Help: Making inferences about the null and alternative hypothesis from t-tests Information about the data
This question is an extension from other questions at Quantile tables and confidence intervals and Quantiles tables and hypothesis testing. This particular question focuses on making inferences about the null and alternative hypothesis from t-tests. 
For this study, I would like to statistically test how the total length of sparrows affects their survival. Consider here one of the characteristics of the sparrows recorded was total length, which is assumed to be normally distributed. 
Data:


*

*In a group of 21 surviving sparrows, the variance of the total length was 11.05 mm2. 

*In a group of 28 sparrows that subsequently died, the variance of the total length was 15.07 mm2.  

*The µ of the total length of sparrows that survived was 157.4 mm2

*The mean µ of the total length of sparrows that subsequently died was 158.4 mm2
Problems to solve: 
I am unsure if I have understood the concepts correctly underpinning how to: -


*

*Make inferences from t-tests


If this is possible, I would be incredibly grateful if anyone could please check the answers to the questions stated below by providing advice. Many thanks in advance for your help. 
Question
The mean total length of sparrows that survived was 157.4 mm, and the mean total length of sparrows that subsequently died was 158.4 mm. Use a t-test to test the hypothesis that the population mean total length of sparrows does not depend on whether or not the sparrow survived. You should calculate the test statistic and state which distribution this test statistic should be compared with; however, you should be 0.327.
Hypothesis testing:


*

*H0: Total mean length of Sparrow = Total mean length of sparrow survival

*H1: Total mean length of Sparrow ≠ Total mean length of sparrow survival


The t-distribution is similar to a normal distribution, however, the variance is based on the degrees of freedom.
Definitions:


*

*Mean Length of Survivors (MLS)

*Mean Length of Non-Survivors (MLNS)


µ of survivors and non-survivors:


*

*µ1 (MLS) = 157.4 mm2

*µ2 (MLNS) = 158.4 mm2


Variance (S²) of survivors and non-survivors:


*

*S²(MLNS) =S1= 15.07 mm2

*S² (MLS) = S2= 11.05 mm2


The assumption of equal variances for MLS and MLNS groups of survival
-N1 = number of non-survivors (n=28)
-N2 = number of survivors (n=21)
Calculation of equal variance:

Calculate the combined variance for survivors and non-survivors using a two sample t-test:
SP = (28-1) * 15.07+ (21-1) * 11.05/(28+21-2)
SP = 406.89 + 221.0/47 = 627.89/47
SP = 13.3593617
Calculate two sample t-test:
T = (158.4-157.4)/ 13.3593617 * √1/28 + 1/21
T = 0.2592965
The test statistic should be compared to the F distribution.
The null hypothesis has not been rejected showing the mean total length of sparrows does not depend on whether or not the sparrow survived. The p-value is 0.327 which is larger than the significance level of 0.05; therefore, there is little evidence to abandon the null hypothesis
 A: First, a correction to your formula for $T$, above:
$T=\dfrac{\mu_2 - \mu_1}{\sqrt{S_p} \cdot \sqrt{\frac{1}{n_2}+\frac{1}{n_1}}}$
If you use this formula, the final value of $T=0.9478$.
You can look up this value in a t-distribution table, as suggested by @MichaelLew. 
In Stata, 
. ttesti 21 158.4 3.324154 28 157.4 3.8820098

Two-sample t test with equal variances
------------------------------------------------------------------------------
         |     Obs        Mean    Std. Err.   Std. Dev.   [95% Conf. Interval]
---------+--------------------------------------------------------------------
       x |      21       158.4    .7253899    3.324154    156.8869    159.9131
       y |      28       157.4    .7336309     3.88201    155.8947    158.9053
---------+--------------------------------------------------------------------
combined |      49    157.8286    .5215957     3.65117    156.7798    158.8773
---------+--------------------------------------------------------------------
    diff |                   1    1.055121               -1.122629    3.122629
------------------------------------------------------------------------------
    diff = mean(x) - mean(y)                                      t =   0.9478
Ho: diff = 0                                     degrees of freedom =       47

    Ha: diff < 0                 Ha: diff != 0                 Ha: diff > 0
 Pr(T < t) = 0.8259         Pr(|T| > |t|) = 0.3481          Pr(T > t) = 0.1741

The p-value for the two-sided null-hypothesis is 0.3481.
Good luck!
