Decide whether the normal sampling distribution can be used. If it can be used, test the claim about the population proportion p at the given level of significance alpha using the given sample statistics. Claim: pnot equals0.26; alphaequals0.10; Sample statistics: ModifyingAbove p with caretequals0.23, nequals100

7/2

(decimal: 3.5)

welcome ; )

step-by-step explanation:

a) 40 more; b) old factory:   10%, new factory:   16%; c) week 4

explanation:

a) at week 0, we see in the graph that the new factory produces 190 specialty items.  

using the function, substituting 0 in for w,

this is a difference of 230-190 = 40.

b) the function for the old factory is written in the form f(x) = a(1+r)ˣ, where a is the original amount, x is the amount of time and r is the growth rate.   the equation for the old factory is ; 230 is in place of a and is the original amount; w is the amount of time; which means 1+r = 1.1.   this means that r must be 0.1 or 10%.

for the new factory, we find the percent of increase to find the growth rate; the formula for this is

between the first two points, the amount of change is 220-190 = 30; this gives us

30/190 = 0.158 ≈ 0.16 = 16%

c) at week 0, the old factory has 230 while the new factory has 190.

at week 1, the old factory has 230(1.1) = 253 while the new factory has 220.

at week 2, the old factory has 230(1.1²) = 278.3 while the new factory has 252.

at week 3, the old factory has 230(1.1³) = 306.13 while the new factory has 290.

at week 4, the old factory has 230(1.1⁴) = 336.743 while the new factory has 337.   this is the first week that the new factory surpasses the old.

6950 is the