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Sagot :
Answer:
t(s) > t(c) then t(s) is in the rejection region for H₀.
We reject H₀ we support the claim that the mean penetration is bigger than that of the specifications
Step-by-step explanation:
Sample Information:
sample size n = 35
sample mean x = 52.7
sample standard deviation s = 4.8
spec. true average = 50
Test Hypothesis:
Null Hypothesis H₀ x = 50
Alternative Hypothesis Hₐ x > 50
The standarddeviation of the population is unknown therefore even though we assume distribution is normal we have to use t-student distribution
The alternative hypothesis indicates that the test is a one-tail test, we will test at significance level α = 5 % CI = 95 %
for α = 0.05 t(c) is from t-student -table and with df = 35 - 1
df = 34 t(c) = 1.69
t(s) = ( x - μ ) / s/√n
t(s) = ( 52.7 - 50 ) / 4.8/√35
t(s) = 2.7 * 5.92 / 4.8
t(s) = 3.33
Comparing t(s) and t(c)
t(s) > t(c) then t(s) is in the rejection region for H₀.
We reject H₀ we support the claim that the mean penetration is bigger than that of the specifications
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