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 1 60.7 1.0 2 60.5 1.0 Calculate the P-value for the appropriate two-sample z test, assuming that the data was based on n

Questions :

a. 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.)

b. 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.)

c. Is the small P-value for n = 500 indicative of a difference that has practical significance

Answer :

0.1574

0.00157

Yes

Step-by-step explanation:

Hypothesis :

H0 : μ1 - μ2 = 0

H1 : μ1 - μ2 ≠ 0

Given :

x1 = 60.7 ; x2 = 60.5 ; s1 = 1 ; s2 = 2

Based on n = 100

Test statistic, Z :

Z = (x1 - x2)/[√(s1²/n1 + s2²/n2 )]

x1 - x2 = 60.7 - 60.5 = 0.2

0.2 / √(1²/100 + 1²/100 )]

0.2 / √0.02

Z = 1.414

The Pvalue from Zscore :

P(Z < 1.414) = 0.1574

B.)

For n = 500

Z = (x1 - x2)/[√(s1²/n1 + s2²/n2 )]

x1 - x2 = 60.7 - 60.5 = 0.2

0.2 / √(1²/500 + 1²/500 )]

0.2 / √0.004

Z = 3.162

The Pvalue from Zscore :

P(Z < 3.162) = 0.00157

Yes, the small Pvalue for n = 500 is indicative of a difference with practical significance ; as the Pvalue are compare with the α to make about a Decison about our hypothesis.

22.83 repeating

step-by-step explanation:

137 ÷ 6 is 22.8333333 therefore we understand that 3 is repeating.

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