc} \text { Player } & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \text { Expenditure } & 14.4 & 12.1 & 14.3 & 14.2 & 15.2 & 15.5 & 17.8 \\ \text { Intake } & 14.6 & 9.2 & 11.8 & 11.6 & 12.7 & 15.0 & 16.3 \end{array} $$ Test to see whether there is a significant difference between intake and expenditure. Does the conclusion depend on whether a significance level of \(.05, .01\), or \(.001\) is used?

An experiment to compare the tension bond strength of polymer latex modified mortar (Portland cement mortar to which polymer latex emulsions have been added during mixing) to that of unmodified mortar resulted in \(\bar{x}=18.12 \mathrm{kgf} / \mathrm{cm}^{2}\) for the modified mortar \((m=40)\) and \(\bar{y}=16.87 \mathrm{kgf} / \mathrm{cm}^{2}\) for the unmodified mortar ( \(n=32\) ). Let \(\mu_{1}\) and \(\mu_{2}\) be the true average tension bond strengths for the modified and unmodified mortars, respectively. Assume that the bond strength distributions are both normal. a. Assuming that \(\sigma_{1}=1.6\) and \(\sigma_{2}=1.4\), test \(H_{0}\) : \(\mu_{1}-\mu_{2}=0\) versus \(H_{\mathrm{a}}: \mu_{1}-\mu_{2}>0\) at level \(.01\). b. Compute the probability of a type II error for the test of part (a) when \(\mu_{1}-\mu_{2}=1\). c. Suppose the investigator decided to use a level \(.05\) test and wished \(\beta=.10\) when \(\mu_{1}-\mu_{2}=1\). If \(m=40\), what value of \(n\) is necessary? d. How would the analysis and conclusion of part (a) change if \(\sigma_{1}\) and \(\sigma_{2}\) were unknown but \(s_{1}=1.6\) and \(s_{2}=1.4\) ?

A student project by Heather Kral studied students on "lifestyle floors" of a dormitory in comparison to students on other floors. On a lifestyle floor the students share a common major, and there are a faculty coordinator and resident assistant from that department. Here are the grade point averages of 30 students on lifestyle floors (L) and 30 students on other floors \((\mathrm{N})\) : L: \(2.00,2.25,2.60,2.90,3.00,3.00,3.00,3.00\), \(3.00,3.20,3.20,3.25,3.30,3.30,3.32,3.50\), \(3.50,3.60,3.60,3.70,3.75,3.75,3.79,3.80\), \(3.80,3.90,4.00,4.00,4.00,4.00\). \(\mathrm{N}: 1.20,2.00,2.29,2.45,2.50,2.50,2.50,2.50\), \(2.65,2.70,2.75,2.75,2.79,2.80,2.80,2.80\), \(2.86,2.90,3.00,3.07,3.10,3.25,3.50,3.54\), \(3.56,3.60,3.70,3.75,3.80,4.00\). Notice that the lifestyle grade point averages have a large number of repeats and the distribution is skewed, so there is some question about normality. a. Obtain a \(95 \%\) confidence interval for the difference of population means using the method based on the theorem of Section 10.2. b. Obtain a bootstrap sample of 999 differences of means. Check the bootstrap distribution for normality using a normal probability plot. c. Use the standard deviation of the bootstrap distribution along with the mean and \(t\) critical value from (a) to get a \(95 \%\) confidence interval for the difference of means. d. Use the bootstrap sample and the percentile method to obtain a \(95 \%\) confidence interval for the difference of means. e. Compare your three confidence intervals. If they are very similar, why do you think this is the case? f. Interpret your results. Is there a substantial difference between lifestyle and other floors? Why do you think the difference is as big as it is?

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