Consider random samples of size 58 drawn from population A with proportion 0.77 and random samples of size 70 drawn from population B with proportion 0.67 . (a) Find the standard error of the distribution of differences in sample proportions, . Round your answer for the standard error to three decimal places. standard error = Enter your answer in accordance to the question statement (b) Are the sample sizes large enough for the Central Limit Theorem to apply?

Step-by-step explanation:

a) The formula for determining the standard error of the distribution of differences in sample proportions is expressed as

Standard error = √{(p1 - p2)/[(p1(1 - p1)/n1) + p2(1 - p2)/n2}

where

p1 = sample proportion of population 1

p2 = sample proportion of population 2

n1 = number of samples in population 1,

n2 = number of samples in population 2,

From the information given

p1 = 0.77

1 - p1 = 1 - 0.77 = 0.23

n1 = 58

p2 = 0.67

1 - p2 = 1 - 0.67 = 0.33

n2 = 70

Standard error = √{(0.77 - 0.67)/[(0.77)(0.23)/58) + (0.67)(0.33)/70}

= √0.1/(0.0031 + 0.0032)

= √1/0.0063

= 12.6

the standard error of the distribution of differences in sample proportions is 12.6

b) the sample sizes are large enough for the Central Limit Theorem to apply because it is greater than 30

send it over

step-by-step explanation:

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8