Question

1. To obtain information on the corrosion-resistance properties of a certain type of steel conduit, 40 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 of 48.7 and a sample standard deviation of 5.6. The specifications require that the true average penetration should be lesser than 50 mils.

1. Write down the null and alternative hypotheses to be tested. Clearly define the terms used.          

2. What is the type of statistical test procedure that should be used to test the hypotheses? Explain.

3. Construct a 95% confidence interval. Test the hypotheses using the confidence interval. Interpret the test result clearly.

4. Test the hypotheses using the Test Statistic / Critical Value method (you must clearly indicate the test statistic, and the critical value(s) use Take α=5%). Do you get the same answer as part (c)? If your answer is the same as (c), you do not have to interpret again. Otherwise, interpret the test findings.

What is the P value of the test? Test the hypotheses using the P value. Take α=5%. Do you get the same answer as parts (c) and (d)? If your answer is the same as (c) and (d), you do not have to interpret again. Otherwise, interpret the test findings.             

Answer 1

Given
X bar	48.7
μ	50
S	5.6
n	40

a)

Hypothesis:

H0: μ = 50

Ha: μ < 50

b)

one tailed t test why because we don't have population standard deviation

c)

95% CI	alpha	0.05
tc	2.022691	T.INV.2T(alpha,df)
ME	1.790967	tc*(S/SQRT(n))
Upper	50.49097	X bar + tc*(S/SQRT(n))
Lower	46.90903	X bar - tc*(S/SQRT(n))

Rejection region:

μ < Lower bound

50 > 46.9, Do not reject H0

There is not enough evidence to conclude that the true average penetration should be lesser than 50 mils.

d)

i)

α=	0.05
df	39	n-1

t Critical Value :
tc	-1.684875122	T.INV(alpha,df)	LEFT
Rejection region:
ts	<=	tc	LEFT	To reject

Test :
ts	-1.468200342	(X bar-μ )/(S/SQRT(n))

Decision:

ts > tc, Do not reject H0

There is not enough evidence to conclude that the true average penetration should be lesser than 50 mils.

ii)

P value :
p	0.075034791	T.DIST(ts,df,TRUE)	LEFT

P value > 0.05, Do not reject Ho

There is not enough evidence to conclude that the true average penetration should be lesser than 50 mils.