Question

It is reported that 230 out of 600 middle school boys plan to see the new Captain Marvel movie. Find a 90% confidence interval estimate for the percentage of all middle school boys planning to see the movie. The movie theater claims that 25% of all middle school boys are planning to see the movie. Does their claim appear to be correct?

Answer 1

Answer:

A 90% confidence interval estimate for the percentage of all middle school boys planning to see the movie is [0.350, 0.416].

Step-by-step explanation:

We are given that it is reported that 230 out of 600 middle school boys plan to see the new Captain Marvel movie.

Firstly, the pivotal quantity for finding the confidence interval for the population proportion is given by;

                           P.Q.  =  $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$  ~ N(0,1)

where, $\hat p$ = sample proportion of middle school boys who plan to see the new Captain Marvel movie = $\frac{230}{600}$ = 0.383

           n = sample of middle school boys = 600

           p = population proportion

Here for constructing a 90% confidence interval we have used a One-sample z-test for proportions.

So, 90% confidence interval for the population proportion, p is ;

P(-1.645 < N(0,1) < 1.645) = 0.90  {As the critical value of z at 5% level

                                                   of significance are -1.645 & 1.645}  

P(-1.645 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 1.645) = 0.90

P( $-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

P( $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.90

90% confidence interval for p = [ $\hat p-1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ , $\hat p+1.645 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ]

   = [ $0.383-1.645 \times {\sqrt{\frac{0.383(1-0.383)}{600} } }$ , $0.383+1.645 \times {\sqrt{\frac{0.383(1-0.383)}{600} } }$ ]

   = [0.350, 0.416]

Therefore, a 90% confidence interval estimate for the percentage of all middle school boys planning to see the movie is [0.350, 0.416].

The claim of the movie theater that 25% of all middle school boys are planning to see the movie is incorrect because in the above interval 25% value is not included, that is between 35% and 42% middle school boys are planning to see the movie.