In a study done in Miami's economically disadvantaged community, it was determined that 38 out of 62 children who attended preschool needed social services later in life compared to 49 out 61 children who did not attend preschool. Test the claim that 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. Use a 0.05 level of significance.

Question

Alden

Mathematics

19 June, 03:25

In a study done in Miami's economically disadvantaged community, it was determined that 38 out of 62 children who attended preschool needed social services later in life compared to 49 out 61 children who did not attend preschool. Test the claim that 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. Use a 0.05 level of significance.

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Miriam David · Answer 1

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.