50 likes | 124 Views
Hotel Example, page 425 Solved Using the Z test for difference in two population proportions Manual Calculations. The Application
E N D
Hotel Example, page 425 Solved Using the Z test for difference in two population proportions Manual Calculations The Application Is there a significant difference in the proportion of guests that would return to the Beachcomber different from the proportion of guest that would return to the Windsurfer? Z +3.009 -3.009 pvalue is 0.0013+0.0013=0.0026
Hotel Example, page 425 Solved Using the Z test for difference in two population proportions NCSS
Hotel Example, page 425 Solved Using the Chi-Squared Test Manual Calculations The Chi-Squared test is a general/broad test that is often used to test whether or not two variables are dependent on each other. One of the the two variables is a row in a table the other variable is a column in a table. The entire table is called a contingency table. Chi-Squared tests take the general form below: Although this looks like a two-tail test, the Chi-squared test is always a one-tail test, because it is based on the normal distribution squared, hence its name. All chi-squared values are positive. The example in this handout shows the equivalent two-tailed Z test with the same result (the pvalue is the same). With the chi-squared test we can test the equivalence of multiple proportions or groups, not just 2 groups, the capacity for the corresponding Z-test. area or pvalue is 0.0026 =CHIDIST(test stat value, df) =CHIDIST(9.0526,1)=0.0026 9.0526 Use the test stat for obtaining the pvalue Notice the chi-squared test stat* of 9.0526 is our Z* test stat* squared!
Hotel Example, page 425 Solved Using the Chi-Squared Test NCSS
Hotel Example The Application Is the proportion of guests that would return to the Beachcomber more than 10% higher than the proportion of guests that would return to the Windsurfer? Beyond BSTAT 5325 course scope pvalue is 0.24 Z 0.71