Question

A childcare agency was interested in examining the monthly amount that family pay for childcare outside the home per child. A random sample of 64 families was selected and the average and standard deviation were computed to be $675 and $40 respectively.  

1. Find a 99% confidence interval for the monthly average amount spent per child.
2. A social worker claims that the average monthly amount spent for childcare outside the home is $700 per child. You suspect it is less. Based on the sample data, can you reject the social worker’s claim?

Answer 1

a)

sample mean, xbar = 675
sample standard deviation, s = 40
sample size, n = 64
degrees of freedom, df = n - 1 = 63

Given CI level is 99%, hence α = 1 - 0.99 = 0.01
α/2 = 0.01/2 = 0.005, tc = t(α/2, df) = 2.656

CI = (675 - 2.656 * 40/sqrt(64) , 675 + 2.656 * 40/sqrt(64))
CI = (661.72 , 688.28)

b)

Below are the null and alternative Hypothesis,
Null Hypothesis: μ = 700
Alternative Hypothesis: μ < 700

Test statistic,
t = (xbar - mu)/(s/sqrt(n))
t = (675 - 700)/(40/sqrt(64))
t = -5

P-value Approach
P-value = 0
As P-value < 0.05, reject the null hypothesis.