Answer: 99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week
//0.20113,0.20887[/tex]
Step-by-step explanation:
<u>step 1:-</u>
Given sample size n=200
of the 200 employed individuals surveyed 41 responded that they did work at home at least once per week
Population proportion of employed individuals who work at home at least once per week P = ![\frac{x}{n} =\frac{41}{200} =0.205](https://tex.z-dn.net/?f=%5Cfrac%7Bx%7D%7Bn%7D%20%3D%5Cfrac%7B41%7D%7B200%7D%20%3D0.205)
Q=1-P= 1-0.205 = 0.705
<u>step 2:-</u>
Now ![\sqrt{\frac{P Q}{n} } =\sqrt{\frac{(0.205)(0.705)}{200} }](https://tex.z-dn.net/?f=%5Csqrt%7B%5Cfrac%7BP%20Q%7D%7Bn%7D%20%7D%20%3D%5Csqrt%7B%5Cfrac%7B%280.205%29%280.705%29%7D%7B200%7D%20%7D)
=0.0015
<u>step 3:-</u>
<u>Confidence intervals</u>
<u>using formula</u>
![(P - Z_∝} \sqrt{\frac{P Q}{n},} (P + Z_∝} \sqrt{\frac{P Q}{n},](https://tex.z-dn.net/?f=%28P%20%20-%20%20Z_%E2%88%9D%7D%20%5Csqrt%7B%5Cfrac%7BP%20Q%7D%7Bn%7D%2C%7D%20%28P%20%20%2B%20%20Z_%E2%88%9D%7D%20%5Csqrt%7B%5Cfrac%7BP%20Q%7D%7Bn%7D%2C)
![(0.205-2.58(0.0015),0.205+2.58(0.0015)\\0.20113,0.20887](https://tex.z-dn.net/?f=%280.205-2.58%280.0015%29%2C0.205%2B2.58%280.0015%29%3C%2Fp%3E%3Cp%3E%5C%5C0.20113%2C0.20887)
=0.20113,0.20887[/tex]
<u>conclusion:</u>-
99% of confidence interval for the population proportion of employed individuals who work at home at-least once per week
//0.20113,0.20887[/tex]
Step-by-step explanation:
line s || line t
r is the transversal,
Therefore,
(7x-2)°= (6x+18)° {exterior alternate angles}
7x-6x=18+2
x=20°
r is also a straight line,
therefore,
7x-2+angle1=180° (straight angle)
7x-2+ angle1=180°
7(20)-2+ angle 1= 180°
140°-2+ angle 1= 180°
angle 1= (180-140+2 )°= 42°
angle 1= 42°
3 (2x + 4y - y) Combine like terms (4y and -y)
3 (2x + 3y) Use the Distributive Property [(3 · 2x) + (3 · 3y)]
6x + 9y
Answer:
Moving average model
Step-by-step explanation: