Answer:
A)5
Step-by-step explanation:
I answered the question already.
Equals the third side length
If there are 3 feet per yard, then all you have to do is take the 96 feet and divide by 3, this gives you 32 yards.
If you have any questions please let me know, other wise take care. (:
Answer:
D. 3 and 116
Step-by-step explanation:
d.f.N = k - 1 (numerator degrees of freedom) = 4 - 1 = 3
N = 4 × 30 = 120
d.f.D = N - k (denominator degrees of freedom) = 120 - 4 =116
Answer:
a) 0.0853
b) 0.0000
Step-by-step explanation:
Parameters given stated that;
H₀ : <em>p = </em>0.6
H₁ : <em>p = </em>0.6, this explains the acceptance region as;
p° ≤
=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 <em>pI</em><em> </em>in the subsequent calculated steps below
= P ![[\frac{p°-p}{\sqrt{\frac{p(1-p)}{n}}} >\frac{0.63-p}{\sqrt{\frac{p(1-p)}{n}}} |p=0.6]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bp%C2%B0-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%20%3E%5Cfrac%7B0.63-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%7D%7D%20%7Cp%3D0.6%5D)
= P ![[\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}}} ]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bp%C2%B0-0.6%7D%7B%5Csqrt%7B%5Cfrac%7B0.6%281-0.6%29%7D%7B500%7D%7D%7D%20%3E%5Cfrac%7B0.63-0.6%7D%7B%5Csqrt%7B%5Cfrac%7B0.6%281-0.6%29%7D%7B500%7D%7D%7D%20%5D)
= P ![[Z>\frac{0.63-0.6}{\sqrt{\frac{0.6(1-0.6)}{500} } } ]](https://tex.z-dn.net/?f=%5BZ%3E%5Cfrac%7B0.63-0.6%7D%7B%5Csqrt%7B%5Cfrac%7B0.6%281-0.6%29%7D%7B500%7D%20%7D%20%7D%20%5D)
= 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 <em>pI</em><em> </em>in the subsequent calculated steps below
= P ![[\frac{p°-p} \sqrt{\frac{p(1-p)}{n} } }\leq \frac{0.63-p}{\sqrt{\frac{p(1-p)}{n} } } | p=0.75]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bp%C2%B0-p%7D%20%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%20%7D%20%7D%5Cleq%20%5Cfrac%7B0.63-p%7D%7B%5Csqrt%7B%5Cfrac%7Bp%281-p%29%7D%7Bn%7D%20%7D%20%7D%20%7C%20p%3D0.75%5D)
= P ![[\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} } } ]](https://tex.z-dn.net/?f=%5B%5Cfrac%7Bp%C2%B0-0.6%7D%20%5Csqrt%7B%5Cfrac%7B0.75%281-0.75%29%7D%7B500%7D%20%7D%20%7D%5Cleq%20%5Cfrac%7B0.63-0.75%7D%7B%5Csqrt%7B%5Cfrac%7B0.75%281-0.75%29%7D%7B500%7D%20%7D%20%7D%20%5D)
= P![[Z\leq\frac{0.63-0.75}{\sqrt{\frac{0.75(1-0.75)}{500} } } ]](https://tex.z-dn.net/?f=%5BZ%5Cleq%5Cfrac%7B0.63-0.75%7D%7B%5Csqrt%7B%5Cfrac%7B0.75%281-0.75%29%7D%7B500%7D%20%7D%20%7D%20%5D)
= P [Z ≤ -6.20]
= Ф (-6.20)
≅ 0.0000 (from Cumulative Standard Normal Distribution Table).