Question

A student polls his school to see if students in the school district are for or against the new legislation regarding school uniforms. She surveys 600 students and finds that 480 are against the new legislation.

Required:
a. Compute a 90% confidence interval for the true percent of students who are against the new legislation and interpret the confidence interval.
b. In a sample of 300 students, 68% said they own an iPod and a smart phone. Compute a 97% confidence interval for the true percent of students who own an iPod and a smartphone.

Answer 1

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
`$\alpha = 1 - 0.97 = 0.03 $`

The area to the left of
`$z_(0.015)$` is 0.015 and the area to the right of
`$z_(0.015)$` is
`$1 - 0.015 = 0.985 $`

Thus using
`$TI \ 83, 83+ \ or \ 84+$` calculator function
`$\text{InvNorm} (0.985, 0.1), z_(0.015) = 2.17$`

`$\text{EBP} = z_(\alpha / 2) * \left( \sqrt{(p'q')/(n)}\right) $`

`$\text{EBP} = 1.645 * \sqrt{(0.68 * 0.32)/(300)} $`

= 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%.