A Bank wishes to determine the mean income of its credit card holders and more...

Step 1
Since, we have to investigate whether there are any differences in the mean income between males and females we perform the t-test for two groups where the test statistics is given by,
${\displaystyle t=\frac{(\overline{x_1}-\overline{x_2})-({\mu }_1-{\mu }_2)}{\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2})}}}$
Where:
${\displaystyle n_1,n_2=}$ sample size for the male female groups
${\displaystyle s_1,s_2=}$ sample standard deviations for the two groups
${\displaystyle \overline{x_1},\overline{x_2}=}$ sample means for the male and female groups.
Step 2
To compute confidence intervals we can choose the confidence levels to be 0.01, 0.05, 0.10, and 0.20.
Thus, we can compute 99%, 95%, 90%, 80% confidence intervals.
Using the above t-test statistic we can compute the confidence interval as below:
${\displaystyle CI=(\overline{x_1}-\overline{x_2})\pm t_{\frac{\alpha }{2},(n_1+n_2-2)}\sqrt{\frac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}(\frac{1}{n_1}+\frac{1}{n_2})}}$
Step 3
Based on the confidence interval the following conclusions can be made:
If the calculated confidence interval does not contain zero in it, then we can conclude that the average males and females’ income differ.
If the calculated confidence interval contains zero in it, then we can conclude that the average males and females’ income do not differ.

Step 1
Suppose, random variables X and Y denote income of male and female credit card holders respectively.
Here, two different groups (male and female) are used to collect data. Further we do not know population standard deviation (or variance). So, we have to use test statistic corresponding to two sample t-test.
Corresponding test statistic is given by
${\displaystyle t=\frac{(\bar{x}-\bar{y})-({\mu }_x-{\mu }_y)}{s\times \sqrt{\frac{1}{n_x}+\frac{1}{n_y}}}}$
Here,
First sample size ${\displaystyle n_x}$
Second sample size ${\displaystyle n_y}$
Sample mean of first sample ${\displaystyle \bar{x}}$
Sample mean of second sample ${\displaystyle \bar{y}}$
Pooled sample standard deviation is given by
${\displaystyle s=\sqrt{\frac{(n_x-1)\times s_x^2+(n_y-1)\times s_y^2}{n_x+n_y-2}}}$
Degrees of freedom ${\displaystyle v=n_x+n_y-2}$
For ${\displaystyle 100(1-\alpha )\mathrm{\%}}$ confidence interval we have,
${\displaystyle P(-t_{\frac{\alpha }{2},v}<t<t_{\frac{\alpha }{2},v})=1-\alpha }$ Using R-code ${\displaystyle \sqrt{1-\frac{1-\alpha }{2},v}}$
${\displaystyle \Rightarrow P(-t_{\frac{\alpha }{2},v}<\frac{(\bar{x}-\bar{y})-({\mu }_x-{\mu }_y)}{s\times \sqrt{\frac{1}{n_x}+\frac{1}{n_y}}}<t_{\frac{\alpha }{2},v})=1-\alpha }$
${\displaystyle \Rightarrow P(-s{t}_{\frac{\alpha }{2},v}\sqrt{\frac{1}{n_x}+\frac{1}{n_y}}<(\bar{x}-\bar{y})-({\mu }_x-{\mu }_y)<s{t}_{\frac{\alpha }{2},v}\sqrt{\frac{1}{n_x}+\frac{1}{n_y}})=1-\alpha }$