Posted by Bettie on Tuesday, April 16, 2013 at 6:05pm.
The president of a bank selected a sample of 1500 applications to check if the approval or rejection of an application depends on which one of the two loan officers handles the application. The information obtained from the sample is summarized in the following 2 way table.
Approved Rejected Total
Joe 750 300 1,050
Bill 250 200 450
Total 1,000 500
Test at the 1% significance level if the approval or rejection of a loan application depends on which loan officer handles the application.

Statistics  MathGuru, Tuesday, April 16, 2013 at 9:10pm
This looks like a chisquare test.
To find expected values for each cell, here is a formula you can use:
E = (row total)(column total)/n
...where E is the expected cell count if the null hypothesis is true.
The null hypothesis states the variables are unrelated in the population. The alternate or alternative hypothesis states the variables are related in the population. The purpose of a chisquare test is to determine if two or more variables are independent of each other (the null hypothesis) or are dependent (the alternate or alternative hypothesis). If the null is rejected, you can conclude that the variables are related in some manner or connected in some way. If the null is not rejected, you cannot conclude that the variables are related.
To calculate the chisquare statistic (after you have your expected values), you can use a formula like the following:
Chi sq = the sum of [(O  E)^2/E]
...where ^2 means squared.
(Take each cell, subtract the expected value from the observed value, square it, then divide by the expected value. Do this for each cell. Then add all the values together for the chisquare statistic.)
Using a table for critical or cutoff values for the chisquare, find the critical value using .01 level with degrees of freedom. Degrees of freedom = (r  1)(c  1) ...where r = number of rows and c = number of columns. If your test statistic exceeds the critical value from the table, then the null is rejected. If your test statistic does not exceed the critical value from the table, then the null is not rejected.
I hope this will help get you started.