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 - 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 - 12, in all beginning ice-skating classes at the Ice Chalet. Assume that the 80 children in the selected class are a random sample of the population. Construct a 92% "Confidence Interval" for the true proportion of girls in the ages 8 - 12 beginning ice-skating classes at Chalet.    **Please show all work / formulas etc. Thank you.

Answer 1

The percentage of girls is 64/80*100=80%.

We can now formulate a null hypothesis: H₀ is that 80% of all beginners’ ice-skating classes for 8-12 year-olds at the Ice Chalet are girls. It is for us to demonstrate whether or not to reject H₀. The alternative hypothesis H₁ is that this percentage is not 80%. The significance level is 100-92=8% and two-tailed, because we are looking at possible figures of greater than or less than 80% rather than one or the other. So ɑ/2=4% or 0.04.

If we assume a binomial distribution (similar to a normal distribution) then p=0.8, n=80, mean=np=64, variance=64(1-p)=64(1-0.8)=12.8 and standard deviation s=√12.8. So we would have the statistic that the mean=64±s=64±√12.8.

The standard error=s/√80=√(12.8/80)=0.4. We need this to work out a Z value=(sample mean-µ)/0.4 where µ is the mean of the population of all 8-12 year-olds at Ice Chalet. But we need to adjust the mean to a percentage, and we know that this is 80%. µ will also be a percentage. We need to find the critical value of Z corresponding to a two-tailed 92% confidence level.  From tables or a calculator the critical value for ɑ=0.04 we get 1.66 approximately.

So Z=1.66=|80-µ|/0.4, from which |80-µ|=0.664, and 79.34%≤µ≤80.66% approx. If the population mean is within these limits we do not reject H₀ otherwise we reject it and maintain H₁.