Download presentation

Presentation is loading. Please wait.

Published byChandler Cheatle Modified over 2 years ago

1
1 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Two Sample Proportions Large Sample Difference of Proportions z Test & Confidence Interval

2
2 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Large-Sample Inferences Difference of Two Population (Treatment) Proportions Some notation:

3
3 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Properties: Sampling Distribution of p 1 - p 2 If two random samples are selected independently of one another, the following properties hold: 3.If both n 1 and n 2 are large [n 1 p 1 10, n 1 (1- p 1 ) 10, n 2 p 2 10, n 2 (1- p 2 ) 10], then p 1 and p 2 each have a sampling distribution that is approximately normal, hence the sampling distribution of p 1 – p 2 is approximately normal. 2. 1.

4
4 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Large-Sample z Tests for 1 – 2 = 0 If we want to test the hypothesis 1 – 2 = 0, we assume that 1 – 2 = 0 is true so the combined estimate of the common population proportion becomes

5
5 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Large-Sample z Tests for 1 – 2 = 0 Null hypothesis: H 0 : 1 – 2 = 0 Assumptions: 1.The samples are independently chosen random samples OR treatments are assigned at random to individuals or objects (or vice versa). 2.Both sample sizes are large: n 1 p 1 10, n 1 (1- p 1 ) 10, n 2 p 2 10, n 2 (1- p 2 ) 10 Test statistic:

6
6 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Large-Sample z Tests for 1 – 2 = 0 Alternate hypothesis and finding the P-value: 1.H a : 1 - 2 > 0 P-value = Area under the z curve to the right of the calculated z 2.H a : 1 - 2 < 0 P-value = Area under the z curve to the left of the calculated z 3.H a : 1 - 2 0 i.2(area to the right of z) if z is positive ii.2(area to the left of z) if z is negative

7
7 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Example - Student Retention A group of college students were asked what they thought the “issue of the day”. Without a pause the class almost to a person said “student retention”. The class then went out and obtained a random sample (questionable) and asked the question, “Do you plan on returning next year?” The responses along with the gender of the person responding are summarized in the following table. Test to see if the proportion of students planning on returning is the same for both genders at the 0.05 level of significance?

8
8 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 1 = true proportion of males who plan on returning 2 = true proportion of females who plan on returning n 1 = number of males surveyed n 2 = number of females surveyed p 1 = x 1 /n 1 = sample proportion of males who plan on returning p 2 = x 2 /n 2 = sample proportion of females who plan on returning Null hypothesis: H 0 : 1 – 2 = 0 Alternate hypothesis: H a : 1 – 2 0 Example - Student Retention

9
9 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Assumptions: The two samples are independently chosen random samples. Furthermore, the sample sizes are large enough since n 1 p 1 = 211 10, n 1 (1- p 1 ) = 64 10 n 2 p 2 = 141 10, n 2 (1- p 2 ) = 41 10 Significance level: a = 0.05 Example - Student Retention Test statistic:

10
10 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Calculations : Example - Student Retention

11
11 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. P-value: The P-value for this test is 2 times the area under the z curve to the left of the computed z = -0.19. P-value = 2(0.4247) = 0.8494 Conclusion: Since P-value = 0.849 > 0.05 = , the hypothesis H 0 is not rejected at significance level 0.05. There is no evidence that the return rate is different for males and females.. Example - Student Retention

12
12 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Example A consumer agency spokesman stated that he thought that the proportion of households having a washing machine was higher for suburban households then for urban households. To test to see if that statement was correct at the 0.05 level of significance, a reporter randomly selected a number of households in both suburban and urban environments and obtained the following data.

13
13 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. 1 = proportion of suburban households having washing machines 2 = proportion of urban households having washing machines 1 - 2 is the difference between the proportions of suburban households and urban households that have washing machines. H 0 : 1 - 2 = 0 H a : 1 - 2 > 0 Example

14
14 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Assumptions: The two samples are independently chosen random samples. Furthermore, the sample sizes are large enough since n 1 p 1 = 243 10, n 1 (1- p 1 ) = 57 10 n 2 p 2 = 181 10, n 2 (1- p 2 ) = 69 10 Significance level: = 0.05 Test statistic: Example

15
15 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Calculations: Example

16
16 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. P-value: The P-value for this test is the area under the z curve to the right of the computed z = 2.39. The P-value = 1 - 0.9916 = 0.0084 Conclusion: Since P-value = 0.0084 < 0.05 = , the hypothesis H 0 is rejected at significance level 0.05. There is sufficient evidence at the 0.05 level of significance that the proportion of suburban households that have washers is more that the proportion of urban households that have washers. Example

17
17 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. Large-Sample Confidence Interval for 1 – 2 When 1.The samples are independently selected random samples OR treatments that were assigned at random to individuals or objects (or vice versa), and 2.Both sample sizes are large: n 1 p 1 10, n 1 (1- p 1 ) 10, n 2 p 2 10, n 2 (1- p 2 ) 10 A large-sample confidence interval for 1 – 2 is

18
18 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. A student assignment called for the students to survey both male and female students (independently and randomly chosen) to see if the proportions that approve of the College’s new drug and alcohol policy. A student went and randomly selected 200 male students and 100 female students and obtained the data summarized below. Use this data to obtain a 90% confidence interval estimate for the difference of the proportions of female and male students that approve of the new policy. Example

19
19 Copyright © 2005 Brooks/Cole, a division of Thomson Learning, Inc. For a 90% confidence interval the z value to use is 1.645. This value is obtained from the bottom row of the table of t critical values (Table III). We use p 1 to be the female’s sample approval proportion and p 2 as the male’s sample approval proportion. Based on the observed sample, we believe that the proportion of females that approve of the policy exceeds the proportion of males that approve of the policy by somewhere between 0.028 and 0.222. Example

Similar presentations

Presentation is loading. Please wait....

OK

“Students” t-test.

“Students” t-test.

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google

Download ppt on indus valley civilization writing Ppt on power generation by speed breakers Ppt on bluetooth based smart sensor networks definition Ppt on different types of dance forms list Ppt on organised retailing in india Ppt on word processing tools By appt only movie Ppt on review of related literature on research Ppt on mobile based attendance tracking system Ppt on international space station