Respuesta :
Solution :
68 % of the students owns iPod and a smart phone. So
p' = 0.68
q' = 1'
= 1 - 0.68
= 0.32
Since CL = 0.97, we know that [tex]$\alpha = 1 - 0.97 = 0.03 $[/tex]
The area to the left of [tex]$z_{0.015}$[/tex] is 0.015 and the area to the right of [tex]$z_{0.015}$[/tex] is [tex]$1 - 0.015 = 0.985 $[/tex]
Thus using [tex]$TI \ 83, 83+ \ or \ 84+$[/tex] calculator function [tex]$\text{InvNorm} (0.985, 0.1), z_{0.015} = 2.17$[/tex]
[tex]$\text{EBP} = z_{\alpha / 2} \times \left( \sqrt{\frac{p'q'}{n}}\right) $[/tex]
[tex]$\text{EBP} = 1.645 \times \sqrt{\frac{0.68 \times 0.32}{300}} $[/tex]
= 0.0269
Therefore, we are ninety seven percent confident that a true proportion of all the students who has iPod as well as a smart phone is between the 0.6531 and 0.7069
a). (0.7731, 0.8269); We estimated that with 90% confidence, the true percentage of all the students in district who are against a new legislation is between 77.31% and 82.69%.
b). Now,
Press the 'STAT' and then arrow over to perform TEST.
Now arrow down to the A:1 - PropZint and then place Enter.
Arrow a down x and then enter the 300 x 0.68
Similarly arrow a down to given n and then enter the 300
Now arrow down to the C-level and the enter the 0.97
Arrow a down for calculating and press the enter button.
The confidence interval is (0.6531, 0.7069)
a). (0.7731, 0.8269); We estimated that with 90% confidence, the true percentage of all the students in district who are against a new legislation is between 77.31% and 82.69%.
