91Ó°ÊÓ

Assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9 - 1 along with "Table" answers based on Table \(A\) - 3 with df equal to the smaller of \(\boldsymbol{n}_{I}-\boldsymbol{I}\) and \(\boldsymbol{n}_{2}-\boldsymbol{I} .\) ) Regular Coke and Diet Coke Data Set 26 "Cola Weights and Volumes" in Appendix B includes weights (b) of the contents of cans of Diet Coke \((n=36, \bar{x}=0.78479 \text { lb, } s=0.00439\) lb) and of the contents of cans of regular Coke \((n=36, \bar{x}=0.81682 \mathrm{lb}, s=0.00751 \mathrm{lb})\) a. Use a 0.05 significance level to test the claim that the contents of cans of Diet Coke have weights with a mean that is less than the mean for regular Coke. b. Construct the confidence interval appropriate for the hypothesis test in part (a). c. Can you explain why cans of Diet Coke would weigh less than cans of regular Coke?

Step by step solution

01

State the Hypotheses

Identify the null hypothesis (\text{H}_0) and the alternative hypothesis (\text{H}_a).For part (a):\( \text{H}_0: \text{μ}_1 = \text{μ}_2 \) (The mean weight of Diet Coke is equal to the mean weight of regular Coke)\( \text{H}_a: \text{μ}_1 < \text{μ}_2 \) (The mean weight of Diet Coke is less than the mean weight of regular Coke)

02

Calculate the Test Statistic

Use the formula for the t-test statistic for two independent samples, assuming unequal variances:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]Here, \( \bar{x}_1 = 0.78479 \) lb, \( s_1 = 0.00439 \) lb, \( n_1 = 36 \) for Diet Coke, and \( \bar{x}_2 = 0.81682 \) lb, \( s_2 = 0.00751 \) lb, \( n_2 = 36 \) for regular Coke.Calculate the test statistic:

03

Calculate the Degrees of Freedom

The degrees of freedom (df) for the t-test with unequal variances can be calculated using the following approximation formula by Satterthwaite’s approximation:\[ df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{( \frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1}} \]Substitute the values and calculate df.

04

Determine the Critical Value and Make a Decision

Using a t-distribution table or technology, find the critical t-value for a one-tailed test at the 0.05 significance level with the calculated degrees of freedom. Compare the test statistic with the critical t-value:- If the test statistic is less than the critical t-value, reject the null hypothesis.- Otherwise, do not reject the null hypothesis.

05

Construct the Confidence Interval

For part (b), construct a (1-\the confidence level)100% confidence interval for the difference between two population means.\[ (\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2, df} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \]Find the appropriate critical value (t_{\alpha/2, df}) from the t-distribution table for a 95% confidence interval. Plug in the values to get the confidence interval.

06

Explain the Reasoning

For part (c), Diet Coke contains artificial sweeteners which generally weigh less than the high fructose corn syrup found in regular Coke. This could result in the overall weight of cans of Diet Coke being less than that of regular Coke.

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Key Concepts

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

independent samples

When conducting a t-test, one key aspect involves independent samples. Independent samples are data sets that do not influence each other. For example, when comparing the weights of Diet Coke and regular Coke, one sample set does not affect the other. This independence ensures that there are no biases or confounding factors that could skew the results.
Understanding the independence of samples is crucial because it justifies the use of certain statistical methods, like the t-test.

• Each sample must be randomly selected from its population.
• No individual data point in one sample should influence any point in the other sample.

t-test

A t-test is a statistical test used to compare the means of two groups. The independent samples t-test is used when each sample is drawn independently from a different population.
For the problem at hand, the t-test helps determine if there is a significant difference between the mean weights of Diet Coke and regular Coke. The formula for the t-test is: \[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] Where:

• \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means for Diet Coke and regular Coke, respectively.
• \(s_1\) and \(s_2\) are the sample standard deviations.
• \(n_1\) and \(n_2\) are the sample sizes.

This computation allows researchers to test hypotheses about the equality of the population means. The resultant t-value indicates whether the difference in sample means is statistically significant.

confidence interval

A confidence interval gives a range of values for an unknown population parameter based on sample data. After performing a t-test, constructing a confidence interval provides additional insight.
For instance, a 95% confidence interval for the difference in means between Diet Coke and regular Coke is computed using the formula: \[ (\bar{x}_1 - \bar{x}_2) \pm t_{\alpha/2, df} \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \] Where:

• \(t_{\alpha/2, df}\) is the critical value from the t-distribution, based on the desired confidence level and degrees of freedom.
• The term \((\bar{x}_1 - \bar{x}_2)\) represents the observed difference in sample means.
• \(\alpha\) is the level of significance (e.g., 0.05 for a 95% confidence interval).

The interval provides a range where the true mean difference likely lies, giving an added layer of interpretation beyond hypothesis testing.

degrees of freedom

Degrees of freedom (df) are a critical component in determining the precise shape of the t-distribution. In the context of a t-test with unequal variances, degrees of freedom can be calculated using Satterthwaite’s approximation:
\[ df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{( \frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1}} \]Calculating degrees of freedom is essential because it affects the critical t-value used for hypothesis tests and confidence intervals. A higher df indicates a more accurate estimate because it reflects more information from the data.

• Degrees of freedom help adjust for the sample size and variability in the data.
• They play a pivotal role in identifying the correct critical value from the t-distribution table.

significance level

The significance level (denoted as \(\alpha\)) is the threshold for deciding whether a null hypothesis should be rejected. In most tests, a common significance level is 0.05, meaning there's a 5% risk of concluding that a difference exists when in fact, it doesn't.
For example, in testing whether the mean weight of Diet Coke is less than regular Coke, a significance level of 0.05 might be chosen. This value sets the stage for comparing the test statistic to the critical value:

• If the test statistic falls within the critical region defined by \(\alpha\), the null hypothesis is rejected.
• If not, there's insufficient evidence to reject the null hypothesis.

Establishing a significance level allows for standardized decision-making in hypothesis testing and helps control the rate of Type I errors (false positives).