Answer:
Step-by-step explanation:
This is a test of 2 population proportions. Let 1 and 2 be the subscript for the children who attended preschool and need social services later in life and children who did not attend preschool and need social services later in life. The population proportion for the children who attended preschool and children who did not attend preschool would be p1 and p2 respectively.
p1 - p2 = difference in the proportion of proportion of the children who attended preschool and children who did not attend preschool .
The null hypothesis is
H0 : p1 = p2
p1 - p2 = 0
The alternative hypothesis is
Ha : p1 < p2
p1 - p2 < 0
it is a left tailed test
Sample proportion = x/n
Where
x represents number of success(number of complaints)
n represents number of samples
For children who attended preschool,
x1 = 38
n1 = 62
p1 = 38/62 = 0.61
For children who did not attend preschool,
x2 = 49
n2 = 61
p2 = 49/61 = 0.8
The pooled proportion, pc is
pc = (x1 + x2)/(n1 + n2)
pc = (38 + 49)/(62 + 61) = 0.71
1 - pc = 1 - 0.71 = 0.29
z = (p1 - p2)/√pc(1 - pc)(1/n1 + 1/n2)
z = (0.61 - 0.8)/√(0.71)(0.29)(1/62 + 1/61) = - 0.19/0.08183139728
z = - 2.32
Since it is a left tailed test, we would determine the probability for the area below the z score from the normal distribution table. Therefore,
p = 0.01
Since alpha, 0.05 > than the p value, 0.01, then we would reject the null hypothesis. Therefore, At a 5% level of significance, the the proportion of children who need social services later in life is greater amongst the children who did not attend preschool compare to the ones who did attend preschool.
