Question

Problem 3 There are60 lamb snacks in a pack of Doggy Meat Snacks". The total number of snacks in the bag is 100, we wish to cala late 90% confidence interal for a population proportion of lamb snacks. a. (396) Which distribution should you use for this problemEreresn Uniform Normal Poisson Binomial b. (3%) The sample proportion is c. (3%) The critical value Zerit of the standard normal distribution for the 90% confidence interval is d. (3%) The standard error fr p is standard error e. (396) The error bound for the proportion (EBP) is f" (3%) The confidence interval is given by (L, U) , where the lower bound is L and the upper bound is U

Answer 1

Given data are x= 60 lamp snacks

And n=100 snacks

a) we use normal distribution

b) the sample proportion is  p'= x/n = 60/100 = 0.60

c) at 90% confidence in 2 tail from z table we have z critical value z crit = 1.645

d) standard error for p is =  √(p'q'/n)

Whwre q'=1-p'=1-0.60=0.40

=sqrt ( 0.60 × 0.40 /100)

=Sqrt(0.0024) = 0.04899

e) the error bound for the proportion EBP = (zα/2)√(p'q'/n)

= 1.645×0.04899

=0.08059= 0.0806

f) Lower bound L= P' - EBP = 0.60- 0.0806 =0.5194

the upper bound is U =P' +EBP= 0.60 + 0.0806 = 0.6806