/*! 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 14 The article "Urban Battery Litte... [FREE SOLUTION] | 91影视

91影视

The article "Urban Battery Litter" cited in Example 8.14 gave the following summary data on zinc mass (g) for two different brands of size D batteries: \begin{tabular}{lccc} Brand & Sample Size & Sample Mean & Sample SD \\ \hline Duracell & 15 & \(138.52\) & \(7.76\) \\ Energizer & 20 & \(149.07\) & \(1.52\) \\ \hline \end{tabular} Assuming that both zinc mass distributions are at least approximately normal, carry out a test at significance level \(.05\) using the \(P\)-value approach to decide whether true average zinc mass is different for the two types of batteries.

Short Answer

Expert verified
The data shows a significant difference in zinc mass between Duracell and Energizer batteries.

Step by step solution

01

Formulate Hypotheses

Define the null and alternative hypotheses. - Null hypothesis ( H鈧 ): The true average zinc mass for Duracell is equal to that for Energizer, i.e., H鈧: 渭鈧 = 渭鈧. - Alternative hypothesis ( H鈧 ): The true average zinc mass is different, i.e., H鈧: 渭鈧 鈮 渭鈧. where 渭鈧 and 渭鈧 are the means for Duracell and Energizer, respectively.
02

Determine Test Statistic

Since the variances of the two samples are not assumed to be equal, use the formula for the test statistic in a two-sample t-test with unequal variances. The test statistic is calculated as:\[ t = \frac{\bar{x}_1 - \bar{x}_2}{\sqrt{\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2}}} \]where \(\bar{x}_1 = 138.52\), \(\bar{x}_2 = 149.07\), \(s_1 = 7.76\), \(s_2 = 1.52\), \(n_1 = 15\), and \(n_2 = 20\).
03

Calculate the Test Statistic

Substitute the given values into the formula:\[ t = \frac{138.52 - 149.07}{\sqrt{\frac{7.76^2}{15} + \frac{1.52^2}{20}}} \]Calculate the components:- Variance for Duracell = \({7.76^2}/{15} = 4.0123\)- Variance for Energizer = \({1.52^2}/{20} = 0.1156\) Combine these to find:\[ \sqrt{4.0123 + 0.1156} = 2.0203 \]Then, compute the t-statistic:\[ t = \frac{-10.55}{2.0203} \approx -5.22 \]
04

Determine Degrees of Freedom and Critical Value

For unequal variances, use the Welch-Satterthwaite equation for degrees of freedom:\[ df = \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1}} \]Substitute the values:\[ df \approx \frac{(4.0123 + 0.1156)^2}{\frac{(4.0123)^2}{14} + \frac{(0.1156)^2}{19}} \approx 14.004 \]The critical t-value for a 2-tailed test with (df = 14) at \(伪 = 0.05\)is approximately 卤2.145.
05

Calculate P-Value

Use the t-distribution table or software to find the p-value associated with \(t = -5.22\) and df = 14.Since \(|t| = 5.22\) is quite large, the p-value will be very small, well below \(0.05\).
06

Conclusion

Since the p-value is less than the significance level (\(0.05\)), reject the null hypothesis. This indicates a statistically significant difference in true average zinc mass between Duracell and Energizer batteries.

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 91影视!

Key Concepts

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

Statistical Hypothesis Testing
Statistical hypothesis testing is a fundamental method used to make decisions or inferences about a population based on sample data. It starts by formulating two complementary hypotheses: the null hypothesis ( H鈧 ) and the alternative hypothesis ( H鈧 ). The null hypothesis usually signifies that there is no effect or no difference, and it's the statement being tested. On the other hand, the alternative hypothesis suggests that there is an effect or a difference.

In the context of our battery scenario, we are checking whether there's a difference in zinc mass between two brands of batteries. The null hypothesis would state that there is no difference in average zinc masses, while the alternative hypothesis suggests that there is a difference.

The outcome of a hypothesis test leads us to either "reject" or "fail to reject" the null hypothesis based on statistical evidence provided by the data. It is crucial to understand that in hypothesis testing, we do not "prove" a hypothesis, but rather determine if there is enough evidence to favor one over the other.
Welch-Satterthwaite Equation
Using the Welch-Satterthwaite equation is key when dealing with a two-sample t-test involving unequal variances. It helps estimate the degrees of freedom (df) for the test.

This equation is given as follows:
\[ df = \frac{(\frac{s_1^2}{n_1} + \frac{s_2^2}{n_2})^2}{\frac{(\frac{s_1^2}{n_1})^2}{n_1-1} + \frac{(\frac{s_2^2}{n_2})^2}{n_2-1}} \]
Here, \(s_1^2\) and \(s_2^2\) represent the sample variances for the two groups, while \(n_1\) and \(n_2\) indicate their sample sizes. Essentially, this equation allows for a more accurate determination of the degrees of freedom when sample variances are not equal, which is important for the validity of the two-sample t-test.

The adjusted degrees of freedom calculated with this method help to determine the critical t-value in the t-distribution, which further aids in making the decision to reject or not reject the null hypothesis. Without this adjustment, assuming equal variances could lead to incorrect conclusions.
T-Distribution
The t-distribution is a probability distribution that is used extensively in hypothesis testing, especially when dealing with small sample sizes. It resembles the standard normal distribution but has heavier tails, which means it is more prone to producing values that fall far from its mean. This characteristic accounts for the variability associated with small sample sizes.

In a two-sample t-test, the t-distribution plays a crucial role. It's used to determine the critical values that define the significance threshold for the test. For instance, if you are using a significance level of 0.05 in a two-tailed test, you refer to the t-distribution table (or software) with the appropriate degrees of freedom to find the critical t-value.

T-statistics derived from sample data are compared against this critical value. If the absolute value of the calculated t-statistic exceeds the critical value, it provides evidence to reject the null hypothesis. This helps researchers make informed decisions based on the statistical analysis of data.
Significance Level
The significance level, denoted as \(\alpha\), is a critical concept in hypothesis testing. It is the probability of rejecting the null hypothesis when it is actually true, known as a Type I error. The most common significance level chosen is 0.05, representing a 5% risk of making this error.

In our battery example, with a significance level set at 0.05, it means we are willing to accept a 5% chance of concluding that there is a difference in zinc mass, even if there is not. The significance level directly influences the critical value of the t-distribution.

This predetermined threshold decides whether the p-value (probability of observing the data, assuming the null hypothesis is true) is considered "small" enough to reject the null hypothesis. When the p-value is less than \(\alpha\), it is seen as evidence against the null hypothesis, indicating that the observed effect or difference is statistically significant.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The article "The Influence of Corrosion Inhibitor and Surface Abrasion on the Failure of Aluminum-Wired Twist-On Connections" (IEEE Trans. on Components, Hybrids, and Manuf. Tech., 1984: 20-25) reported data on potential drop measurements for one sample of connectors wired with alloy aluminum and another sample wired with EC aluminum. Does the accompanying SAS output suggest that the true average potential drop for alloy connections (type 1) is higher than that for \(\mathrm{EC}\) connections (as stated in the article)? Carry out the appropriate test using a significance level of \(.01\). In reaching your conclusion, what type of error might you have committed? [Note: SAS reports the \(P\)-value for a two-tailed test.] \(\begin{array}{lrrrr}\text { Type } & \text { N } & \text { Mean } & \text { Std Dev } & \text { Std Error } \\ 1 & 20 & 17.49900000 & 0.55012821 & 0.12301241 \\ 2 & 20 & 16.90000000 & 0.48998389 & 0.10956373 \\ & & & & \\\ & & T & \text { DF } & \text { Prob> }|T| \\ \text { Variances } & & \text { T } & 0.0008 \\ \text { Unequal } & 3.6362 & 37.5 & 0.0008 .0008 \\ \text { Equal } & 3.6362 & 38.0 & 0.0000\end{array}\)

Two different types of alloy, A and B, have been used to manufacture experimental specimens of a small tension link to be used in a certain engineering application. The ultimate strength (ksi) of each specimen was determined, and the results are summarized in the accompanying frequency distribution. \begin{tabular}{lrr} & \(\mathbf{A}\) & \(\mathbf{B}\) \\ \hline \(26-<30\) & 6 & 4 \\ \(30-<34\) & 12 & 9 \\ \(34-<38\) & 15 & 19 \\ \(38-<42\) & 7 & 10 \\ & \(m=40\) & \(m=42\) \\ \hline \end{tabular} Compute a \(95 \%\) CI for the difference between the true proportions of all specimens of alloys \(\mathrm{A}\) and \(\mathrm{B}\) that have an ultimate strength of at least \(34 \mathrm{ksi}\).

An article in the November 1983 Consumer Reports compared various types of batteries. The average lifetimes of Duracell Alkaline AA batteries and Eveready Energizer Alkaline AA batteries were given as \(4.1\) hours and \(4.5\) hours, respectively. Suppose these are the population average lifetimes. a. Let \(\bar{X}\) be the sample average lifetime of 100 Duracell batteries and \(\bar{Y}\) be the sample average lifetime of 100 Eveready batteries. What is the mean value of \(\bar{X}-\bar{Y}\) (i.e., where is the distribution of \(\bar{X}-\bar{Y}\) centered)? How does your answer depend on the specified sample sizes? b. Suppose the population standard deviations of lifetime are \(1.8\) hours for Duracell batteries and \(2.0\) hours for Eveready batteries. With the sample sizes given in part (a), what is the variance of the statistic \(\bar{X}-\bar{Y}\), and what is its standard deviation? c. For the sample sizes given in part (a), draw a picture of the approximate distribution curve of \(\bar{X}-\bar{Y}\) (include a measurement scale on the horizontal axis). Would the shape of the curve necessarily be the same for sample sizes of 10 batteries of each type? Explain.

Recent incidents of food contamination have caused great concern among consumers. The article "How Safe Is That Chicken?" (Consumer Reports, Jan. 2010: 19-23) reported that 35 of 80 randomly selected Perdue brand broilers tested positively for either campylobacter or salmonella (or both), the leading bacterial causes of food-borne disease, whereas 66 of 80 Tyson brand broilers tested positive. a. Does it appear that the true proportion of non- contaminated Perdue broilers differs from that for the Tyson brand? Carry out a test of hypotheses using a significance level.01 by obtaining a \(P\)-value. b. If the true proportions of non-contaminated chickens for the Perdue and Tyson brands are . 50 and \(.25\), respectively, how likely is it that the null hypothesis of equal proportions will be rejected when a, 01 significance level is used and the sample sizes are both 80 ?

It is thought that the front cover and the nature of the first question on mail surveys influence the response rate. The article "The Impact of Cover Design and First Questions on Response Rates for a Mail Survey of Skydivers" (Leisure Sciences, 1991: 67-76) tested this theory by experimenting with different cover designs. One cover was plain; the other used a picture of a skydiver. The researchers speculated that the return rate would be lower for the plain cover. \begin{tabular}{lcc} Cover & Number Sent & Number Returned \\ \hline Plain & 207 & 104 \\ Skydiver & 213 & 109 \\ \hline \end{tabular} Does this data support the researchers' hypothesis? Test the relevant hypotheses using \(\alpha=.10\) by first calculating a \(P\)-value.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.