Question

The Ice Chalet offers dozens of different beginning ice-skating classes. All of the class names are put into a bucket. The 5 P.M., Monday night, ages 8 to 12, beginning ice-skating class was picked. In that class were 64 girls and 16 boys. Suppose that we are interested in the true proportion of girls, ages 8 to 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the children in the selected class are a random sample of the population.

Required:
Construct a 92% Confidence Interval for the true proportion of girls in the ages 8-12 beginning ice-skating classes at the Ice Chalet.

Answer 1

The answer is "(0.73,0.87)".

Explanation:

Given:

number of girls
`= 64`

number of boys
`= 16`

Total number of children
`= 64+16= 80`

So,
`n=80`

Calculating the value of the proportion which is given by girls:

`\hat{p}= \frac{\text{number of girls}}{\text{total number of childrens}}`

`=(64)/(80)\\\\=(8)/(10)\\\\= 0.8`

Therefore the confidence interval is:

`\to \hat{p}\pm z_{(1-(\alpha )/(2))}\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\\\\\to 0.8 \pm 1.75\sqrt{(0.8(1-0.8))/(80)}\\\\\to 0.8 \pm 1.75 \sqrt{(0.8(0.2))/(80)}\\\\\to 0.8 \pm 1.75 \sqrt{(0.16)/(80)}\\\\\to 0.8 \pm 1.75 √(0.002)\\\\\to 0.8 \pm 1.75 (0.04)\\\\\to 0.8 \pm 0.07\\\\\to (0.73,0.87)`

`\therefore \\\\\text{The lower confidence limit} = 0.73\\\\\text{The upper confidence limit} = 0.87\\`