Besides optimism, there are other benefits associated with exercise. A doctor claims the proportion of those who exercise and got sick in the past year is smaller than the proportion of those who do not exercise.

To investigate, an analyst selects independent random samples: 50 adults who exercise regularly and 75 adults who do not exercise regularly. Of those who exercise regularly, 18 got sick in the past year, and of those who do not exercise regularly, 56 got sick in the past year. Do these data provide convincing evidence that these two population proportions differ?

The random and [tex]10\%[/tex] conditions for this problem are met, but what about the large counts condition? Calculate [tex]\hat{p}_c = \frac{X_1 + X_2}{n_1 + n_2}[/tex].

Enter your answer to 3 decimal places.

[tex]\hat{p}_c = \square[/tex]

To calculate the pooled proportion, we use the formula

[tex]$$
\hat{\rho}_c = \frac{X_1 + X_2}{n_1 + n_2},
$$[/tex]

where
- [tex]$X_1$[/tex] is the number of individuals who got sick in the exercise group,
- [tex]$X_2$[/tex] is the number of individuals who got sick in the non-exercise group,
- [tex]$n_1$[/tex] is the size of the exercise group, and
- [tex]$n_2$[/tex] is the size of the non-exercise group.

Given:
- [tex]$X_1 = 18$[/tex],
- [tex]$X_2 = 56$[/tex],
- [tex]$n_1 = 50$[/tex], and
- [tex]$n_2 = 75$[/tex],

first, add the number of individuals who got sick from both groups:

[tex]$$
X_1 + X_2 = 18 + 56 = 74.
$$[/tex]

Next, add the sample sizes from both groups:

[tex]$$
n_1 + n_2 = 50 + 75 = 125.
$$[/tex]

Now, the pooled proportion is

[tex]$$
\hat{\rho}_c = \frac{74}{125} \approx 0.592.
$$[/tex]

Rounding to three decimal places gives

[tex]$$
\hat{p}_t = 0.592.
$$[/tex]

Thus, the pooled proportion is [tex]$0.592$[/tex].