answer:
we reject null and conclude that this manufacturers claim is false
Step-by-step explanation:
p = 30% = 0.30
p^ = 93/150 = 0.62
we state the hypothesis
H0: p = 0.30
h1: p not equal to 0.30
we find the z test stattistics
[tex]z=\frac{p^--p}{\sqrt{\frac{p(1-p)}{n} } }[/tex]
[tex]z=\frac{0.62-0.30}{\sqrt{\frac{0.30(1-0.30)}{150} } }[/tex]
[tex]z=\frac{0.32}{\sqrt{\frac{0.30*0.70}{150} } }[/tex]
[tex]z=\frac{0.32}{0.03741}[/tex]
z = 8.5538
at alpha = 0.05
z-critical = Z₀.₀₅/₂ = Z₀.₀₂₅
= 1.96
we compare z critical with the test statistic
z statistic > z critical so we have to reject H₀ and conclude that the manufacturers claim is not valid at 0.05 level of significance.
