Respuesta :
Answer:
There is not enough evidence to support the claim that the true average penetration is at most 50 mils.
Step-by-step explanation:
We are given the following in the question:
Population mean, μ = 50
Sample mean, [tex]\bar{x}[/tex] = 52.8
Sample size, n = 45
Alpha, α = 0.05
Sample standard deviation, s = 4.5
First, we design the null and the alternate hypothesis
[tex]H_{0}: \mu \leq 50\text{ mils}\\H_A: \mu > 50\text{ mils}[/tex]
We use one-tailed t test to perform this hypothesis.
Formula:
[tex]t_{stat} = \displaystyle\frac{\bar{x} - \mu}{\frac{s}{\sqrt{n}} }[/tex]
Putting all the values, we have
[tex]t_{stat} = \displaystyle\frac{52.8 - 50}{\frac{4.5}{\sqrt{45}} } = 4.1739[/tex]
Now, [tex]t_{critical} \text{ at 0.05 level of significance, 44 degree of freedom } = 1.6802[/tex]
Since,
The calculated test statistic is greater than the critical value, we fail to accept the null hypothesis and reject it. We accept the alternate hypothesis.
Conclusion:
There is not enough evidence to support the claim that the true average penetration is at most 50 mils.
Answer:
We conclude that specification has not been met as the true average penetration comes out to be greater than 50 mils.
Step-by-step explanation:
We are given that to obtain information on the corrosion-resistance properties of a certain type of steel conduit, 45 specimens are buried in soil for a 2-year period. The maximum penetration (in mils) for each specimen is then measured, yielding a sample average penetration of x = 52.8 and a sample standard deviation of s = 4.5.
We have to test if the true average penetration would be at most 50 mils.
Let, NULL HYPOTHESIS, [tex]H_0[/tex] : [tex]\mu \leq[/tex] 50 mils {means that the true average penetration is at most 50 mils}
ALTERNATE HYPOTHESIS, [tex]H_a[/tex] : [tex]\mu[/tex] > 50 mils {means that the true average penetration is greater than 50 mils}
The test statistics that will be used here is One-sample t-test;
T.S. = [tex]\frac{\bar X - \mu}{\frac{s}{\sqrt{n} } }[/tex] ~ [tex]t_n_-_1[/tex]
where, [tex]\bar X[/tex] = sample average penetration = 52.8
s = sample standard deviation = 4.5
n = sample of specimens = 45
So, test statistics = [tex]\frac{52.8-50}{\frac{4.5}{\sqrt{45} } }[/tex] ~ [tex]t_4_4[/tex]
= 4.174
Now, at 0.05 significance level, t table gives a critical value of 1.6808 at 44 degree of freedom. Since our test statistics is way higher than the critical value of t so we have sufficient evidence to reject null hypothesis as it will fall in the rejection region.
Therefore, we conclude that the true average penetration is greater than 50 mils.
