Inferences about Two Proportions

Inferences about Two Proportions

• 1.
  Elementary Statistics Chapter 9:Inferences from Two Samples 9.1 Two Proportions 1
• 2.
  Chapter 9: Inferencesfrom Two Samples 9. 1 Two Proportions 9.2 Two Means: Independent Samples 9.3 Two Dependent Samples (Matched Pairs) 9.4 Two Variances or Standard Deviations 2 Objectives: • Test the difference between two proportions. • Test the difference between sample means, using the z test. • Test the difference between two means for independent samples, using the t test. • Test the difference between two means for dependent samples. • Test the difference between two variances or standard deviations.
• 3.
  Synopsis: 1. Testing aclaim made about two population proportions 2. Constructing a confidence interval estimate of the difference between two population proportions. (in forms of probabilities or the decimal equivalents of percentages) Objectives 1. Hypothesis Test: Conduct a hypothesis test of a claim about two population proportions. 2. Confidence Interval: Construct a confidence interval estimate of the difference between two population proportions. 9. 1 Two Proportions 3 1. The sample proportions are from two simple random samples. 2. The two samples are independent. (Samples are independent if the sample values selected from one population are not related to or somehow naturally paired or matched with the sample values from the other population.) 3. 𝑛𝑝 & 𝑛𝑞 ≥ 5 𝑓𝑜𝑟 𝑒𝑎𝑐ℎ 𝑠𝑎𝑚𝑝𝑙𝑒
• 4.
  For population 1& 2: 4 9. 1 Inferences about Two Proportions 𝑝2 = 𝑥2 𝑛2 , 𝑞2 = 1 − 𝑝2 𝑝1 = 𝑥1 𝑛1 , 𝑞1 = 1 − 𝑝1 𝑝 = 𝑥1 + 𝑥2 𝑛1 + 𝑛2 𝑞 = 1 − 𝑝 The pooled sample proportion combines the two samples proportions into one proportion & Test Statistic : 𝑇𝑆: 𝑧 = (𝑝1 − 𝑝2) − (𝑝1 − 𝑝2) 𝑝𝑞 1 𝑛1 + 1 𝑛2 Or (𝑝1 − 𝑝2) − (𝑝1 − 𝑝2) 𝑝𝑞 𝑛1 + 𝑝𝑞 𝑛2 Confidence Interval Estimate of p1 − p2 (𝑝1 − 𝑝2) − 𝑧𝛼 2 𝑝1𝑞1 𝑛1 + 𝑝2𝑞2 𝑛2 < 𝑝1 − 𝑝2 < (𝑝1 − 𝑝2) + 𝑧𝛼 2 𝑝1𝑞1 𝑛1 + 𝑝2𝑞2 𝑛2 (𝑝1 − 𝑝2) ± 𝐸 The P-value method and the critical value method are equivalent, but the confidence interval method is not equivalent to the P-value method or the critical value method. 𝜎𝑝1−𝑝2 = 𝑝1𝑞1 𝑛1 + 𝑝2𝑞2 𝑛2 , 𝐸 = 𝑧𝛼/2 𝑝1𝑞1 𝑛1 + 𝑝2𝑞2 𝑛2 TI Calculator: 2 - Proportion Z - test 1. Stat 2. Tests 3. 2 ‒ PropZTest 4. Enter Data or Stats 𝒏𝟏 , 𝒏𝟐, 𝒙𝟏, 𝒙𝟐 5. Choose RTT, LTT, or 2TT TI Calculator: Confidence Interval: 2 proportion 1. Stat 2. Tests 3. 2-prop ZINT 4. Enter:𝒏𝟏 , 𝒏𝟐, 𝒙𝟏, 𝒙𝟐 & CL
• 5.
  Step 4: Decision: a.Reject H0 b. The claim is True c. There is enough evidence to say that the proportion of drivers who send text messages is larger today than it was last year. 5 A survey of 1000 drivers this year showed that 29% of the people send text messages while driving. Last year a survey of 1000 drivers showed that 17% of those send text messages while driving. At α = 0.01, test the claim that there has been an increase in the number of drivers who text while driving? Example 1 Step 3: CV: α = 0.01 → 𝑧 = 2.33, Or 2.325 Step 1: H0: p1 = p2 & H1: p1 > p2 (claim), RTT Given: BD, α = 0.01 𝒏𝟏 = 𝟏𝟎𝟎𝟎, 𝒑𝟏 = 𝟎. 𝟐𝟗, 𝒏𝟐 = 𝟏𝟎𝟎𝟎, 𝒑𝟐 =0.17 Step 1: H0 , H1, claim & Tails Step 2: TS Calculate (TS) Step 3: CV using α Step 4: Make the decision to a. Reject or not H0 b. The claim is true or false c. Restate this decision: There is / is not sufficient evidence to support the claim that… Step 2: 𝑇𝑆: 𝑧 = 0.29−0.17 0.23(0.77) 1000 + 0.23(0.77) 1000 = 6.38 𝑝 = 1000(0.29) + 1000(0.17) 1000 + 1000 = 0.23 → 𝑞 = 1 − 𝑝 = 0.77 𝑧 = (𝑝1 − 𝑝2) − (𝑝1 − 𝑝2) 𝑝𝑞 1 𝑛1 + 1 𝑛2 𝑜𝑟 (𝑝1 − 𝑝2) − (𝑝1 − 𝑝2) 𝑝𝑞 𝑛1 + 𝑝𝑞 𝑛2 𝑝 = 𝑥1 + 𝑥2 𝑛1 + 𝑛2
• 6.
  6 Proportions of Carswith Rear License Plates Only: Are the Proportions the Same in Connecticut and New York? Use a 0.05 significance level to test the claim that Connecticut and New York have the same proportion of cars with rear license plates only. Example 2 a: CV Method CV: α = 0.05 → 𝑧 = ±1.96 H0: p1 = p2 , Claim & H1: p1 ≠ p2, 2TT Given: BD, 𝒙𝟏 = 𝟐𝟑𝟗, 𝒏𝟏 = 𝟐𝟎𝟒𝟗, 𝒙𝟐 = 𝟗, 𝒏𝟐 = 𝟓𝟓𝟎, α = 0.05 𝑇𝑆: 𝑧 = 0.117 − 0.016 0.0954(0.9046) 2049 + 0.0954(0.9046) 550 = 7.11 𝑝1 = 𝑥1 𝑛1 = 239 2049 = 0.1166 𝑝2 = 𝑥2 𝑛2 = 9 550 = 0.0164 𝑝 = 𝑥1 + 𝑥2 𝑛1 + 𝑛2 = 239 + 9 2049 + 550 = 0.09542132 → 𝑞 = 0.90457868 Blank Connecticut New York Cars with rear license plate only 239 9 Cars with front and rear license plates 1810 541 Total 2049 550 Decision: a. Reject H0 b. The claim is False c. There is not sufficient evidence to support the claim that p1 = p2. That is, there is sufficient evidence to conclude that Connecticut and New York have different proportions of cars with rear license plates only. It’s reasonable to speculate that enforcement of the license plate laws is much stricter in New York than in Connecticut, and that is why Connecticut car owners are less likely to install the front license plate. 𝑧 = 𝑝1 − 𝑝2 𝑝𝑞 𝑛1 + 𝑝𝑞 𝑛2 TI Calculator: 2 - Proportion Z - test 1. Stat 2. Tests 3. 2 ‒ PropZTest 4. Enter Data or Stats 𝒏𝟏 , 𝒏𝟐, 𝒙𝟏, 𝒙𝟐 5. Choose RTT, LTT, or 2TT 𝑝 = 𝑥1 + 𝑥2 𝑛1 + 𝑛2
• 7.
  7 Proportions of Carswith Rear License Plates Only: Are the Proportions the Same in Connecticut and New York? Use a 0.05 significance level to test the claim that Connecticut and New York have the same proportion of cars with rear license plates only. a. By the Traditional (CV) Method b. By The P-value Method c. By Confidence Interval (90%) . Example 2b: P-value Method Given: BD, 𝒙𝟏 = 𝟐𝟑𝟗, 𝒏𝟏 = 𝟐𝟎𝟒𝟗, 𝒙𝟐 = 𝟗, 𝒏𝟐 = 𝟓𝟓𝟎, , 𝒑𝟏=𝒑𝟐=0.8, α = 0.05 𝑝1 = 0.1166 𝑝2 == 0.0164 𝑝 = 0.09542132 𝑞 = 0.90457868 Blank Connecticut New York Cars with rear license plate only 239 9 Cars with front and rear license plates 1810 541 Total 2049 550 2TT: P-value = twice the area to the right of the test statistic z = 7.11 The P-value: 2(0.0001) = 0.0002. Technology: P-value = 0.00000000000119 which is often expressed as 0.0000 or “P-value < 0.0001.” The P-value = 0.0000 < α = 0.05 Reject the null hypothesis of p1 = p2
• 8.
  8 Proportions of Carswith Rear License Plates Only: Are the Proportions the Same in Connecticut and New York? Use a 0.05 significance level to test the claim that Connecticut and New York have the same proportion of cars with rear license plates only. a. By the Traditional (CV) Method b. By The P-value Method c. By Confidence Interval (95%) . Example 2c: CI Method Given: BD, 𝒙𝟏 = 𝟐𝟑𝟗, 𝒏𝟏 = 𝟐𝟎𝟒𝟗, 𝒙𝟐 = 𝟗, 𝒏𝟐 = 𝟓𝟓𝟎, , α = 0.05 𝐸 = 𝑧𝛼 2 𝑝1𝑞1 𝑛1 + 𝑝2𝑞2 𝑛2 𝐸 = 1.96 0.117(0.883) 2049 + 0.016(0.984) 550 = 1.96(0.008890763) = 0.017482 Interpretation: The confidence interval limits do not contain 0, suggesting that there is a significant difference between the two proportions. The confidence Interval suggests that the value of p1 is greater than the value of p2, so there appear to be sufficient evidence to warrant rejection of the claim that “Connecticut and New York have the same proportion of cars with rear license plates only.” 0.1166 − 0.0164 ± 0.0175 0.0828 < ( p1– p2) < 0.1178 (𝑝1 − 𝑝2) ± 𝐸 = 0.1003 ± 0.0175 TI Calculator: Confidence Interval: 2 proportion 1. Stat 2. Tests 3. 2-prop ZINT 4. Enter:𝒏𝟏 , 𝒏𝟐, 𝒙𝟏, 𝒙𝟐 & CL 𝑝1 = 0.1166 𝑝2 == 0.0164 𝑝 = 0.09542132 𝑞 = 0.90457868
• 9.
  9 A researchers foundthat 12 out of 34 inner-city nursing homes had a flue vaccination rate of less than 80%, while 17 out of 24 countryside nursing homes had a flue vaccination rate of less than 80%. At α = 0.05, test the claim that there is no difference in the proportions of the two nursing homes with a flue vaccination rate of less than 80%. Example 3: CV Method Step 3: CV: α = 0.05 → 𝑧 = ±1.96 Step 1: H0: p1 – p2 = 0, Claim, H1: p1 – p2  0, 2TT Given: BD, 𝒙𝟏 = 𝟏𝟐, 𝒏𝟏 = 𝟑𝟒, 𝒙𝟐 = 𝟏𝟕, 𝒏𝟐 = 𝟐𝟒, , 𝒑𝟏=𝒑𝟐=0.8, α = 0.05 Step 4: Decision: a. Reject H0 b. The claim is False c. There is not sufficient evidence to support the claim that there is no difference in the proportions of inner-city and countryside nursing homes with a resident vaccination rate of less than 80%. Step 2: 𝑇𝑆: 𝑧 = 0.35−0.71 0.5(0.5)( 1 34 + 1 24 ) = −2.7 𝑝1 = 𝑋1 𝑛1 = 12 34 = 0.35 & 𝑝2 = 𝑋2 𝑛2 = 17 24 = 0.71 𝑝 = 𝑥1 + 𝑥2 𝑛1 + 𝑛2 = 12 + 17 34 + 24 = 29 58 = 0.5 → 𝑞 = 0.5 Step 1: H0 , H1, claim & Tails Step 2: TS Calculate (TS) Step 3: CV using α Step 4: Make the decision to a. Reject or not H0 b. The claim is true or false c. Restate this decision: There is / is not sufficient evidence to support the claim that… 𝑧 = 𝑝1 − 𝑝2 𝑝𝑞 𝑛1 + 𝑝𝑞 𝑛2 TI Calculator: 2 - Proportion Z - test 1. Stat 2. Tests 3. 2 ‒ PropZTest 4. Enter Data or Stats 𝒏𝟏 , 𝒏𝟐, 𝒙𝟏, 𝒙𝟐 5. Choose RTT, LTT, or 2TT 𝑝 = 𝑥1 + 𝑥2 𝑛1 + 𝑛2
• 10.
  10 A researchers foundthat 12 out of 34 inner-city nursing homes had a flue vaccination rate of less than 80%, while 17 out of 24 countryside nursing homes had a flue vaccination rate of less than 80%. At α = 0.05, test the claim that there is no difference in the proportions of the small and large nursing homes with a flue vaccination rate of less than 80%. Find the 95% confidence interval for the difference of the proportions. Example 3: CI Method Given: BD, 𝒙𝟏 = 𝟏𝟐, 𝒏𝟏 = 𝟒, 𝒙𝟐 = 𝟏𝟕, 𝒏𝟐 = 𝟐𝟒, , 𝒑𝟏=𝒑𝟐=0.8, α = 0.05 𝑝1 = 0.3529 & 𝑞1 = 0.6471 𝑝2 = 0.7083 & 𝑞2 = 0.2917 (𝑝1 − 𝑝2) − 𝑧𝛼 2 𝑝1𝑞1 𝑛1 + 𝑝2𝑞2 𝑛2 < 𝑝1 − 𝑝2 < (𝑝1 − 𝑝2) + 𝑧𝛼 2 𝑝1𝑞1 𝑛1 + 𝑝2𝑞2 𝑛2 𝐸 = 1.96 0.35(0.65) 34 + 0.71(0.29) 24 −0.36 − 0.242 < 𝑝1 − 𝑝2 < −0.36 + 0.242 −0.602 < 𝑝1 − 𝑝2 < −0.118 = 0.242 Since 0 is not contained in the interval, the decision is to reject the null hypothesis H0: p1 = p2. (𝑝1 − 𝑝2) ± 𝐸 TI Calculator: Confidence Interval: 2 proportion 1. Stat 2. Tests 3. 2-prop ZINT 4. Enter:𝒏𝟏 , 𝒏𝟐, 𝒙𝟏, 𝒙𝟐 & CL
• 11.
  11 For the sampledata listed in, use a 0.05 significance level to test the claim that the proportion of minority drivers stopped by the police is greater than the proportion of white drivers who are stopped. Example 4 Given: BD, 𝒏𝟏 = 𝟐𝟎𝟎, 𝒙𝟏 = 𝟐𝟒, 𝒑𝟏 = 𝟎. 𝟏𝟐, 𝒏𝟐 = 𝟏𝟒𝟎𝟎, 𝒙𝟐 = 𝟏𝟒𝟕, 𝒑𝟐 =0.105, α = 0.05 𝑝 = 𝑋1 + 𝑋2 𝑛1 + 𝑛2 = 24 + 147 200 + 1400 = 171 1600 = 0.106875 → 𝑞 = 0.893125 Racial Profiling data Minority White & Non-Hispanic Drivers stopped by police 24 147 Total 200 1400 % 24/200 = 12% 147/1400 = 10.5%
• 12.
  Step 4: Decision: a.Fail to Reject H0 b. The claim is False c. There is not sufficient evidence to support the claim that the proportion of minority drivers stopped by police is greater than that for white drivers. (This does not mean that racial profiling has been disproved. The evidence might be strong enough with more data.) 12 For the sample data listed in, use a 0.05 significance level to test the claim that the proportion of minority drivers stopped by the police is greater than the proportion of white drivers who are stopped. Example 4: CV Method Step 3: CV: α = 0.05 → 𝑧 = 1.645 Step 1: H0: p1 = p2 & H1: p1 > p2 (claim), RTT Step 2: 𝑇𝑆: 𝑧 = 0.12−0.105 0.1069(0.8931) 200 + 0.1069(0.8931) 1400 = 0.6422 𝑝 = 0.106875 → 𝑞 = 0.893125 Given: BD, 𝒏𝟏 = 𝟐𝟎𝟎, 𝒙𝟏 = 𝟐𝟒, 𝒑𝟏 = 𝟎. 𝟏𝟐, 𝒏𝟐 = 𝟏𝟒𝟎𝟎, 𝒙𝟐 = 𝟏𝟒𝟕, 𝒑𝟐 =0.105, α = 0.05 Step 1: H0 , H1, claim & Tails Step 2: TS Calculate (TS) Step 3: CV using α Step 4: Make the decision to a. Reject or not H0 b. The claim is true or false c. Restate this decision: There is / is not sufficient evidence to support the claim that… 𝑧 = (𝑝1 − 𝑝2) − (𝑝1 − 𝑝2) 𝑝𝑞 1 𝑛1 + 1 𝑛2 𝑜𝑟 (𝑝1 − 𝑝2) − (𝑝1 − 𝑝2) 𝑝𝑞 𝑛1 + 𝑝𝑞 𝑛2
• 13.
  13 For the sampledata listed in, use a 0.05 significance level to test the claim that the proportion of minority drivers stopped by the police is greater than the proportion of white drivers who are stopped. a. By the Traditional (CV) Method b. By The P-value Method c. By Confidence Interval (90%) . Example 4: P-value & CI Method TS: z = 0.64 RTT: the P-value = The area to the right of the test statistic: z = 0.64. The P-value = 0.2611. P-value = 0.2611 >  = 0.05 Fail to reject the null hypothesis. 𝐸 = 𝑧𝛼 2 𝑝1𝑞1 𝑛1 + 𝑝2𝑞2 𝑛2 𝐸 = 1.645 0.12(0.88) 200 + 0.105(0.895) 1400 = 0.0404 (0.120 – 0.105) – 0.0404 < ( p1– p2) < (0.120 – 0.105) + 0.0404 –0.025 < ( p1– p2) < 0.055 (𝑝1 − 𝑝2) ± 𝐸 Given: BD, 𝒏𝟏 = 𝟐𝟎𝟎, 𝒙𝟏 = 𝟐𝟒, 𝒑𝟏 = 𝟎. 𝟏𝟐, 𝒏𝟐 = 𝟏𝟒𝟎𝟎, 𝒙𝟐 = 𝟏𝟒𝟕, 𝒑𝟐 =0.105, α = 0.05
• 14.
  Why Do theProcedures of This Section Work? 14 The distribution of 𝑝1can be approximated by a normal distribution with mean p1, and standard deviation: 𝑝1𝑞1/𝑛1 The difference 𝑝1 − 𝑞1 can be approximated by a normal distribution with mean p1 – p2 and variance: The variance of the differences between two independent random variables is the sum of their individual variances which leads to: Therefore, the z test statistic has the form given earlier. When constructing the confidence interval estimate of the difference between two proportions, we don’t assume that the two proportions are equal, and we estimate the standard deviation as 𝜎(𝑝1−𝑝2) 2 = 𝜎𝑝1 2 + 𝜎𝑝2 2 = 𝑝1𝑞1 𝑛1 + 𝑝2𝑞2 𝑛2 𝜎(𝑝1−𝑝2) = 𝑝1𝑞1 𝑛1 + 𝑝2𝑞2 𝑛2 𝜎 = 𝑝1𝑞1 𝑛1 + 𝑝2𝑞2 𝑛2 𝑧 = (𝑝1 − 𝑝2) − (𝑝1 − 𝑝2) 𝑝1𝑞1 𝑛1 + 𝑝2𝑞2 𝑛2