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 \(\boldsymbol{n}_{2}-\boldsymbol{I} .\) ) Are Quarters Now Lighter? Weights of quarters are carefully considered in the design of the vending machines that we have all come to know and love. Data Set 29 "Coin Weights" in Appendix B includes weights of a sample of pre-1964 quarters \((n=40, \bar{x}=6.19267 \mathrm{g}\) \(s=0.08700 \mathrm{g}\) ) and weights of a sample of post- 1964 quarters \((n=40, \bar{x}=5.63930 \mathrm{g}\) \(s=0.06194 \mathrm{g}\) a. Use a 0.05 significance level to test the claim that pre-1964 quarters have a mean weight that is greater than the mean weight of post-1964 quarters. b. Construct a confidence interval appropriate for the hypothesis test in part (a). c. Do post-1964 quarters appear to weigh less than before \(1964 ?\) If so, why aren't vending machines affected very much by the difference?

Step by step solution

01

State the Hypothesis

Identify the null and alternative hypotheses. The null hypothesis (\text{H}_0) is that there is no difference in the mean weights of pre-1964 and post-1964 quarters. The alternative hypothesis (\text{H}_1) is that the mean weight of pre-1964 quarters is greater than that of post-1964 quarters: \[ \text{H}_0: \bar{x}_{pre-1964} = \bar{x}_{post-1964} \] \[ \text{H}_1: \bar{x}_{pre-1964} > \bar{x}_{post-1964} \]

02

Determine the Test Statistic

Using the formula for the test statistic for two means where population standard deviations are not assumed equal: \[ t = \frac{(\bar{x}_1 - \bar{x}_2) - (\text{hypothesized difference})}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \] Given, \( \bar{x}_1 = 6.19267 \, \text{g} \), \( s_1 = 0.087 \, \text{g} \), \( n_1 = 40 \), \( \bar{x}_2 = 5.6393 \, \text{g} \), \( s_2 = 0.06194 \, \text{g} \), \( n_2 = 40 \), and applying the formula: \[ t = \frac{(6.19267 - 5.6393) - 0}{\sqrt{\frac{0.087^2}{40} + \frac{0.06194^2}{40}}} \]

03

Calculate Degrees of Freedom

Using the formula for degrees of freedom for unequal variances: \[ df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1 - 1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2 - 1}} \] After calculating, \( df \approx 76 \) is used.

04

Find the Critical Value

At a significance level \( \alpha = 0.05 \), and using a one-tailed t-test, the critical t-value can be found using a t-distribution table or technology: \[ t_{critical} = t_{0.05, 76} \approx 1.665 \]

05

Compare Test Statistic and Critical Value

Calculate the test statistic from step 2: \[ t \approx 33.59 \] Since \( t \approx 33.59 \) is greater than \( t_{critical} = 1.665 \), reject the null hypothesis \( H_0 \).

06

Construct the Confidence Interval

To construct the confidence interval for the difference of means with a 95% confidence level, use the formula: \[ \text{Difference} ± t_{critical} \cdot \sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}} \] Substituting the correct values: \[ (6.19267 - 5.6393) ± 1.665 \cdot \sqrt{\frac{0.087^2}{40} + \frac{0.06194^2}{40}} \] Leads to: \[ (0.55337) ± 0.02255 \] Thus confidence interval: \[ {0.531} \text{ to } {0.576} \]

07

Interpret the Confidence Interval and Conclusion

Since the confidence interval for the difference in means does not include 0 and lies entirely above 0, it supports the conclusion that pre-1964 quarters are heavier than post-1964 quarters. Post-1964 quarters do weigh less. Vending machines are not significantly affected because the weight difference is small enough that machines can still recognize the coin as a valid quarter.

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Key Concepts

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

independent samples t-test

An independent samples t-test compares the means of two independent groups to determine if there is a statistically significant difference between them. This test assumes that the two samples are independent and are drawn from normally distributed populations. In our problem with coin weights, we're comparing the weights of pre-1964 quarters with post-1964 quarters. We use the t-test to evaluate whether the difference in means is due to random chance or a true difference.
This specific t-test does not assume equal variances between the groups, which makes it suitable for our data where the standard deviations of the two groups differ.
It involves:

• Calculating the test statistic using the formula for independent samples.
• Determining the degrees of freedom using a more complex formula when variances are unequal.
• Finding the critical t-value for a given significance level.
• Comparing the test statistic to the critical value to make a decision about the hypotheses.

confidence interval

A confidence interval provides a range of values within which we are reasonably certain the true population parameter lies. In hypothesis testing, a 95% confidence interval means that if we were to take 100 different samples and compute a confidence interval for each, approximately 95 of them would contain the true mean difference.
To construct a confidence interval for the difference in means:

• Calculate the point estimate, which is the difference in sample means.
• Determine the margin of error using the critical t-value and standard error.
• Add and subtract the margin of error from the point estimate to get the interval.

In the coin weight problem, we calculated a confidence interval for the difference between pre-1964 and post-1964 quarters' weights, giving us a range where the true mean difference likely falls.

degrees of freedom

Degrees of freedom (df) refer to the number of independent values or quantities which can be assigned to a statistical distribution. They are crucial in determining the critical value of t for hypothesis testing.
When working with two samples and not assuming equal variances, degrees of freedom are calculated using the formula:
\[ df = \frac{\left( \frac{s_1^2}{n_1} + \frac{s_2^2}{n_2} \right)^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1 - 1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2 - 1}} \]
In our exercise, after substituting the values, we found the degrees of freedom to be approximately 76.
Degrees of freedom help us find the correct critical t-value from the t-distribution table for our hypothesis test.

significance level

The significance level (\(\alpha\)) is the threshold for deciding whether a hypothesis test result is statistically significant. It's the probability of rejecting the null hypothesis when it is actually true (Type I error). Common values for \(\alpha\) are 0.05, 0.01, or 0.10.
In this problem, we use a 0.05 significance level. This means that we are willing to accept a 5% chance of incorrectly rejecting the null hypothesis.
If our calculated p-value is less than 0.05, we reject the null hypothesis. Conversely, if it is greater than 0.05, we fail to reject it.
The significance level thus acts as a decision rule that helps us determine the strength of our evidence.

vending machines and coin weight

The weight of coins is vital in the design and operation of vending machines, as they need to correctly identify and validate coins to function properly. Even slight differences in weight can potentially cause issues.
In our exercise, pre-1964 quarters were found to be heavier than post-1964 quarters. However, vending machines have a tolerance range to accommodate slight variations in coin weight. This tolerance range ensures that despite the small decrease in weight for post-1964 quarters, they are still recognized and accepted as valid quarters by vending machines.
Understanding this tolerance helps explain why vending machines are not significantly affected by the weight change, maintaining their functionality and reliability.