Answer:
a) 0.0853
b) 0.0000
Step-by-step explanation:
Parameters given stated that;
H₀ : p = 0.6
H₁ : p = 0.6, this explains the acceptance region as;
p° ≤ [tex]\frac{315}{500}[/tex]=0.63 and the region region as p°>0.63 (where p° is known as the sample proportion)
a).
the probability of type I error if exactly 60% is calculated as :
∝ = P (Reject H₀ | H₀ is true)
= P (p°>0.63 | p=0.6)
where p° is represented as pI in the subsequent calculated steps below
= P [tex][\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6][/tex]
= P [tex][\frac{p°-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} >\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500}}} ][/tex]
= P [tex][Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ][/tex]
= P [Z > 1.37]
= 1 - P [Z ≤ 1.37]
= 1 - Ф (1.37)
= 1 - 0.914657 ( from Cumulative Standard Normal Distribution Table)
≅ 0.0853
b)
The probability of Type II error β is stated as:
β = P (Accept H₀ | H₁ is true)
= P [p° ≤ 0.63 | p = 0.75]
where p° is represented as pI in the subsequent calculated steps below
= P [tex][\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75][/tex]
= P [tex][\frac{p°-0.6} \sqrt{\frac{0.75(1-0.75)}{500} } }\leq \frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ][/tex]
= P[tex][Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ][/tex]
= P [Z ≤ -6.20]
= Ф (-6.20)
≅ 0.0000 (from Cumulative Standard Normal Distribution Table).
