Question

A social worker was interested in determining whether there is a significance difference in the average monthly cost per child for childcare outside the home between state supported facilities and privately owned facilities. Two independent random samples yielded the following information:

	State Supported Facilities (population 1)	Privately Owned Facilities (population 2)
Sample Size	64	64
Sample Mean	705	675
Standard Deviation	95	80

Which Excel will calculate the test statistic?

	a.	=(705-675)/(SQRT(95/64+80/64))
	b.	=(705-675)/(SQRT(95^2/8+80^2/8))
	c.	=(705-675)/(SQRT(95/8+80/8))
	d.	=(705-675)/(SQRT(95^2/64+80^2/64))

Answer 1

Given that,
mean(x)=705
standard deviation , sigma1 =95
number(n1)=64
y(mean)=675
standard deviation, sigma2 =80
number(n2)=64
null, Ho: u1 = u2
alternate, H1: μ1 != u2
level of significance, alpha = 0.05
from standard normal table, two tailed z alpha/2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
zo=705-675/sqrt((9025/64)+(6400/64))
zo =1.932
| zo | =1.932
critical value
the value of |z alpha| at los 0.05% is 1.96
we got |zo | =1.932 & | z alpha | =1.96
make decision
hence value of | zo | < | z alpha | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.932 ) = 0.05331
hence value of p0.05 < 0.05331,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: μ1 != u2
test statistic: 1.932
critical value: -1.96 , 1.96
decision: do not reject Ho
p-value: 0.05331
we do not have enough evidence to support the claim that there is a significance difference in the average monthly cost per child for childcare outside the home between state supported facilities and privately owned facilities
Answer:
option:D
test statistic: 1.932