Answer to Question #108817 in Statistics and Probability for harry

Question

In a survey carried out at a famous waterpark in Toronto, 28 children out of a random
sample of 80 said that they used the water slide regularly. Find a 95 % confidence
interval for the true proportion of all children at the waterpark who uses the water slide
regularly.
The owner of the water park found that 45 children out of a random sample of 100 said
that they used the pool regularly. Find a 98% confidence interval for the true proportion
of all children in the water park who uses the pool.

Answer 1

Proportion of children who said they were using a water slide is p= 28/80 = 0.35 (35%)

and that of children who do not use a water slide is q = 1-p = 1-0.35 = 0.65 (65%).

A 95 % confidence interval for the true proportion of all children at the waterpark who uses the water slide regularly:

$p-\Delta_p \le\ p \le p+\Delta_p$

$\Delta_p =l*1.96,$

$l=\sqrt{\smash[b]{p(1-p)/n}}=\sqrt{\smash[b]{0.35 *0.65/80}}=0.05332,$

$\Delta_p =$ 0.05332 *1.96= 0.1045,

0.35 - 0.1045 $\le\ p \le$ 0.35+0.1045

0.2455 $\le\ p \le$ 0.4545

Proportion of children who said they were using the pool regularly is p= 45/100 = 0.45 (45%)

and that of children who do not use is q = 1-p = 1-0.45 = 0.55 (55%)

A  98% confidence interval for the true proportion of all children in the water park who uses the pool:

$p-\Delta_p \le\ p \le p+\Delta_p$

$\Delta_p =l*2.326,$

$l=\sqrt{\smash[b]{p(1-p)/n}} = \sqrt{\smash[b]{0.45*0.55/100}}=0.04975$

$\Delta_p$ = 2.326*0.04975 =0.1157,

0.45-0.1157 $\le\ p \le$ 0.45+0.1157,

0.3343 $\le\ p \le$ 0.5657