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To decide whether two different types of steel have the same true average fracture toughness values, n specimens of each type are tested, yielding the following results:

\(\begin{array}{l}\underline {\begin{array}{*{20}{c}}{ Type }&{ Sample Average }&{ Sample SD }\\1&{60.1}&{1.0}\\2&{59.9}&{1.0}\\{}&{}&{}\end{array}} \\\end{array}\)

Calculate the P-value for the appropriate two-sample \(z\) test, assuming that the data was based on \(n = 100\). Then repeat the calculation for \(n = 400\). Is the small P-value for \(n = 400\) indicative of a difference that has practical significance? Would you have been satisfied with just a report of the P-value? Comment briefly.a

Short Answer

Expert verified

the solution is

The slight difference in the genuine average has been amplified statistically due to the huge sample size \(n = 400\). In practice, the little difference in sample average values would lead one to believe that there is no difference between them. The P- value would be insufficient - it would be unsatisfactory

Step by step solution

01

calculate p value

when the null hypothesis is

\({H_0}:{\mu _1} - {\mu _2} = {\Delta _0}\)

the test statistic value

\(z = \frac{{(\bar x - \bar y) - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }}\)

The \(P\) value is derived for the acceptable alternative hypothesis by calculating the adequate area under the standard normal curve. When both \(m > 40 and n > 40\), this test can be employed.

The sample size is the same and exceeds \(40(100,400 > 40)\). When \(n = 100\), the value of the test statistic is.

\(z = \frac{{(\bar x - \bar y) - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }} = \frac{{60.1 - 59.9}}{{\sqrt {\frac{1}{{100}} + \frac{1}{{100}}} }} = 1.41.\)

02

calculated phi

the alternative hypothesis is two sided \(\left( {{H_a}:{\mu _1} - {\mu _2} \ne 0} \right)\) the p value

\(\begin{array}{l}2.P(Z > 1.41) = 2 \times (1 - P(Z \le 1.41) = 2.(1 - \Phi (1.41))\\ = 0.1586\end{array}\)

where \(\Phi (1.41)\) was calculated using the appendix table.

When \(n = 400\)

\(z = \frac{{(\bar x - \bar y) - {\Delta _0}}}{{\sqrt {\frac{{s_1^2}}{m} + \frac{{s_2^2}}{n}} }} = \frac{{60.1 - 59.9}}{{\sqrt {\frac{1}{{400}} + \frac{1}{{400}}} }} = 2.83.\)

The alternative hypothesis is two sided \(\left( {{H_a}:{\mu _1} - {\mu _2} \ne 0} \right)\), the \(P\) value is

\(\begin{array}{l}2.P(Z > 2.83) = 2 \times (1 - P(Z \le 2.83) = 2 \times (1 - \Phi (2.83))\\ = 0.0046\end{array}\)

where \(\Phi (2.83)\) was calculated using the appendix table.

The slight difference in the genuine average has been amplified statistically due to the huge sample size \(n = 400\). In practice, the little difference in sample average values would lead one to believe that there is no difference between them. The P- value would be insufficient - it would be unsatisfactory.

03

conclusion

The slight difference in the genuine average has been amplified statistically due to the huge sample size\(n = 400\). In practice, the little difference in sample average values would lead one to believe that there is no difference between them. The P- value would be insufficient - it would be unsatisfactory

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Most popular questions from this chapter

The National Health Statistics Reports dated Oct. \(22,2008\), included the following information on the heights (in.) for non-Hispanic white females:

Sample sample Std. Error

Age Size Mean Mean

\(\begin{array}{*{20}{l}}{20 - 39}&{866}&{64.9}&{.09}\\{60 and older }&{934}&{63.1}&{.11}\\{}&{}&{}&{}\end{array}\)

  1. Calculate and interpret a confidence interval at confidence level approximately \(95\% \) for the difference between population mean height for the younger women and that for the older women.
  2. Let \({\mu _1}\) denote the population mean height for those aged \(20 - 39\) and \({\mu _2}\) denote the population mean height for those aged 60 and older. Interpret the hypotheses \({H_0}:{\mu _1} - {\mu _2} = 1 and {H_a}:{\mu _1} - {\mu _2} > 1,\) and then carry out a test of these hypotheses at significance level \(.001\)
  3. Based on the \(p\)-value calculated in (b) would you reject the null hypothesis at any reasonable significance level? Explain your reasoning.
  4. What hypotheses would be appropriate if \({\mu _1}\) referred to the older age group, \({\mu _2}\) to the younger age group, and you wanted to see if there was compelling evidence for concluding that the population mean height for younger women exceeded that for older women by more than \(1\)in.?

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House

\(\begin{array}{*{20}{l}}{\;\;\;\;\;\;\;\;\;\;\;\;1\;\;\;\;\;\;2\;\;\;\;\;3\;\;\;\;\;\;\;4\;\;\;\;\;5\;\;\;\;\;\;\;6\;\;\;\;\;\;\;7\;\;\;\;\;\;\;8\;\;\;\;\;\;\;9}\\{Indoor\;\;\;\;\;\;\;\;\;.07\;\;\;\;.08\;\;\;\;.09\;\;\;\;.12\;\;\;\;.12\;\;\;\;.12\;\;\;\;.13\;\;\;\;.14\;\;\;\;.15}\\{Outdoor\;\;\;\;\;\;.29\;\;\;\;.68\;\;\;\;.47\;\;\;\;.54\;\;\;\;.97\;\;\;\;.35\;\;\;\;.49\;\;\;\;.84\;\;\;\;.86}\end{array}\)

House

\(\begin{array}{*{20}{l}}{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;10\;\;\;\;\;11\;\;\;\;\;12\;\;\;\;\;13\;\;\;\;\;14\;\;\;\;\;15\;\;\;\;\;16\;\;\;\;\;17}\\{Indoor\;\;\;\;\;\;\;\;\;.15\;\;\;\;.17\;\;\;\;.17\;\;\;\;.18\;\;\;\;.18\;\;\;\;.18\;\;\;\;.18\;\;\;\;.19}\\{Outdoor\;\;\;\;\;\;.28\;\;\;\;.32\;\;\;\;.32\;\;\;\;1.55\;\;\;.66\;\;\;\;.29\;\;\;\;.21\;\;\;\;1.02}\end{array}\)

House

\(\begin{array}{*{20}{c}}{}&{18}&{19}&{20}&{21}&{22}&{23}&{24}&{25}\\{Indoor}&{.20}&{.22}&{.22}&{.23}&{.23}&{.25}&{.26}&{.28}\\{Outdoor}&{1.59}&{.90}&{.52}&{.12}&{.54}&{.88}&{.49}&{1.24}\end{array}\)

House

\(\begin{array}{*{20}{c}}{}&{26}&{27}&{28}&{29}&{30}&{31}&{32}&{33}\\{Indoor}&{.28}&{.29}&{.34}&{.39}&{.40}&{.45}&{.54}&{.62}\\{Outdoor}&{.48}&{.27}&{.37}&{1.26}&{.70}&{.76}&{.99}&{.36}\end{array}\)

a. Calculate a confidence interval for the population mean difference between indoor and outdoor concentrations using a confidence level of\(95\% \), and interpret the resulting interval.

b. If a\(34\)th house were to be randomly selected from the population, between what values would you predict the difference in concentrations to lie?region.

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YF:\(29,34,33,27,28,32,31,34,32,27\)

OF:\(18,15,23,13,12\)

Does the data suggest that true average maximum lean angle for older females is more than\(10\)degrees smaller than it is for younger females? State and test the relevant hypotheses at significance level . \(10\).

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a. Calculate an upper confidence bound for the true average time that blackbirds spend on a single visit at the experimental location.

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