Performing a Chi-Square test for statistical significance of independence between two

 a. For the Sleep apnea problem (Question 1), indicate the specific null and alternative hypotheses that will be used with a Chi-Square Test. Write the null hypothesis and alternative hypothesis for this 2 x 2 table.

b. Calculate the expected counts using the method shown in the example in the online notes 11.2 . Show all work.  (Hint: Expected Count = n_row*n_column/n_total)

c. Interpret all four expected counts in the context of the problem.

(We would expect ____, if in fact there is no relationship between _____ and _____ in the population)   You will have four answers--one for each expected count. 

d. Calculate the Chi-Square Statistics by hand using the formula    

e. Determine a p-value associated with the test statistics you calculated in previous question. Write down the degree of freedom and p-value from TABLE.

DF = (# rows -1) (# columns - 1)

p-value =         

f. Confirm your p-value using Minitab. Copy and Paste your output.

[Hint: Minitab Express User: Statistics > Distribution Plots > Display Probability > Distribution > Chi-Square distribution > putting degree of freedom in the box > select “ a specified x value” > Right tail > Putting the chi-square statistics you calculated by hand > OK]

g. what is your conclusion based on your chi-square test? Reasoning?