Presentation is loading. Please wait.

Presentation is loading. Please wait.

WARM – UP A company is criticized because only 9 of 43 randomly choosen people in executive-level positions are women. The company explains that although.

Similar presentations


Presentation on theme: "WARM – UP A company is criticized because only 9 of 43 randomly choosen people in executive-level positions are women. The company explains that although."— Presentation transcript:

1 WARM – UP A company is criticized because only 9 of 43 randomly choosen people in executive-level positions are women. The company explains that although this proportion is lower that what it might wish, it not surprising given that only 40% of their employees are women. a.) Perform the appropriate significance test to determine whether there is evidence that the proportion of women in executive-level positions is statistically different from that of the all employees in the company. b.) A Confidence Interval reveals that we can be 95% confident that the true proportions of women in executive-level positions is between (0.087, 0.331). Does this support your conclusions in (a)?

2 WARM – UP p=The true proportion of female executives. H0 : p = 0.40
A company is criticized because only 9 of 43 SRS people in executive-level positions are women. The company explains that although this proportion is lower that what it might wish, it not surprising given that only 40% of their employees are women. Is there evidence of a difference? One Proportion z – Test p=The true proportion of female executives. H0 : p = 0.40 HA : p ≠ 0.40 SRS – stated 2. Appr. Normal: 43 (0.4)=17.2 ≥ 10 AND (1 – (0.4))=25.8 ≥ 10 Since p-val.< 0.05, we will Reject H0. There is strong evidence to suggest the proportion of female executives is different from the overall proportion of female employees at the company.

3 YES, 40% is NOT in the Interval.
WARM – UP Since p-val.< 0.05, we will Reject H0. There is strong evidence to suggest the proportion of female executives is different from the overall proportion (40%) of female employees at the company. b.) A 95% confidence interval reveals that we can be 95% confident that the true proportions of women in executive- level positions is between (0.087, 0.331). Does this support your conclusions in (a)? YES, 40% is NOT in the Interval.

4

5

6

7 Ch. 22: Comparing TWO Proportions
In a two sample problem we want to compare two populations or responses to two treatments based on independent samples. To compare two proportions of categorical successes in two groups we use Two-Proportion Z-Procedures. TWO PROPORTION Z – CONFIDENCE INTERVAL Statistic ± Critical Value · Standard Deviation/Error With z*= upper critical value |invNorm(1 – C)/2| or from TABLE

8 TWO PROPORTION ASSUMPTIONS / CONDITIONS Both SRS
n1p1 ≥ 10 and n1(1 - p1) ≥ 10 AND n2p2 ≥ 10 and n2(1 – p2) ≥ 10 3. Population ≥ 10 · n1 AND Population ≥ 10 · n2

9 The Conclusion for a 2 –Prop Z Conf. Interval:
We can be 95% Confident that the true difference in the proportion of … {context} … is between “a” and “b”

10 TWO PROPORTION Z – SIGNIFICANCE TEST
H0: p1 = p2 Ha: p1 >, <, or ≠ p2 The Null Hypothesis states that there is NO Difference between the population proportions. If this is true then the observations really come from a single population, so instead of p1 and p2 separately we use pp = (p-pooled) in the test. TWO PROPORTION ASSUMPTIONS / CONDITIONS Same as Conf. Intervals

11 EXAMPLE Is there evidence of a difference in the proportion of “A’s” earned on the Quiz in Period 1 vs. Period 4? pi = The true proportion of “A’s” earned in… p1 = Period 1 and p2 = Period 4 TWO Proportion z – Test H0: p1 = p2 Ha: p1 ≠ p2 SRS –NOT collected randomly Appr. Normal: 19(.421)=8 ≥ (1 –.421)=11 ≥ 10 18(.722)=13 ≥ (1 –.722)=6 ≥ 10 Since the P-Value > we Fail to REJECT H0 . There is NOT enough evidence to suggest a difference in A’s earned among the classes. -PWC

12 Eye Dominance! (Page 11)

13 Finding which eye is dominant
Take a scratch piece of paper and fold it into quarters. Tear out a small section where all sides fold together making a small hole in the unfolded paper. Hold the paper at arm’s length and with both eyes open, focus on something through the hole in the paper. Then shut your left eye and see if the image moves. If it does, you are left eye dominant. Then shut your right eye and see if the image moves. If it does, you are right eye dominant. Note: a very few people have no dominant eye Gather the data. (male – right eye dominant, female – right eye dominant, etc.)

14

15 Is there a significant difference in the proportion of males who are right eye dominant and the proportion of females who are right eye dominant? (α = 0.05) pi = The true proportion of each gender that is right eye dominant. pm = Men and pw = Women TWO Proportion z – Test 1. Random Selection/Assignment – Both sample data were collected randomly. Appr. Normality n· pm ______ ≥ AND n· (1 – pm) _______≥ 10 n· pw ______ ≥ AND n· (1 – pw) _______≥ 10 H0: pm = pw Ha: pm ≠ pw Since the P-Value is less than α = REJECT H0 . The proportion of males who are right eye dominant differs from the proportion of females who are right eye dominant Since the P-Value is greater than α = FAIL to REJECT H0 . There is insufficient evidence that the proportion of males who are right eye dominant differs from the proportion of females who are right eye dominant

16 z – Confidence Interval
HW Page 508: 3, 4, 11 TWO Proportion z – Confidence Interval SRS – The data was collected randomly for both ·(0.406)=411 ≥ 10 AND 1012·(1 – 0.406)=601 ≥ 10 1062·(0.504)=535 ≥ 10 AND 1062·(1 – 0.504)=527 ≥ 10 We can be 95% Confident that the true difference in the proportion of men and women suffering from arthritis is between and pm – pw = Negative # So men suffer less than women

17

18

19 EXAMPLE A random poll conducted in 2002 asked if we should have or should not have gone to war in Iraq. 57 of 99 Independents said “Should have” while 75 of 98 Republicans said “Should have”. Is there evidence that there exists a significant difference in the proportion of Independent and Republicans that favored the war at that time? pi = The true proportion of the political party that supported the war. pI = Independents and pR = Republicans TWO Proportion z – Test H0: pI = pR Ha: pI ≠ pR SRS – The data was collected randomly 99 · (0.5758) ≥ AND · (1 – ) ≥ 10 98 · (0.7653) ≥ AND · (1 – ) ≥ 10 3. Population of Independents is ≥ 10 · (99) Population of Republicans is ≥ 10 · (98)

20 H0: pI = pR Ha: pI ≠ pR Since the P-Value is less than α = there is strong evidence to REJECT H0 . There is evidence to suggest that Independents and Republicans supported the war differently.

21 EXAMPLE Does Belin’s AP Statistics Period 3 class have a higher proportion of A’s than 4th Period? Test using appropriate hypothesis and technique. Per. 3 = 6/21 Per. 4 = 8/30 pi = The true proportion of A’s in Belin’s AP Statistics class. p3 = Period 3 and p4 = Period 4 TWO Proportion z – Test H0: p3 = p4 Ha: p3 > p4 SRS – The data was not randomly Population of Period 3 students ≥ 10 · (21) Population of Period 4 students ≥ 10 · (30) 21 · (0.2857) ≥ AND · (1 – ) ≥ 10 30 · (0.2667) ≥ AND · (1 – ) ≥ 10 X -- Fail X -- Fail X -- Fail X -- Fail X -- Fail Since the P-Value is NOT less than α = the data IS NOTsignificant. There is NO evidence to REJECT H0 . There is NOT evidence to suggest that the proportion of A’s in Period 3 is higher than that of Period 4.


Download ppt "WARM – UP A company is criticized because only 9 of 43 randomly choosen people in executive-level positions are women. The company explains that although."

Similar presentations


Ads by Google