Answer:
a) To show that we can use the 1 proportion z-interval, we know that 250 ( size sample) is big enough to approximate the binomial distribution ( tolerance-intolerance) to a normal distribution.
b) See step by step explanation
CI 90 % = ( 0,296 ; 0,392)
c) We can support that with 90 % of confidence we will find the random variable between this interval ( 0,296 ; 0,392)
d) the CI 95 % will be wider
Step-by-step explanation:
Sample Information:
Sample size n = 250
number of people with milk intolerance x = 86
p₁ = 86 / 250 p₁ = 0.344 and q₁ = 1 - p₁ q₁ = 0,656
To calculate 90 % of Confidence Interval α = 10% α/2 = 5 %
α/2 = 0,05 z(c) from z-table is: z(c) = 1.6
Then:
CI 90 % = ( p₁ ± z(c) * SE )
SE = √ (p₁*q₁)/n = √ 0,225664/250
SE = 0,03
CI 90 % = ( 0,344 ± 1,6* 0,03 )
CI 90 % = ( 0,344 - 0,048 ; 0,344 + 0,048)
b) CI 90 % = ( 0,296 ; 0,392)
a) To show that we can use the 1 proportion z-interval, we know that 250 ( size sample) is big enough to approximate the binomial distribution ( tolerance-intolerance) to a normal distribution.
c) We can support that with 90 % of confidence we will find the random variable between this interval ( 0,296 ; 0,392) .
d) CI 95 % then significance level α = 5 % α/2 = 2.5 %
α/2 = 0,025 z(c) = 1.96 from z-table
SE = 0,03
And as 1.96 > 1.6 the CI 95 % will be wider
CI 95% = ( 0,344 ± 1.96*0,03 )
CI 95% = ( 0,344 ± 0,0588 )
CI 95% = ( 0,2852 ; 0,4028 )
