Homogeneity or heterogeneity of variance The international comparative school performance study PIRLS raised in 2011 the reading competences of the fourth graders in more than $40$ countries, among others in Germany. 
They would now like to investigate how the cultural capital of the parents affects the reading competencies of the children in the fourth class.
They operationalize the cultural capital as the volume of the books in the parents' house. 
They want to find out if there is a significant difference between children whose parents have over $100$ books, and children whose parents have a maximum of one hundred books.
A table of statistic is given: 
 
where Mittelwert=mean value, Standardabweichung=standard deviation, Standardfehler des Mittelswertes=standard error of mean value. 
We have to compute a T-Test to find out, if we can confirm out hypothesis and to get a statistic significant relation. 
First we have to prove with a F-Test, if we have to apply a double T-Test  or a Test of Welch. 
At the F-Test and the T-Test we have a significance level of $5 \%$. 
$$$$ 
How can we check what T-Test we have to apply?  
The null hypothesis is that when the parents have more than $100$ books then the children are better in reading than others, or not? 
Do we get from that that we don't need a double T-Test? 
Also using the F-Test how can we check if there is an homogeneity or a heterogeneity of variance? Do  we have to use this formula ?
 A: Welch t test. Unless you have good reason from prior experience with such data, you should
not assume that the population variances for Strong and Weak are equal.
In your case, I think you should begin with a Welch ('separate variances')
two-sample t test. I assume the formula is in your book. (Including, a
somewhat complicated additional formula for finding degrees of freedom.)
Here is output from Mintab 17 statistical software. I have rounded means
and SDs, but that should not make a consequential difference in the results.
 Two-Sample T-Test and CI 

 Sample     N   Mean  StDev  SE Mean
 1       1541  555.3   64.9      1.7
 2       1598  512.9   70.8      1.8

 Difference = μ (1) - μ (2)
 Estimate for difference:  42.37
 95% CI for difference:  (37.62, 47.12)
 T-Test of difference = 0 (vs ≠): 
     T-Value = 17.49  P-Value = 0.000  DF = 3129

The T-value 17.49 is quite large. If $|T| > 1.96,$ then you would
reject $H_0: \mu_1 = \mu_2$ against $H_a: \mu_1 \ne \mu_2$ at the 5%
level of significance. 
The P-value is the probability that a value
as far from 0 or farther than 17.49 would occur due to sampling
error if $H_0$ were true. Here it is $P(|T| \ge 17.49) < 0.0005,$
computed using $T \sim \mathsf{T}(\nu = 3129).$
One-sided alternative. One might assume that people from homes with many books would
score generally higher on PIRLS. In that case, one might want
to do a one-sided test of $H_0: \mu_1 = \mu_2$ against $H_a: \mu_1 > \mu_2.$
Then the P-value would be half the size as for a two-sided test (but
still essentially $0$, reported as 0.000 in the software).
I think by 'double t test' you mean two-sided t test. If the researchers
anticipated before seeing data, that the scores would be higher among
students from homes with more books, then they should use a one-sided test.
F-test for equal variances. Because you are explicitly asked to do an F-test to determine whether
the data are consistent with equal variances in the two populations, you
should do that. The test statistic F is the ratio of the two sample variances.
For convenience using tables, I would put the larger sample variance in the numerator:
$F \approx 70.75^2/64.93^2 = 1.187304.$
The critical value for an F-test with such large degrees of freedom
will not be shown in most printed tables, but might guess from looking
at the largest available numerator and denominator DF than the critical
value is around 1.10. So (even though the SDs seem close together)
you have enough data to detect a difference in population variances.
[The reason for putting the larger sample variance in the numerator
is that F-tables do not usually give information for F-values smaller
than 1.]
Here is Minitab output: I entered sample SDs (smaller one first).
 Test and CI for Two Variances 

 Method

 Null hypothesis         σ(First) / σ(Second) = 1
 Alternative hypothesis  σ(First) / σ(Second) ≠ 1
 Significance level      α = 0.05

 F method was used. This method is accurate for normal data only.

 Ratio of standard deviations = 0.918
 Ratio of variances = 0.842

 Test

                          Test
 Method   DF1   DF2  Statistic  P-Value
 F       1540  1597       0.84    0.001

The test statistic given here is the reciprocal of the one I gave above:
$1/1.187 \approx 0.84.$ The P-value  0.001 says you can reject $H_0$ at
the 0.1% level--or any greater level such as 5%.
Note:  I believe it is now established statistical practice to use the
Welch t test (instead of the 'pooled' test, which assumes equal population
variances) unless there is excellent prior evidence, not just based on
the data at hand, that the populations have equal variances.
There are several reasons for not letting an T-test "decide" which t test to do. (1) The F-test has very low power. That is,
it frequently does not detect population variances are unequal even when
they are. (2) Doing a pooled t test when population variances are
unequal can lead to making the wrong decision extraordinarily often. (3)
If you do the F-test at the 5% level and then do one of the t tests at
the 5% level, the significance level of the 'hybrid' combination of
tests is not clear.
Many simulations have been done under various circumstances comparing the
Welch test by itself against doing an F-test to decide between Welch and
pooled t tests. Doing the Welch test straightaway is better, sometimes
much better. You can google for papers on this on the Internet. I have
repeated some of these simulations for myself to verify that I get the
same results, because nowadays we have the computer power to do more accurate simulations.
