Question

A recent article in The Wall Street Journal compared the cost of adopting children from China with that of Russia. For a sample of 16 adoptions from China, the mean cost was $11,045, with a standard deviation of $835. For a sample of 18 adoptions from Russia, the mean cost was $12,840, with a standard deviation of $1,545. Can we conclude the mean cost is larger for adopting children from Russia

Answer 1

Answer:

We can conclude the mean cost is larger for adopting children from Russia.

Step-by-step explanation:

To answer this question we can perform an hypothesis testing on the differences of the means with the following null and alternative hypothesis:

$H_0: \mu_1\leq\mu_2\\\\H_1: \mu_1>\mu_2$

The significance level we assume is 0.05.

First we calculate the mean standard error:

$MSE=\frac{SSE}{df}=\frac{n_1\sigma_1^2+n_2\sigma_2^2}{(n_1-1)+(n_2-1)} =\frac{16*835^2+18*1545^2}{16-1+18-1} =1,691,314$

The harmonic mean of the sample sizes is:

$n_h=\frac{2}{1/n_1+1/n_2}=\frac{2}{0.0625+0.0556}=\frac{2}{0.1181}=16.94$

Then we can calculate the estimated standard deviation as:

$\sigma=\sqrt{\frac{2MSE}{n_h} } =\sqrt{\frac{2*1691314}{16.94} } =447$

The t-statistic can be calculated as:

$t=\frac{\Delta M-\Delta \mu}{\sigma}=\frac{(11045-12840)-0}{447} =\frac{-1795}{447}= -4.015$

The degrees of freedom are $df=n_1+n_2-2=18+16-2=32$.

We can look up the P-value in a table or applet. For t=-4.015 and df=32, the P-value is 0.00017.

The P-value is smaller than the significance level, so the effect is significant and the null hyphotesis (China cost is equal or less than in Russia) is rejected.

We can conclude the mean cost is larger for adopting children from Russia.