Question

A 2003 New York Times CBS news poll sampled 523 adults who were planning a vacation during the next 6 months and found that 141 were expecting to travel by airplane. A similar survey question in May 1993 New York Times CBS news poll found that of 477 adults who were planning to take a vacation in the next 6 months, 81 were planning to travel by airplane. State the hypotheses that can be used to determine whether a significant change occurred in the population proportion planning to travel by airplane over 10 years.

Answer 1

Answer:

The solution can be defined as follows:

Step-by-step explanation:

$H_0 : p_1 = p_2\\\\H_a : p_1 \neq p_2$

Testing the statistics values:

$p_1 = \frac{141}{523} = 0.2696 \\\\ n_1 = 523\\\\p_2 = \frac{81}{477} = 0.1698\\\\ n_2 = 477$

$p = \frac{(p_1 \times n_1 + p_2 \times n_2)}{(n_1 + n_2)}$

  $= \frac{(0.2696 \times 523 + 0.1698 \times 477)}{( 523 + 477)}\\\\= \frac{(141.0008 + 80.9946)}{(1000)}\\\\= \frac{(221.99)}{(1000)}\\\\= \frac{(222.0)}{(1000)}\\\\= 0.2220$

$SE = \sqrt{ p \times ( 1 - p ) \times [ (\frac{1}{n_1}) + (\frac{1}{n_2}) ] }$

     $= \sqrt{( 0.222 \times (1-0.222) \times((\frac{1}{523}) + (\frac{1}{477})))}\\\\= 0.0263$

$z = \frac{(p_1 - p_2)}{SE}$

  $= \frac{( 0.2696 - 0.1698)}{0.0263}\\\\= 3.7947$

p = 0.0001

It reject all the null hypothesis values