# Performing a Chi-Square test for statistical significance of independence

August 30, 2017

estion
.9px;=”” currentcolor;=”” 0px=”” 12px=”” sans-serif;=”” sans”,=”” “open=”” 24px=”” 14px=””>

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 =

[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]

Get a 30 % discount on an order above \$ 100
Use the following coupon code:
RESEARCH
Positive SSL