Answer: (3,2)
Step-by-step explanation: <u>Diagonal</u> is a line connecting two opposite points of a form. In parallelograms, there two diagonals and they bisect each other, i.e., the point where the diagonals meet divide them into two segments with the same measure. In short, diagonals meet at their midpoint.
The way to determine midpoint is given by
(x,y) = ![{\frac{x_{1}+x_{2}}{2} },\frac{y_{1}+y_{2}}{2}](https://tex.z-dn.net/?f=%7B%5Cfrac%7Bx_%7B1%7D%2Bx_%7B2%7D%7D%7B2%7D%20%7D%2C%5Cfrac%7By_%7B1%7D%2By_%7B2%7D%7D%7B2%7D)
(x,y) = ![{\frac{1+5}{2} },\frac{3+1}{2}](https://tex.z-dn.net/?f=%7B%5Cfrac%7B1%2B5%7D%7B2%7D%20%7D%2C%5Cfrac%7B3%2B1%7D%7B2%7D)
(x,y) = (3,2)
The coordinates where diagonals of parallelogram GHJK intersect are (3,2).
The difference (adult - teenager) in the sample proportions of those who would recommend this action movie varies about 0.091 from the true difference in proportions.
Answer:
a)![0.0299\leq \widehat{p}\leq 0.0640](https://tex.z-dn.net/?f=0.0299%5Cleq%20%5Cwidehat%7Bp%7D%5Cleq%200.0640)
b)Yes
Step-by-step explanation:
n = 593
x = 28
![\widehat{p}=\frac{x}{n}](https://tex.z-dn.net/?f=%5Cwidehat%7Bp%7D%3D%5Cfrac%7Bx%7D%7Bn%7D)
![\widehat{p}=\frac{28}{593}](https://tex.z-dn.net/?f=%5Cwidehat%7Bp%7D%3D%5Cfrac%7B28%7D%7B593%7D)
![\widehat{p}=0.047](https://tex.z-dn.net/?f=%5Cwidehat%7Bp%7D%3D0.047)
Confidence level = 95%
So,
at 95% = 1.96
Formula of confidence interval of one sample proportion:
=![\widehat{p}-Z_\alpha \sqrt{\frac{\widehat{p}(1-\widehat{p}}{n}}\leq \widehat{p}\leq \widehat{p}+Z_\alpha \sqrt{\frac{\widehat{p}(1-\widehat{p}}{n}}](https://tex.z-dn.net/?f=%5Cwidehat%7Bp%7D-Z_%5Calpha%20%5Csqrt%7B%5Cfrac%7B%5Cwidehat%7Bp%7D%281-%5Cwidehat%7Bp%7D%7D%7Bn%7D%7D%5Cleq%20%5Cwidehat%7Bp%7D%5Cleq%20%5Cwidehat%7Bp%7D%2BZ_%5Calpha%20%5Csqrt%7B%5Cfrac%7B%5Cwidehat%7Bp%7D%281-%5Cwidehat%7Bp%7D%7D%7Bn%7D%7D)
=![0.047-(1.96)\sqrt{\frac{0.047(1-0.047)}{593}}\leq \widehat{p}\leq 0.047+(1.96)\sqrt{\frac{0.047(1-0.047}{593}}](https://tex.z-dn.net/?f=0.047-%281.96%29%5Csqrt%7B%5Cfrac%7B0.047%281-0.047%29%7D%7B593%7D%7D%5Cleq%20%5Cwidehat%7Bp%7D%5Cleq%200.047%2B%281.96%29%5Csqrt%7B%5Cfrac%7B0.047%281-0.047%7D%7B593%7D%7D)
=![0.0299\leq \widehat{p}\leq 0.0640](https://tex.z-dn.net/?f=0.0299%5Cleq%20%5Cwidehat%7Bp%7D%5Cleq%200.0640)
Hence a 95 percent confidence interval for the proportion of all new websites that were anonymous is ![0.0299\leq \widehat{p}\leq 0.0640](https://tex.z-dn.net/?f=0.0299%5Cleq%20%5Cwidehat%7Bp%7D%5Cleq%200.0640)
b) May normality of p be assumed?
Condition for normality : np>10 and np(1-p)>10.
and ![0.047\cdot593(1-0.047)>10](https://tex.z-dn.net/?f=0.047%5Ccdot593%281-0.047%29%3E10)
27.871 and 26.561063
Hence p is assumed to be normal since the condition is satisfied