91Ó°ÊÓ

Aluminum Bottles The aluminum bottle, first introduced in 1991 by CCL Container for mainly personal and household items such as lotions, has become popular with beverage manufacturers. Besides being lightweight and requiring less packaging, the aluminum bottle is reported to cool faster and stay cold longer than typical glass bottles. A small brewery tests this claim and obtains the following information regarding the time (in minutes) required to chill a bottle of beer from room temperature \(\left(75^{\circ} \mathrm{F}\right)\) to serving temperature \(\left(45^{\circ} \mathrm{F}\right) .\) Construct and interpret a \(90 \%\) confidence interval for the mean difference in cooling time for clear glass versus aluminum. $$ \begin{array}{lcc} & \text { Clear Glass } & \text { Aluminum } \\ \hline \text { Sample size } & 42 & 35 \\ \hline \begin{array}{l} \text { Mean time to } \\ \text { chill (minutes) } \end{array} & 133.8 & 92.4 \\ \hline \begin{array}{l} \text { Sample standard } \\ \text { deviation (minutes) } \end{array} & 9.9 & 7.3 \\ \hline \end{array} $$

Step by step solution

01

Define the Variables

Let \( \bar{X}_1 \) and \( \bar{X}_2 \) be the mean cooling times for clear glass and aluminum respectively. We know \( \bar{X}_1 = 133.8 \) minutes, \( \bar{X}_2 = 92.4 \) minutes, \( s_1 = 9.9 \) minutes, \( s_2 = 7.3 \) minutes, \( n_1 = 42 \), and \( n_2 = 35 \).

02

State the Formula for Confidence Interval

The formula for the confidence interval for the difference between two means is \[ (\bar{X}_1 - \bar{X}_2) \pm t \times \text{SE} \]. Where \( \text{SE} = \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \).

03

Calculate the Standard Error (SE)

Calculate the standard error using the given sample standard deviations and sample sizes. \[ \text{SE} = \sqrt{\frac{9.9^2}{42} + \frac{7.3^2}{35}} \approx 1.850 \].

04

Determine the t-Score for a 90% Confidence Interval

For a 90% confidence interval with degrees of freedom approximately equal to the smaller of \( n_1-1 \) and \( n_2-1 \), the value of \( t \) is about 1.68.

05

Compute the Confidence Interval

Now, compute the confidence interval using \[ (\bar{X}_1 - \bar{X}_2) \pm t \times \text{SE} \] \[ = (133.8 - 92.4) \pm 1.68 \times 1.850 \] \[ = 41.4 \pm 3.108 \] \[ = (38.292, 44.508) \].

06

Interpret the Confidence Interval

The 90% confidence interval for the mean difference in cooling time is (38.292, 44.508) minutes. This means we are 90% confident that the mean difference in chilling time between clear glass and aluminum bottles lies between approximately 38.3 and 44.5 minutes.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

cooling time comparison

In this exercise, we will compare the cooling times of clear glass bottles and aluminum bottles. Cooling time refers to the duration it takes for a bottle to chill a beer from room temperature \(75^{\circ} \mathrm{F}\) to serving temperature \(45^{\circ} \mathrm{F}\).
The data collected includes the sample size, mean cooling times, and the standard deviations for each type of bottle.
The clear glass bottles have a mean cooling time of 133.8 minutes with a standard deviation of 9.9 minutes, while the aluminum bottles have a mean cooling time of 92.4 minutes with a standard deviation of 7.3 minutes.

By comparing these two mean cooling times, we aim to construct a 90% confidence interval for the mean difference. This will help us understand if aluminum bottles indeed cool faster than glass bottles, as claimed.

This comparison is useful for manufacturers and consumers alike, as it provides insight into the efficiency of different bottle materials in chilling beverages.

standard error calculation

To construct a confidence interval for the mean difference in cooling times, we first need to calculate the Standard Error (SE). The SE measures the amount of variability in the sampling distribution of the difference between two means.

The formula for calculating the SE for two independent samples is given by:
\[ \text{SE} = \sqrt{\frac{\text{s}_1^2}{\text{n}_1} + \frac{\text{s}_2^2}{\text{n}_2}} \]
where \text{s}_1 and \text{s}_2 are the sample standard deviations, and \text{n}_1 and \text{n}_2 are the sample sizes.

Plugging in the values from our data:
\[ \text{SE} = \sqrt{\frac{9.9^2}{42} + \frac{7.3^2}{35}} \approx 1.850 \]
The calculated SE comes out to approximately 1.850 minutes.

This value of SE will be used in calculating the confidence interval, as it indicates how much the difference between sample means might vary due to random sampling error.

t-score for confidence interval

Once we have calculated the Standard Error (SE), the next step is to determine the critical t-score for a 90% confidence interval. The t-score corresponds to how far, in terms of standard errors, we would need to go to achieve the desired level of confidence.

For our case with degrees of freedom approximately equal to the smaller of \(\text{n}_1 - 1 \) and \(\text{n}_2 - 1\), we find a t-score for a 90% confidence interval to be approximately 1.68.

The confidence interval formula is:
\[ (\bar{X}_1 - \bar{X}_2) \pm t \times \text{SE} \]
Using our data, this becomes:
\[ (133.8 - 92.4) \pm 1.68 \times 1.850 \]
Which simplifies to:
\[ 41.4 \pm 3.108 \]
Thus, the confidence interval is:
\[ (38.292, 44.508) \]
This interval means we are 90% confident that the true mean difference in chilling time between clear glass and aluminum bottles lies between approximately 38.3 and 44.5 minutes.
Understanding this confidence interval helps in making informed decisions about the effectiveness of aluminum bottles compared to glass bottles.