Aresearcher claims that the average age of a woman before she has her first child is greater than the 1990 mean age of 26.4 years, on the basis of data from national vital statistics report, vol. 48, no. 14. she wants to obtain a simple random sample of women who gave birth to their first child in 1999. assuming that the standard deviation is 6.4, what sample size is required to get a margin of error of 0.5 years for a 95% confidence interval?

the sample size must be 629.

Step-by-step explanation:

In order to determine the sample size n, we must use the formula:

n = (z_(α/2) × δ / ε)²

where

Step 1:

Determine the z-value:

Since this is at confidence level of 95%, therefore,

α = 1 - 95%

  = 0.05

Thus,

z_(α/2) = z_(0.05/2) = z_(0.025)

Looking at the table of standard normal probabilities, we find the z_(0.025) = 1.96

Step 2:

Calculate n:

Therefore,

n = (z_(α/2) × δ / ε)²

n = ( (1.96×6.4) / 0.5 )²

n = 629

Therefore,  the sample size must be 629.

tigers

step-by-step explanation:

cougars 12/10 (12-10 = 2)

bobcats 20/10 (20-10 =10)

bulldogs 8/5 (8-5 =3)

tigers 3/5 (3-5 =-2)

725 parts per hour.

step-by-step explanation:

hi there, this answer is actually really simple.

you can set it up as 1450/2 or 3625/5 to find the answer.

they want to know how much per hour is being made, so you just divide.

1450/2=725.

and that's how many are produced per hour.

hope this . : )