Question

please write legible and explain the answer . i do not want spss answes that do not explain the problem.

A researcher conducts a study of white and black attitudes toward the police in her community. The percentage of a random sample of white respondents (N-300) who say they have a he percentage of a random sample of black favorable attitude toward the police is 51%. T respondents (N-400) who say they have a favorable attitude toward the police is 47%. You are asked if there is a real difference between the percentage of whites and blacks who have a positive attitude toward the police in the larger population, or is this sample difference likely to have occurred by random chance or sampling error. How do you respond? Explain your answer. Construct a 95% confidence interval for the proportion of Blacks in the population who . have a favorable attitude toward the police

Answer 1

ANSWER:

n1 = Size of sample 1 = 300

p1 = Proportion of Sample 1 = 0.51

n2 = Size of Sample 2 = 400

p2 = Proportion of Sample 2 = 0.47

H0: P1 = P2 (There is no real difference between the percentage of whites and blacks)

HA: P1 is not equal to P2 (There is real difference between the percentage of whites and blacks)

Estimate of population proportion is:

nln2

=  (300 X 0.51 + 400 X 0.47​) / (300+400)

​​​​​​   = 0.487

Q = 1 - P= 0.512

SE -VPQ(1/nl1/n2)

=SQRT(0.487X0.512(1/300 + 1/400)

= 0.03813

The test statistic is:

Z = (p1 - p2)/SE = (0.51 - 0.47)/0.3813 = 0.1049

$\alpha$ = 0.05.

From Table, Critical Value of Z = 1.96.

Since the calculated value of Z = 0.4132 is less than Table value of Z = 1.96, H0 is accepted.

Concluson:

There is no real difference between the percentage of whites and blacks.

To construct 95% Confidence Limits of Whites:

$\alpha$=0.05

So, Critical Z = 1.96

p1 = 0.51

So,

q1 = 1 - p1 = 0.49

So,

SE = =SQRT(0.487 X 0.512 / 300) = 0.0288

u/bd

So,

Confidence Interval:

p1 $\pm$ Z SE

= 0.51 $\pm$ 1.96 X 0.0288

= 0.51 $\pm$ 0.05644

= ( 0.5664, 0.453)