Question

A study to compare the mean level of LDL cholesterol in male and female students who do not exercise regularly gave the data shown in the table. We may assume that the populations of LDL level are normally distributed.
n

LaTeX: \overline{X} X ¯
S

Male

35

109

31

Female

60

96

26

A 90% confidence interval for the difference in mean LDL cholesterol level between mean and women is

13 ± 8.3
13 ±11.6
13 ± 9.8
13 ± 10.5
13 ± 7.6

Answer 1

Female:                                     Male:
60                                              35
96                                             109
26                                              31
n₁=no. of entries=3                    n₂=no. of entries=3
s₁=sample standard                   s₂=sample standard 
      deviation=35.005                       deviation=43.924
x₁=sample mean=60.666           x₂=sample mean=58.333

Here s and x are calculated from calculator.
Now,
for significance level
1-α=90%
1-α=0.9
α=0.1
α/2=0.05

As population variances are unknown for both Male and female so the best suitable formula is

    (x₁ - x₂) ± tα/2,v₁ * $\sqrt{ \frac{s_1^2 }{n_1} + \frac{s_2^2}{n_2} }$
   ........1

where v(degree of fredom)=(s₁²/n₁+s₂²/n₂)² / ((s₁²/n₁)²/n₁-1)+(s₂²/n₂)²/n₂-1))
    by putting values 
           v=3.81026
From Statistics table,
      See the t table at α/2=0.05  and v=3.81026
use interpolation as v value is not in the table,
      3.81026-3/4-3 = x-3.182/2.776-3.182
                          x= 2.853
by putting values in eq. 1 we get
  
(60.666-58.333) ± (2.853)*$\sqrt{ \frac{35.005^2 }{3} + \frac{43.924^2}{3} }$      

  2.333 ± 92.516 is the correct answer.