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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:

Brand Sample Size Sample Mean Sample SD

Duracell 15 138.52 7.76

Energizer 20 149.07 1.52

Assuming that both zinc mass distributions are at least approximately normal, carry out a test at significance level .05 to decide whether true average zinc mass is different for the two types of batteries.

Short Answer

Expert verified

There is sufficient evidence to support the claim that the true average zinc mass is different from the two types of batteries.

Step by step solution

01

To finding the zinc mass is different for the two types of batteries.

Given in the output:

\(\begin{array}{l}{{\bar x}_1} = 138.52\\{s_1} = 7.76\\{{\bar x}_2} = 149.07\\{s_2} = 1.52{n_1} = 15\\{n_2} = 20\\\alpha = 0.05\end{array}\)

Given claim: Differs

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.

\(\begin{array}{l}{H_0}:{\mu _1} = {\mu _2}\\{H_a}:{\mu _1} \ne {\mu _2}\end{array}\)

02

To finding the zinc mass is different for the two types of batteries.

Determine the test statistic:

\(t = \frac{{\left( {{{\bar x}_1} - {{\bar x}_2}} \right) - \left( {{\mu _1} - {\mu _2}} \right)}}{{\sqrt {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} }} = \frac{{138.52 - 149.07 - 0}}{{\sqrt {\frac{{7.7{6^2}}}{{15}} + \frac{{1.5{2^2}}}{{20}}} }} \approx - 5.191\)

Determine the degrees of freedom (rounded down to the nearest integer):

\(\Delta = \frac{{{{\left( {\frac{{s_1^2}}{{{n_1}}} + \frac{{s_2^2}}{{{n_2}}}} \right)}^2}}}{{\frac{{{{\left( {s_1^2/{n_1}} \right)}^2}}}{{{n_1} - 1}} + \frac{{{{\left( {s_2^2/{n_2}} \right)}^2}}}{{{n_2} - 1}}}} = \frac{{{{\left( {\frac{{7.7{6^2}}}{{15}} + \frac{{1.5{2^2}}}{{20}}} \right)}^2}}}{{\frac{{{{\left( {7.7{6^2}/15} \right)}^2}}}{{15 - 1}} + \frac{{{{\left( {1.5{2^2}/20} \right)}^2}}}{{20 - 1}}}} \approx 14\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme. The P-value is the number (or interval) in the column title of Student's T distribution in the appendix containing the t-value in the row df=14 :

\(P < 2 \times 0.0005 = 0.001\)

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

\(P < 0.01 \Rightarrow Reject {H_0}\)

There is sufficient evidence to support the claim that the true average zinc mass is different from the two types of batteries.

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Most popular questions from this chapter

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.

\(A\)

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\(\begin{array}{*{20}{l}}{\;Subject\;\;\;\;\;1\;\;\;2\;\;3\;\;\;4\;\;\;5\;\;\;6\;\;\;\;\;7}\\{\;Slide\;\;\;\;\;\;\;\;30\;35\;40\;25\;20\;30\;\;35}\\{\;Digital\;\;\;\;\;\;\;25\;16\;15\;15 10\;20\;\;\;7}\\{\;Difference\;\;5\;19\;25\;10\;10\;10\;28}\\{\;Subject\;\;\;\;\;8\;\;9\;\;10\;11\;12\;13}\\{\;Slide\;\;\;\;\;\;\;\;\;62\;40\;51\;25\;42\;33\;}\\{Digital\;\;\;\;\;\;\;16\;15\;13\;11\;19\;19}\\{Difference\;46\;25\;38\;14\;23\;14}\end{array}\)

a. Construct a comparative boxplot of times for the two types of retrieval, and comment on any interesting features.

b. Estimate the difference between true average times for the two types of retrieval in a way that conveys information about precision and reliability. Be sure to check the plausibility of any assumptions needed in your analysis. Does it appear plausible that the true average times for the two types of retrieval are identical? Why or why not?

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\(\begin{array}{l}BeverageSampleSizeSampleMeanSampleSD\\Strawberrydrink1554021\\Cola1555415\end{array}\)

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