Can you help with another statistics and significance level question?

Question

I posted a statistics problem a few minutes ago. Here is another one. Same deal, I need step by step instructions on how to solve the problem and why the test is chosen.

A sample of employees at a large chemical plant was asked to indicate a preference for one of three pension plans. The results are given in the following table. Does it seem that there is a relationship between the pension plan selected and the job classification of the employees? Use the .01 significance level.

Job Class --- Plan A --- Plan B --- Plan C
Supervisor --- 10 -------- 13 -------- 29
Clerical -------- 19 -------- 80 ------- 19
Labor ---------- 81 -------- 51 -------- 22

Thanks for the help to all who respond.

Answer 1

The place to get step-by-step instructions is from your notes or textbook or by visiting with your instructor. Each class covers this somewhat differently, and you'll want to be sure you're doing it the way you should.

To get you started, though, this will be a chi-square test often called a "test for independence" or a "matrix test". The fact that the information is arranged in a table showing two different variables (pension plan and job classification) is the giveaway to what test it is.

Most people today use either a graphing calculator or a software package to compute a value of chi-square for their data. Alternatelly you can use the formula X^2 = SIGMA( (O - E)^2 / E), where "O" is each observed value, and "E" is each associated expected value.

You will compare your calculated value with a value of chi-square you'd find in a table. To find that value you need to know a number of degrees of freedom. The formula is d.f. = (# of rows - 1) times (# of columns - 1). Since there are 3 rows and 3 columns in this problem, you'd take 2 x 2, which gives you 4 degrees of freedom. The critical value for 4 d.f. and a .01 level of significance is 13.28. If your calculated value is more than 13.28, you'd have a significant result; otherwise you won't.

Answer 2

a million) For the two a 2-tail and a million-tail try the p-cost could desire to be <0.01 as a results of fact P(Z>=4.6) <0.00003. 2) The z-cost could be adverse for the alternative to be mu<5 and it relatively is not any longer. 3) The null is very nearly certainly to be rejected as a results of fact the p-cost is so small. 4)that's right, see above.