Question

	Asian	Black	Latino	White
Sample Mean	83.70	73.50	80.60	79.00
Sample Variance	26.30	27.50	28.50	24.80
Sample Size	20	20	20	20

Part A.

According to The Boston Globe (August 16, 2010), Asian residents in Boston have the highest average life expectancy of any racial or ethnic group. The report shows, in particular, that Asians live a decade longer than black residents. Suppose the sample results below are indicative of the overall results. Calculate the F statistic.

1. 79.20
2. 13.67
3. 2.72
4. 26.70
5. None of the above

Part B.

Suppose the p-value for the above study = 0.00. Which of the following conclusions with respect to the study is correct?

1. At the 5% level, do not reject the null hypothesis. Average life expectancy differs across the four racial groups
2. At the 1% level, reject the null hypothesis. Average life expectancy is the same across the four racial groups
3. At the 1% level, reject the null hypothesis. Average life expectancy differs across the four racial groups
4. None of the above

Answer 1

Solution:

Part A.

According to The Boston Globe (August 16, 2010), Asian residents in Boston have the highest average life expectancy of any racial or ethnic group. The report shows, in particular, that Asians live a decade longer than black residents. Suppose the sample results below are indicative of the overall results. Calculate the F statistic.

Answer: d. None of the above.

Part B.

Suppose the p-value for the above study = 0.00. Which of the following conclusions with respect to the study is correct?

Answer: c. At the 1% level, reject the null hypothesis. Average life expectancy diers across the four racial groups

Explanation:

We have to find the ANOVA table to find the F test statistic:

Correction factor:

$\dpi{100} \small G=\sum T_{i}=n_{i}\sum \bar{X_{i}}=20 \times 83.70+20 \times 73.50+20 \times 80.60+20 \times 79=6336$

$\dpi{100} \small CF=\frac{G^2}{N}=\frac{6336^2}{80}=501811.2$

$\dpi{100} \small RSS=\sum n_{i}\left ( s_{i}^2+\bar{X_{i}}^2 \right )$

$\dpi{100} \small =20(26.3-83.7^2)+20(27.5-73.5^2)+20(28.5-80.6^2)+20(24.8-79^2)$

$\dpi{100} \small =505048$

Therefore, the total sum of squares is:

$\dpi{100} \small TSS=RSS-CF=505048-501811.2=3236.8$

The Grand mean is given below:

$\dpi{100} \small \bar{\bar{X}}=\frac{83.7+73.5+80.6+79}{4}=79.2$

Now the sum of squares between groups is given below:

$\dpi{100} \small TrSS=\sum n_{i}(\bar{x_{i}}-\bar{\bar{x}})^2$

$\dpi{100} \small =20(83.7-79.2)^2+20(73.5-79.2)^2+20(80.6-79.2)^2+20(79-79.2)^2$

$\dpi{100} \small =1094.8$

And now Error sum of squares is given below:

$\dpi{100} \small ESS=TSS-TrSS$

  $\dpi{100} \small =3236.8-1094.8$

  $\dpi{100} \small =2142$

Now the degrees of freedom:

The degrees of freedom for the total is $\dpi{100} \small N-1=80-1=79$

The degrees of freedom for the Between groups is $\dpi{100} \small k-1=4-1=3$

The degrees of freedom for the within groups is $\dpi{100} \small (N-1)-(k-1)=79-3=76$

Mean sum of squares:

$\dpi{100} \small MSTr=\frac{TrSS}{df}=\frac{1094.8}{3}=364.9333$

$\dpi{100} \small MSE=\frac{TrSS}{df}=\frac{2142}{76}=28.1842$

Now the F statistic:

$\dpi{100} \small F-statistic = \frac{MSTr}{MSE}=\frac{364.9333}{28.1842}=12.948$

Therefore, none of the options is correct