Performing a Chi-Square test for statistical significance of independence

.9px;=”” currentcolor;=”” 0px=”” 12px=”” sans-serif;=”” sans”,=”” “open=”” 24px=”” 14px=””>GenderSleep Apnea? yesnoMen40360400Women12288300Total52648700here 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?

 

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