A special cable has a breaking strength of 800 pounds. The researcher variable i

Question

A special cable has a breaking strength of 800 pounds. The researcher variable is normally distributed and alpha-0.01. Is there sufficient strength is less than 800 pounds O Yes there is sufficient evide d deviation of the population is 12 pounds. A a sample of 20 cables and finds that the average breaking strength is 793 pounds. Assume the evidence to support the claim that the breaking nce to support the claim that the breaking strength is less than 800 pounds because the sample was less than 800 pounds No there is not sufficient evidence to support the claim that the breaking strenght is less than 800 pounds because because the test value of -2.61 is not in the critical region. the test value of -2.61 is in the critical region. breaking strength was given as 800 pounds. O Yes there is sufficient evidence to support the claim that the breaking strength is less than 800 pounds because O No there is not sufficient evidence to support the claim the breaking strength is less than 800 pounds because the QUESTION 9 s points The average farm size in the United States is 444 acres with a population standard deviation of 52 acres. A random sample of 40 farms in Oregon indicated a mean size of 430 acres. Can it be concluded that the average farm in Oregon differs from the national mean? Alpha -0.05 O Yes, it can be concluded that the average farm in Oregon is different from 444 acres acres. O Yes it can be concluded that the average farm in Oregon differs from 444 acres because the test value of -1.71 Is O No it cannot be concluded that the average farm size in Oregon differs from 444 acres because the sample size of e No it cannot be concluded that the average farm in Oregon differs from 444 acres because the test value of -1.7 is in the critical region. 40 is too small to make any conclusions. not in the critical region.

Explanation / Answer

Hypothesis:

H0 : mu = 800
HA ; mu < 800

Test statistics;

t = ( x -mean)/(s/sqrt(n))
= ( 793 - 800)/(12/sqrt(20))
= -2.6087

p value = .0086

Critical value = -2.5395
Reject the null hypothesis

Option C)

9)

H0 : mu = 444
Ha : mu not equals to 444

test statistics:

z = ( x -mean)/(s/sqrt(n))
= ( 430 - 444)/(52/sqrt(40))
= -1.7028

Critical value = +/- 1.96

Option D)

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