Question

Suppose a random sample of n measurement is selected from a population with mean My=100, and variance oy2=100. For each of the following values of n, calculate the mean and standard erro of the sampling distribution of the sample mean y.

A) n=64

B) n=81

C) n=100

D) n=1000

Book, 4,8 Supplementary problems. 1. Suppose a Hy -100, and variance o,2100. For each of the following values of n, calculate the mean and standard error of the sampling distribution of the sample mean-. a) n-64 b) n-81 c) n-100 d) n= 1000 random sample of n measurements is selected from a population with mea random sample of n-100 observations is selected from a population with Hy-30 2. A and ơ,-16. Find the following proportions: a) P(y> 28) b) P(28.2< y < 32) Due: 3/27.

Answer 1

Solution:-
1) Given that information μy = 100 , σ^2y = 100 , σy = sqrt(100) = 10

a) n = 64

mean = 100

standard deviation = 10/sqrt(64) = 1.25

b) n = 81

mean = 100

standard error = 10/sqrt(81) = 1.11

c) n = 100

mean = 100

standardard error = 10/sqrt(100) = 1

d) n = 1000

mean = 100

standard error = 10/sqrt(1000) = 0.3162

2) GIven information n = 100 , μy = 30 , σy = 16

a) P(y > 28) = P(Z > (28 - 30)/(16/sqrt(100))

= P(Z > -1.25)

=  0.8944

b) P(28.2 < y < 32) = P((28.2 - 30)/(16/sqrt(100) < Z < (32 - 30)/(16/sqrt(100))

= P(-1.125 < Z < 1.25)

= 0.7652

Formula Z = (y - μ)/(σ/sqrt(n))