Solution :
It is a left tailed test.
Given : n = 1700
[tex]$\hat P = 0.73$[/tex]
α = 0.05
We want to test,
the null hypothesis, [tex]$H_0:P=0.70$[/tex]
the alternate hypothesis, [tex]$H_1: P> 0.70 $[/tex]
Test statistics is
[tex]$z=\frac{\hat P - P_0}{\sqrt{\frac{P_0(1-P_0 )}{n}}} $[/tex]
[tex]$z=\frac{0.73 - 0.70}{\sqrt{\frac{0.70(1-0.70 )}{1700}}} $[/tex]
[tex]$z=2.70 $[/tex]
Thus z - critical value = 1.64
Since, z calculated ----- z-critical value
(2.70) (1.64)
Thus we reject the null hypothesis.
So there is sufficient evidence at 0.05 level to support the company's claim that more the 70% do not fail in the first 1000 hours of their use.
