Question

Building specifications in a city require that they mean breaking strength of a sewer pipe be greater than 2,500 pounds per foot of length. A manufacturer wants to sell pipes to the city and needs to show that their pipes exceed the specification. If a sample of 25 pipes is chosen and it is found that the sample has a mean breaking strength of 2,565 pounds with a standard deviation of 225 pounds, is there enough evidence at a 0.05 significance level to convince the city that their pipes exceed the requirement?

Answer 1

Solution :

Given that ,

= 2500  

= 2565

= 225

n = 25

The null and alternative hypothesis is ,

H₀ :   = 2500

H_a : > 2500

This is the right tailed test .

Test statistic = z

= ( - ) / / n

= ( 2565 - 2500) / 225 / 25

= 1.44

The test statistic = 1.44

P - value = P(Z > 1.44 ) = 1 - P (Z < 1.44)

= 1 - 0.9251

= 0.0749

P-value = 0.0749

= 0.05  

0.0749 > 0.05

P-value >

Fail to reject the null hypothesis .

Conclusion :- There is not sufficient evidence to test the claim.  that the mean breaking strength of a sewer pipe be greater than 2,500 pounds per foot of length. at = 0.05 significance level.