Question

Many individuals over the age of 40 develop intolerance for milk and milk-based products. A dairy has developed a line of lactose-free products that are more tolerable to such individuals. To assess the potential market for these products, the dairy commissioned a market research study of individuals over 40 years old in its sales area. A random sample of 250 individuals showed that 86 of them suffer from milk intolerance. Based on the sample results, calculate a 90% confidence interval for the population proportion that suffers milk intolerance. Interpret this confidence interval. a) First, show that it is okay to use the 1-proportion z-interval. b) Calculate by hand a 90% confidence interval. c) Provide an interpretation of your confidence interval. d) If the level of confidence was 95% instead of 90%, would the resulting interval be narrower or wider

Answer 1

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 )