Question

Suppose building specifications in a certain city require that the average breaking strength of residential
sewer pipe be more than 3570 kilograms per meter of length (i.e.per linear meter). Each manufacturer
who wants to sell pipe in that city must demonstrate that its product meets the specification. . Suppose
the sample mean breaking strength for the 50 sections of sewer pipe turned out to be 3620 kilograms per
linear foot. Assuming that the sample standard deviation is s = 200, does the city have enough evidence
exists to conclude that the manufacturer’s pipe meets specifications? What is the observed significance
level?

Answer 1

Answer)

Ho : u <= 3570

Ha : u > 3570

As the population s.d is unknown, we will use t distribution to conduct the test

Test statistics = (Sample mean - claimed mean)/(s.d/√n)

Test statistics t = (3620 - 3570)/(200/√50)

t = 1.768

Degrees of freedom is = n-1 = 50 -1 = 49

For df 49 and test statistics 1.768, p-value from t distribution is 0.041063

Now if p-value is greater than the given significance level, then we fail to reject the null hypothesis

If significance level is 0.05 (5%)

We will reject the null hypothesis

As 0.041063 < 0.05

As we have rejected the null hypothesis, we have enough evidence to conclude that u > 3570