In: Statistics and Probability
You are doing a chi-square test with a 5 by 9 table of counts (that is, 5 rows and 9 columns).
What is the degrees of freedom?
Suppose you get ?^2 = 45.23. Calculate the p-value.
Suppose that you want to see if there is association between income level (high/medium/low) and gender.
a)Write the null hypothesis.
b)Use StatCrunch to calculate the test statistic (that is, the x?2 value) and the p-value for this contingency table, which shows the income level for a random sample of adults cross classified by gender.
Low | Middle | High | Total | |
Male | 23 | 30 | 29 | 82 |
Female | 26 | 28 | 20 | 74 |
Total | 49 | 56 | 49 | 156 |
3.( YES / NO ) Based on the data in problem #2, there is a significant association between gender and income level (use significance level 0.05).
4.( YES / NO ) Based on the data in problem #2, there is a significant difference between the income levels for the two groups (males, females) (use significance level 0.01).
5.Use StatCrunch to find a 90% confidence interval to estimate the proportion of all females who have a high income level.
a)Write the null hypothesis.
Ans: Null hypothesis: There is no a significant association between row and column.
b)Use StatCrunch to calculate the test statistic (that is, the x 2 value) and the p-value for this
Ans: Suppose you get ?^2 = 45.23 and has 5 by 9 table of counts (that is, 5 rows and 9 columns). The estimated p-value for ?^2 = 45.23 at (5-1)(9-1)=32 degree of freedom is 0.0606.
2.
Null hypothesis: There is no a significant association between gender and income level
Alternative hypothesis: There is a significant association between gender and income level.
3)
Observed (O) | ||||
Low | Middle | High | Total | |
Male | 23 | 30 | 29 | 82 |
Female | 26 | 28 | 20 | 74 |
Total | 49 | 56 | 49 | 156 |
Expected ( E) | ||||
Low | Middle | High | ||
Male | 25.7564103 | 29.4359 | 25.75641 | |
Female | 23.2435897 | 26.5641 | 23.24359 | |
(O-E)^2/E | ||||
Low | Middle | High | ||
Male | 0.29498666 | 0.01081 | 0.408476 | |
Female | 0.32687711 | 0.077616 | 0.452636 |
Chi=square= Sum{(O-E)^2/E}= 1.5714
p-value =0.4558.
Ans: NO: Based on the data in problem #2, there is a significant association between gender and income level (use significance level 0.05).
4)
Ans: NO: Based on the data in problem #2, there is a significant difference between the income levels for the two groups (males, females) (use significance level 0.01).
5)