Question

In a survey, 498 out of 1,397 us adults said they drink at least 4 cups of coffee a day. find a point estimate (p) for the population proportion of us adults who drink at least 4 cups of coffee a day, then construct a 90% confidence interval for the proportion of adults who drink at least 4 cups of coffee a day.

Answer 1

Let X be the number of adults who drink at least 4 cups of coffee a day. Let n be the sample size and p be the proportion of adults who drink at least 4 cups of coffee a day.

Given : n=1397 and x =498

a. Point estimate p for the population proportion is sample proportion

p = x/n = 498/1397 = 0.3564

p= 0.3564

b .90% confidence interval for population proportion.

The distribution of sample proportion is normal with parameters

mean =p =0.3564

standard deviation = $\sqrt{\frac{p*(1-p)}{n}}$

= $\sqrt{\frac{0.3564*(1-0.3564)}{1397}}$

standard deviation = 0.0128

The 90% confidence interval for population proportion is

(p - Margin of error, p+ margin of error)

where Margin of error = $z _{\alpha/2}\sqrt{\frac{p*(1-p)}{n} }$

where $z_{\alpha/2}$ is critical z score value for 90% confidence interval

alpha = 1- c = 1 - 0.9 = 0.1

$z_{\alpha/2}$ = $z_{0.1/2}$ = z (0.05)

This is z score value for which probability below -z is 0.05 and probability above z is 0.05

Using excel function to find z score for probability 0.05

P(Z < z) = 0.05

z = NORM.S.INV(0.05) = -1.645

For calculating confidence interval we consider positive z score value

z = 1.645

Margin of error = $z _{\alpha/2}\sqrt{\frac{p*(1-p)}{n} }$

= $1.645\sqrt{\frac{0.3564*(1-0.3564)}{1397} }$

= 1.645 * 0.0128

Margin of error = 0.021

The 90% confidence interval for population proportion is

(p - Margin of error, p+ margin of error)

(0.3564 - 0.021, 0.3564 + 0.021)

(0.3354 , 0.3774)

a 90% confidence interval for the proportion of adults who drink at least 4 cups of coffee a day is (0.3354, 0.3774)