![Null & Alternative
Hypothesis
Step 1: H0: P = 0.80
[Not more than 80% of the customers are satisfied ]
Step 2: H1: P > 0.80
[More than 80% of the customers are satisfied ]
(One-tailed test)](https://image.slidesharecdn.com/proptestusingchi-square-220124171947/85/Proportion-test-using-Chi-square-7-320.jpg)
![Null & Alternative
Hypothesis
Step 1: H0: 𝑃1 = 𝑃1
[There is no significant difference in likeness of the two brands]
Step 2: H1: 𝑃1 ≠ 𝑃1
[There is significant difference in likeness of the two brands]
(Two-tailed test)](https://image.slidesharecdn.com/proptestusingchi-square-220124171947/85/Proportion-test-using-Chi-square-16-320.jpg)
Proportion test using Chi square
- 1. TEST FOR PROPORTION(S) USING SPSS
- 2. TEST FOR PROPORTION(S) There are mainly two tests for proportion • Test for single proportion • Test for significant difference between proportion of two samples SPSS does not have Z test to test the proportion(s). But Chi-Square test can be used to the test the proportion in SPSS.
- 3. Chi- Square Statistic If X~𝑁(𝜇, 𝜎2) then 𝑍 = 𝑋−𝜇 𝜎 ~𝑁(0,1) and 𝑍2 = 𝑋−𝜇 𝜎 2 ~𝜒2 with 1 d.f. and 𝑋−𝜇 𝜎 2 ~𝜒2 with n d.f.
- 4. Test of Single Proportion Step 1: Null hypothesis H0: 𝑃 = 𝑃0 Step 2: Alternative hypothesis H1: 𝑃 ≠ 𝑃0 or 𝑃 > 𝑃0 or 𝑃 < 𝑃0 Step 3: Test statistics 𝑧 = 𝑝 −𝑃0 𝑃0 ∗𝑄0 𝑛 Denominator is the Standard Error of sample proportion i.e. S.E.(𝑝) Step 4: Table value of z at 𝛼 % level of significance Step 5: If z ≤ z table value, H0 is Accepted If z > z table value, H0 is Rejected
- 5. Test of Single Proportion Step 1: Null hypothesis H0: 𝑃 = 𝑃0 Step 2: Alternative hypothesis H1: 𝑃 ≠ 𝑃0 or 𝑃 > 𝑃0 or 𝑃 < 𝑃0 Step 3: Check Assumptions Step 4: Test statistic – Chi-square and p value Step 5: Conclusion • If p ≤ Level of significance (∝), We Reject Null hypothesis • If p > Level of significance (∝), We fail to Reject Null hypothesis
- 6. Case Study The CEO of a large water pump manufacturing company claims that more than 80 percent of his customers are satisfied with the water pump. To test this claim, the local newspaper surveyed 100 customers, using simple random sampling. Among the sampled customers, 73 percent say they are very satisfied. Based on these findings, what can we say about the manufacturer’s claim ? Use a 0.05 level of significance.
- 7. Null & Alternative Hypothesis Step 1: H0: P = 0.80 [Not more than 80% of the customers are satisfied ] Step 2: H1: P > 0.80 [More than 80% of the customers are satisfied ] (One-tailed test)
- 8. Data in SPSS The local newspaper surveyed 100 customers, using simple random sampling. Among the sampled customers, 73 percent say they are very satisfied
- 10. Chi-Square test
- 11. Chi-Square test
- 12. Output The Chi-square value is 3.063 and p value = (0.080)/ 2 = 0.04 As p < 0.05, We reject the Null hypothesis. ∴ More than 80% of the customers are satisfied 𝑧 = 𝑝 −𝑃 𝑃∗𝑄 𝑛 = 0.73 −0.80 0.80∗0.20 100 = 1.75 Chi-Square = Square of z value
- 13. Test of significance of difference between two proportions Step 1: Null hypothesis H0: 𝑃1= 𝑃2 Step 2: Alternative hypothesis H1: 𝑃1 ≠ 𝑃2 or 𝑃1 > 𝑃2 or 𝑃1 < 𝑃2 Step 3: Test statistics 𝑧 = 𝑝1 − 𝑝2 𝑃1∗𝑄1 𝑛1 + 𝑃2∗𝑄2 𝑛2 or 𝑝1 − 𝑝2 𝑃∗𝑄 ( 1 𝑛1 + 1 𝑛2 ) where 𝑃 = 𝑛1𝑝1+𝑛2𝑝2 𝑛1+𝑛2 Denominator is the Standard Error of difference in sample proportion i.e. S.E.(𝑝1 − 𝑝2) Step 4: Table value of z at 𝛼 % level of significance Step 5: If z ≤ z table value, H0 is Accepted If z > z table value, H0 is Rejected
- 14. Z test for two proportions Step 1: Null hypothesis H0: 𝑃1 = 𝑃2 Step 2: Alternative hypothesis H1: 𝑃1≠ 𝑃2 or 𝑃1 > 𝑃2 or 𝑃1 < 𝑃2 Step 3: Check Assumptions Step 4: Test statistic – Chi-square and p value Step 5: Conclusion • If p ≤ Level of significance (∝), We Reject Null hypothesis • If p > Level of significance (∝), We fail to Reject Null hypothesis Assumptions Tests 𝑛1 × 𝑝1 > 10 and 𝑛1 × (1 − 𝑝1) > 10 𝑛2 × 𝑝2 > 10 and 𝑛2 × (1 − 𝑝2) > 10 ____
- 15. Case Study Two brands of coffee - Starbucks and CCD were to be compared. A sample of 100 under- graduate students were selected and of them randomly 50 students were given Starbucks coffee and remaining 50 were given CCD coffee. Coffee were served to them without brand labels. 42 students liked the taste of Starbucks coffee while 45 students liked the taste of CCD coffee. Is there a significant difference in likeness of the two brands for the under-graduate students?
- 16. Null & Alternative Hypothesis Step 1: H0: 𝑃1 = 𝑃1 [There is no significant difference in likeness of the two brands] Step 2: H1: 𝑃1 ≠ 𝑃1 [There is significant difference in likeness of the two brands] (Two-tailed test)
- 17. Data in SPSS 50 students were given Starbucks coffee and remaining 50 were given CCD coffee. Coffee were served to them without brand labels. 42 students liked the taste of Starbucks coffee while 45 students liked the taste of CCD coffee
- 19. Chi-Square test
- 20. Chi-Square test
- 21. Chi-Square test
- 22. Output The Chi-square value is 0.796 and p value = 0.372 > 0.05. We fail to reject the Null hypothesis. ∴ There is no significant difference in likeness of the two brands among under-graduate students Z= 𝑝1 − 𝑝2 𝑃∗𝑄 ( 1 𝑛1 + 1 𝑛2 ) = 0.84−0.90 0.87∗0.13 ( 1 50 + 1 50 ) = 0.8955 Chi-Square = Square of z value
- 23. THANK YOU Dr Parag Shah | M.Sc., M.Phil., Ph.D. ( Statistics) pbshah@hlcollege.edu www.paragstatistics.wordpress.com