Question

In a study of the relationship between pet ownership and physical activity in older adults, 619 subjects reported that they owned a pet, while 1904 reported that they did not. Give a 90% confidence interval for the proportion of older adults in this population who are pet owners. (Round your answers to three decimal places.)

Answer 1

Answer: (0.230,0.260)

Step-by-step explanation:

The confidence interval for proportion is given by :-

$p\pm z_{\alpha/2}\sqrt{\dfrac{p(1-p)}{n}}$

Given : Number of subjects reported that they owned a pet = 619

Number of subjects reported that they did not own a pet =1904

Total subjects = $n=619+1904=2523$

Significance level : $\alpha=1-0.9=0.1$

The proportion that of older adults in this population who are pet owners :-

$p=\dfrac{619}{2523}\approx0.245$

Now, the 90% confidence interval for the proportion of older adults in this population who are pet owners is given by :-

$0.245\pm z_{0.05}\sqrt{\dfrac{0.245(0.755)}{2523}}\\\\=0.245\pm(1.645)(0.009)\\\\=0.245\pm0.014805\\\\=(0.230195,0.259805)\approx(0.230,0.260)$