Question

An engineer designed a valve that will regulate water pressure on an automobile engine. The engineer designed the valve such that it would produce a mean pressure of 5.9 pounds/square inch. It is believed that the valve performs above the specifications. The valve was tested on 160 engines and the mean pressure was 6.0 pounds/square inch. Assume the variance is known to be 0.36. A level of significance of 0.05 will be used. Determine the decision rule.

Answer 1

Answer:

Calculated value Z = 2.1097 > 1.96 at 0.05 level of significance

There is a difference between the means

Step-by-step explanation:

Step(i):-

Given that mean of the Population = 5.9pounds/square inch

Given mean of the sample = 6.0pounds/square inch

Given that variance of the Population = 0.36

The standard deviation of the Population = √0.36 =0.6

critical value (Z₀.₀₅)= 1.96

Step(ii):-

Null Hypothesis:H₀: x⁻ = μ

Alternative Hypothesis:H₁:  x⁻ ≠ μ

Test statistic

              $Z = \frac{x^{-}-mean }{\frac{S.D}{\sqrt{n} } }$

               $Z = \frac{6.0-5.9 }{\frac{0.6}{\sqrt{160} } }$

              Z  =  2.1097

Final answer:-

Calculated value Z = 2.1097 > 1.96 at 0.05 level of significance

The null hypothesis is rejected

There is a difference between the means