Question

Belsky, Weinraub, Owen, and Kelly (2001) reported on the effects of preschool childcare on the development of young children. One result suggests that children who spend more time away from their mothers are more likely to show behavioral problems in kindergarten. Using a standardized scale, the average rating of behavioral problems for kindergarten children is µ = 35. A sample of n = 16 kindergarten children who had spent at least 20 hours per week in child care during the previous year produced a mean score of M=42.7 with a standard deviation of s=6. (a) Are the data sufficient to conclude that children with a history of child care show significantly more behavioral problems than the average kindergarten child? Use a one-tail test with α = .01. (b) Compute the 90% confidence interval for the mean rating of behavioral problems for the population of kindergarten children who have a history of child care.

Answer 1

Explanation:

a) We would set up the hypothesis test. This is a test of a single population mean since we are dealing with mean

For the null hypothesis,

µ = 35

For the alternative hypothesis,

µ > 35

It is a right tailed test

Since the number of samples is small and no population standard deviation is given, the distribution is a student's t.

Since n = 16,

Degrees of freedom, df = n - 1 = 16 - 1 = 15

t = (x - µ)/(s/√n)

Where

x = sample mean = 42.7

µ = population mean = 32

s = samples standard deviation = 6

t = (42.7 - 32)/(6/√16) = 7.13

We would determine the p value using the t test calculator. It becomes

p = 0.00001

Since alpha, 0.01 > than the p value, 0.00001, then we would reject the null hypothesis. Therefore, At a 1% level of significance, there is sufficient data to conclude that children with a history of child care show significantly more behavioral problems than the average kindergarten child

b) Confidence interval is written in the form,

(Sample mean - margin of error, sample mean + margin of error)

The sample mean, x is the point estimate for the population mean.

Margin of error = z × s/√n

the information given, the from population standard deviation is unknown and the sample size is small, hence, we would use the t distribution to find the z score

In order to use the t distribution,

Since confidence level = 90% = 0.95, α = 1 - CL = 1 – 0.90 = 0.1

α/2 = 0.1/2 = 0.05

the area to the left of z0.05 is 0.05 and the area to the right of z0.05 is 1 - 0.05 = 0.95

Looking at the t distribution table for t.95 and df = 15

z = 1.753

Margin of error = 1.753 × 6/√16

= 2.63

Confidence interval = 35 ± 2.63