Question

A poll shows that 50% of students play sports

Answer 1

Answer:

The chance of getting a sample  proportion of 70% or greater is 0.026.

Step-by-step explanation:

We are given that a poll shows that 50% of students play sports .

A random sample of 20 students showed that  70% of them play sports.

Let $\hat p$ = sample proportion of students who play sports

The z-score probability distribution for the sample proportion is given by;

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

where, $\hat p$ = sample proportion of students who play sports = 70%

           p = population proportion of students who play sports = 50%

           n = sample of students = 20

Now, the chance of getting a sample  proportion of 70% or greater is given by = P($\hat p$ $\geq$ 70%)

   P($\hat p$ $\geq$ 70%) = P( $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ $\geq$ $\frac{0.70-0.50}{\sqrt{\frac{0.70(1-0.70)}{20} } }$ ) = P(Z $\geq$ 1.95) = 1 - P(Z < 1.95)

                                                                 = 1 - 0.97441 = 0.026

The above probability is calculated by looking at the value of x = 1.95 in the z-table which has an area of 0.97441.

Hence, the chance of getting a sample  proportion of 70% or greater is 0.026.