Respuesta :
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 = [tex] \sqrt{\frac{p*(1-p)}{n}} [/tex]
= [tex] \sqrt{\frac{0.3564*(1-0.3564)}{1397}} [/tex]
standard deviation = 0.0128
The 90% confidence interval for population proportion is
(p - Margin of error, p+ margin of error)
where Margin of error = [tex] z _{\alpha/2}\sqrt{\frac{p*(1-p)}{n} } [/tex]
where [tex] z_{\alpha/2} [/tex] is critical z score value for 90% confidence interval
alpha = 1- c = 1 - 0.9 = 0.1
[tex] z_{\alpha/2} [/tex] = [tex] z_{0.1/2} [/tex] = 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 = [tex] z _{\alpha/2}\sqrt{\frac{p*(1-p)}{n} } [/tex]
= [tex] 1.645\sqrt{\frac{0.3564*(1-0.3564)}{1397} } [/tex]
= 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)
