To decide whether two different types of steel have the same true average fractu

Question

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 Type Sample Average Sample SD 60.4 60.2 1.0 1.0 Calculate the P-value for the appropriate two-sample z test, assuming that the data was based on n-100. (Round your answer to four decimal places.) Calculate the P-value for the appropriate two-sample z test, assuming that the data was based on n-500. (Round your answer to four decimal places.) Is the small P-value for n = 500 indicative of a difference that has practical significance? would you have been satisfied with just a report of the P-value? Comment briefly.

Explanation / Answer

Solution.

(a) Z test statistic when n = 100

= (60.4-60.2)/(1/100 + 1/100)0.5 = 1.4142

P(Z>1.4142) = 0.1573 (two tailed)

(b) Z test statistic when n = 500

= (60.4-60.2)/(1/500 + 1/500)0.5 = 3.1623

P(Z>3.1623) = 0.0016

(c) Yes, small p value indicates that there is significant difference between the two samples and as n gets large it will approximately lead to the true poulation parameters.

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