Chapter 8: Q. 85 (page 482)

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, for ages $8$ to $12$ , the 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.
State the estimated distribution of $P'$ . Construct a $92 %$ Confidence Interval for the true proportion of girls in the ages $8 - 12$ beginning ice-skating classes at the Ice Chalet.

Short Answer

Expert verified

With a mean of $0.80$ and a variance of $0.002$ , follows the normal distribution. is the confidence interval for the given condition.

Step by step solution

Given information

Total number of children $= 80$

Total number of girls $= 64$

Total number of boys $= 16$

Explanation

There are $80$ children in this question, $64$ of them are girls and $16$ of whom are boys.

$p' = 0.80$ and $n = 80$ are the values we have.

Will follow a Normal distribution with a mean of $μ$ equal to $' p'$ and

a variance of $σ^{2} = \frac{p' (1 - p')}{n}$

Now, according to the specifications of the query, there is a normal distribution with a mean of $0.80$ and a variance of $\frac{0.80 (1 - 0.20)}{80} = 0.002$ .

Confidence interval $= [p \pm Z_{\frac{a}{2}} \times \sqrt{\frac{p (1 - p)}{n}}]$

$= [0.80 \pm Z_{\frac{0.08}{2}} \times \sqrt{\frac{0.80 \times 0.20}{80}}] = [0.80 \pm 1.7507 \times \sqrt{\frac{0.80 \times 0.20}{80}}] = [0.7217, 0.8783]$

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