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

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Gender

Sleep Apnea?

 

 yes

no

Men

40

360

400

Women

12

288

300

Total

52

648

700

 


 

here is data from question 1. Need #2

 

Performing a Chi-Square test for statistical significance of independence between two categorical variables based on Question 1 (Using the same data as Question1).

 

 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?

 


 

  • 10 years ago
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