Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.

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.

Sagot :

Answer:

The answer is "(0.73,0.87)".

Step-by-step explanation:

Given:

number of girls[tex]= 64[/tex]

number of boys[tex]= 16[/tex]

Total number of children[tex]= 64+16= 80[/tex]

So,[tex]n=80[/tex]

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

[tex]\hat{p}= \frac{\text{number of girls}}{\text{total number of childrens}}[/tex]

  [tex]=\frac{64}{80}\\\\=\frac{8}{10}\\\\= 0.8[/tex]

Therefore the confidence interval is:  

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

[tex]\therefore \\\\\text{The lower confidence limit} = 0.73\\\\\text{The upper confidence limit} = 0.87\\[/tex]