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Confidence Interval for Propo Given binomial random variable with Y success •M ean (Y)or Expected value of Y= 𝑛∗? •Variance (Y) = 𝑛∗? ∗(1−?) •Standard deviation (Y) = √𝑣𝑎𝑟𝑖𝑎𝑛𝑐? = √(𝑛∗? ∗(1 The Central Limit Theorem applies. Recall that variable can be approximated by a normal rand For a given sample of n trials and Y successes, •The sample proportion ? ̂ = 𝑌/𝑛 •The sampling distribution of the sample prop distributed. (? ̂−?)/√((? ̂(1−? ̂))/𝑛) is approximately a 95% CI for estimation of ? = ? ̂ ± Z 𝛼⁄2 * √((? This is predicated on the binomial random approximated by a random normal variabl when the following conditions are met: 1. n ? ̂ -5 ≥ 0 2. n ? ̂ +5 ≤ n If E(Y) is too close to 0 or n, the normal distribu because there is too much area to the left of 0
ortion ses in n trials 1−?) ) a binomial random dom variable. portion ? ̂ is normally standard normal ? ̂(1−? ̂))/𝑛) m variable being le. This is only valid ution cannot be used 0 or right of n.
Estim n (sample size) 125 Number that meet the outcome studied 84 -hat ? 0.672 Does it meet conditions? Value n -5>/0 *? 79 n +5</n *? 89 Sample Calculations Point estimate hat) (? 0.672 Standard Error= sqrt hat(1 hat)/n] [? -? 0.04 Confidence interval Confidence interval (enter proportion) 0.95 Z multiple/Critical Z 1.96 95%Confidence Interval Lower Limit 0.5897 Upper Limit 0.7543 Confidence Interval Length 0.1646 Hypothesis Testing 0.6 Z statistic (Zsample) 1.643 Expected proportion (p o )
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Test (One Sample) Ha p≠0.6 p<0.6 p>0.6 Test (Choose from drop down list) Right-tailed test Critical Z value 1.645 Z statistic (from your sampling) 1.643 p-value 0.0502 Decision cannot reject Ho Two sided test( H 0 : p=p o , H a : p≠p o ) Left tailed test (H o : p≥p o , H a : p<p o ) Right tailed test (H o : p≤p o , H a : p>p o )
mating Proportions (One Sample) Enter Enter Meets Condition? yes yes Enter l Enter To use this template, only en will autopopulate. Rememb conditions to be able to cons proportion. To confirm that the template C12: 125 C13: 84 C25: .95 If done correctly, the 95% CI Test Statistic Z= ((𝑝 ̂−𝑝 0 ))/√((𝑝 0 (1−𝑝 0 ))/𝑛)
95% Conf Interval Z Critical Value Lower Bound Upper Bound p-value +/- 1.960 0.5897 0.7543 0.1003 -1.645 - ∞ 0.7411 0.9498 1.645 0.6029 + ∞ 0.0502
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nter values into the gray box. The rest ber that you have to meet the stated struct a confidence interval for e is working enter: I should be (0.59,0.75).
Decision cannot reject Ho cannot reject Ho cannot reject Ho
Sample Size Calculation for p ⍺ 0.05 Half-length of interval (margin of error, accuracy) 0.10 Proportion (estimate, hat) ? 0.40 z multiple 1.960 Sample size 93 Sample Size F 𝑛=(? 𝑚𝑢𝑙𝑡𝑖𝑝𝑙?/( 𝑀𝑎𝑟𝑔𝑖𝑛
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Formula for Estimating a Proportion 𝑛 𝑜𝑓 𝐸𝑟𝑟𝑜𝑟))^2 𝑝 ̂(1−𝑝 ̂ )
Estim n (sample size) Number that meet the outcome studied -hat ? #DIV/0! Does it meet conditions? Value n -5>/0 *? #DIV/0! n +5</n *? #DIV/0! Sample Calculations Point estimate hat) (? #DIV/0! Standard Error= sqrt hat(1 hat)/n] [? -? #DIV/0! Confidence interval Confidence interval (enter proportion) Z multiple/Critical Z 0.00 0%Confidence Interval Lower Limit #DIV/0! Upper Limit #DIV/0! Confidence Interval Length #DIV/0! Hypothesis Testing Z statistic (Zsample) #DIV/0! Expected proportion (p o )
Test (One Sample) Ha p≠ p< p> Test (Choose from drop down list) Right-tailed test Critical Z value #VALUE! Z statistic (from your sampling) #DIV/0! p-value #DIV/0! Decision #DIV/0! Two sided test( H 0 : p=p o , H a : p≠p o ) Left tailed test (H o : p≥p o , H a : p<p o ) Right tailed test (H o : p≤p o , H a : p>p o )
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mating Proportions (One Sample) Enter Enter Meets Condition? #DIV/0! #DIV/0! Enter l Enter Test Statistic Z= ((𝑝 ̂−𝑝 0 ))/√((𝑝 0 (1−𝑝 0 ))/𝑛) In a random sampling of 1 Calculate a 95% confidenc VISA. Suppose you expected tha sampling support or refut to include what testing wo
0% Conf Interval Z Critical Value Lower Bound Upper Bound p-value +/- 0.000 #DIV/0! #DIV/0! #DIV/0! #VALUE! - ∞ #VALUE! #DIV/0! #VALUE! #VALUE! + ∞ #DIV/0!
180 credit card purchase slips, 54 were made with VISA. ce interval for the proportion of purchase slips made with at the 25% of purchase slips are made with VISA, does your te your expected proportion (5% significance level)? Be sure ould you perform and why?
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Decision #DIV/0! #DIV/0! #DIV/0!
Estim n (sample size) 180 Number that meet the outcome studied 54 -hat ? 0.300 Does it meet conditions? Value n -5>/0 *? 49 n +5</n *? 59 Sample Calculations Point estimate hat) (? 0.300 Standard Error= sqrt hat(1 hat)/n] [? -? 0.03 Confidence interval Confidence interval (enter proportion) 0.95 Z multiple/Critical Z 1.96 95%Confidence Interval Lower Limit 0.2331 Upper Limit 0.3669 Confidence Interval Length 0.1339 Hypothesis Testing 0.25 Z statistic (Zsample) 1.549 Expected proportion (p o )
Test (One Sample) Ha p≠0.25 p<0.25 p>0.25 Test (Choose from drop down list) 2-tailed test Critical Z value +/- 1.960 Z statistic (from your sampling) 1.549 p-value 0.1213 Decision cannot reject Ho Two sided test( H 0 : p=p o , H a : p≠p o ) Left tailed test (H o : p≥p o , H a : p<p o ) Right tailed test (H o : p≤p o , H a : p>p o )
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mating Proportions (One Sample) Enter Enter Meets Condition? yes yes Enter l Enter Test Statistic Z= ((𝑝 ̂−𝑝 0 ))/√((𝑝 0 (1−𝑝 0 ))/𝑛)
95% Conf Interval Z Critical Value Lower Bound Upper Bound p-value +/- 1.960 0.2331 0.3669 0.1213 -1.645 - ∞ 0.3562 0.9393 1.645 0.2438 + ∞ 0.0607 Friendly reminder that critical values for 2 sided test are +/- Critical Z
In a random sampling of 180 credit card Calculate a 95% confidence interval for t with VISA. 2-sided 95% CI: (0.233, 0.367) 95% confident that the interval 23% to 3 credit card purchases made with VISA. Suppose you expected that the 25% of p your sampling support or refute your exp level)? Because we are interested in whether th would do a 2-sided Z test with Ha: p≠0.2 and the Z test statistic for the sample is 1 hypothesis that the proportion is 0.25 be more extreme than the critical Z value. T the p-value of 12% is over 5% and that th contains 0.25. If one were only interested in exploring o sided test. For example, if you think that right sided test with Ha:p>0.25. The criti Z value is 1.549 (remember that your tes of whether it is a 1-sided or 2-sided test, value(s)). Because the sample Z is not m be rejected. The p-value of 6.1% is above includes 0.25. The decision will be the sa
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Decision cannot reject Ho cannot reject Ho cannot reject Ho
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purchase slips, 54 were made with VISA. the proportion of purchase slips made 37% contains the true proportion of purchase slips are made with VISA, does pected proportion (5% significance he proportion is above or below 25%, we 25. The critical value for =0.05 is 1.96 ⍺ 1.549. We cannot reject the null ecause the sample test statistic is not This is also demonstrated by the fact that he 2 sided 95% confidence interval one direction, you would perform a one- t the proportion is >25%, you would do a tical Z value is now 1.645 and the sample st statistic does NOT change regardless , what changes are your critical Z more extreme than critical Z, Ho cannot e alpha (5%) and the 95% CI (0.244,+ ∞) ame no matter which method you use.
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Estim n (sample size) Number that meet the outcome studied -hat ? #DIV/0! Does it meet conditions? Value n -5>/0 *? #DIV/0! n +5</n *? #DIV/0! Sample Calculations Point estimate hat) (? #DIV/0! Standard Error= sqrt hat(1 hat)/n] [? -? #DIV/0! Confidence interval Confidence interval (enter proportion) Z multiple/Critical Z 0.00 0%Confidence Interval Lower Limit #DIV/0! Upper Limit #DIV/0! Confidence Interval Length #DIV/0! Hypothesis Testing Z statistic (Zsample) #DIV/0! Expected proportion (p o )
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Test (One Sample) Ha p≠ p< p> Test (Choose from drop down list) Right-tailed test Critical Z value #VALUE! Z statistic (from your sampling) #DIV/0! p-value #DIV/0! Decision #DIV/0! Two sided test( H 0 : p=p o , H a : p≠p o ) Left tailed test (H o : p≥p o , H a : p<p o ) Right tailed test (H o : p≤p o , H a : p>p o )
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mating Proportions (One Sample) Enter Enter Meets Condition? #DIV/0! #DIV/0! Enter l Enter Test Statistic Z= ((𝑝 ̂−𝑝 0 ))/√((𝑝 0 (1−𝑝 0 ))/𝑛)
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0% Conf Interval Z Critical Value Lower Bound Upper Bound p-value +/- 0.000 #DIV/0! #DIV/0! #DIV/0! #VALUE! - ∞ #VALUE! #DIV/0! #VALUE! #VALUE! + ∞ #DIV/0!
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In the past, 60% of all undergraduate studen University earned their degrees within four matriculation. A random sample of 95 stude that matriculated in the fall of 2012 was rec test whether there has been a change in the students who graduate within four years. Ad that 40 of these 95 students graduated in th (i.e., four academic years after matriculation a. Given the sample outcome, calculate a 9 interval for the relevant population propor interval estimate suggest that there has be proportion of students who graduate withi or why not? b. Suppose now that State University admi test the claim made by faculty that the pro who graduate within four years at State Un below the historical value of 60% this year. proportion to test their claim. Report a p-v it.
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Decision #DIV/0! #DIV/0! #DIV/0!
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nts enrolled at State years of ents from the class cently selected to e proportion of dministrators found he spring of 2016 n). 95% confidence rtion. Does this een a change in the in four years? Why inistrators want to oportion of students niversity has fallen . Use this sample value and interpret
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Estim n (sample size) 95 Number that meet the outcome studied 40 -hat ? 0.421 Does it meet conditions? Value n -5>/0 *? 35 n +5</n *? 45 Sample Calculations Point estimate hat) (? 0.421 Standard Error= sqrt hat(1 hat)/n] [? -? 0.05 Confidence interval Confidence interval (enter proportion) 0.95 Z multiple/Critical Z 1.96 95%Confidence Interval Lower Limit 0.3218 Upper Limit 0.5203 Confidence Interval Length 0.1986 Hypothesis Testing 0.6 Z statistic (Zsample) -3.560 Expected proportion (p o )
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Test (One Sample) Ha p≠0.6 p<0.6 p>0.6 Test (Choose from drop down list) 2-tailed test Critical Z value +/- 1.960 Z statistic (from your sampling) -3.560 p-value 0.0004 Decision reject Ho Two sided test( H 0 : p=p o , H a : p≠p o ) Left tailed test (H o : p≥p o , H a : p<p o ) Right tailed test (H o : p≤p o , H a : p>p o )
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mating Proportions (One Sample) Enter Enter Meets Condition? yes yes Enter l Enter Test Statistic Z= ((𝑝 ̂−𝑝 0 ))/√((𝑝 0 (1−𝑝 0 ))/𝑛)
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95% Conf Interval Z Critical Value Lower Bound Upper Bound p-value +/- 1.960 0.3218 0.5203 0.0004 -1.645 - ∞ 0.5044 0.0002 1.645 0.3377 + ∞ 0.9998 Friendly reminder that critical values for 2 sided test are +/- Critical Z
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In the past, 60% of all undergraduate stude State University earned their degrees within matriculation. A random sample of 95 stude that matriculated in the fall of 2012 was rec test whether there has been a change in th students who graduate within four years. A found that 40 of these 95 students graduate 2016 (i.e., four academic years after matricu a. Given the sample outcome, calculate a 9 interval for the relevant population propor interval estimate suggest that there has be proportion of students who graduate with or why not? 95% CI: (0.3218,0.5203) Yes, the 2-sided 95% CI suggests that the p students who graduate within 4 years has because the CI does NOT contain the value suggests that fewer students are graudatin since all of the values are below 0.60. b. Suppose now that State University admi test the claim made by faculty that the pro students who graduate within four years a has fallen below the historical value of 60% this sample proportion to test their claim. and interpret it.
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Decision reject Ho reject Ho cannot reject Ho and interpret it. Ho: proportion that graduate within 4 year Ha: proportion that graduate within 4 year 0.60. Test statistic is -3.560 and the p-value is 0.0 test. This p-value is lower than 0.05 ( , 95% ⍺ thus, we can reject Ho in favor of Ha. In fac low p-value, we can even reject Ho in favo of 0.01 (99% confidence).
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ents enrolled at n four years of ents from the class cently selected to he proportion of Administrators ed in the spring of ulation). 95% confidence rtion. Does this een a change in the hin four years? Why proportion of changed from 0.60 e 0.60. The 95% CI ng within 4 years inistrators want to oportion of at State University % this year. Use Report a p-value
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rs = 0.60 rs is not equal to 004 for a 2 sided % confidence) ct, given this very or of Ha with an ⍺
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Estim n (sample size) Number that meet the outcome studied -hat ? #DIV/0! Does it meet conditions? Value n -5>/0 *? #DIV/0! n +5</n *? #DIV/0! Sample Calculations Point estimate hat) (? #DIV/0! Standard Error= sqrt hat(1 hat)/n] [? -? #DIV/0! Confidence interval Confidence interval (enter proportion) Z multiple/Critical Z 0.00 0%Confidence Interval Lower Limit #DIV/0! Upper Limit #DIV/0! Confidence Interval Length #DIV/0! Hypothesis Testing Z statistic (Zsample) #DIV/0! Expected proportion (p o )
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Test (One Sample) Ha p≠ p< p> Test (Choose from drop down list) 2-tailed test Critical Z value +/- 0.000 Z statistic (from your sampling) #DIV/0! p-value #DIV/0! Decision #DIV/0! Two sided test( H 0 : p=p o , H a : p≠p o ) Left tailed test (H o : p≥p o , H a : p<p o ) Right tailed test (H o : p≤p o , H a : p>p o )
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mating Proportions (One Sample) Enter Enter Meets Condition? #DIV/0! #DIV/0! Enter l Enter Test Statistic Z= ((𝑝 ̂−𝑝 0 ))/√((𝑝 0 (1−𝑝 0 ))/𝑛)
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0% Conf Interval Z Critical Value Lower Bound Upper Bound p-value +/- 0.000 #DIV/0! #DIV/0! #DIV/0! #VALUE! - ∞ #VALUE! #DIV/0! #VALUE! #VALUE! + ∞ #DIV/0! Friendly reminder that critical values for 2 sided test are +/- Critical Z
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Newborn babies are more likely to be boys th A random sample found 13,173 boys were bo newborn children. The sample proportion of b Is this sample evidence that the birth of boys than the birth of girls in the entire population
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Decision #DIV/0! #DIV/0! #DIV/0!
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han girls. orn among 25,468 boys was 0.5172. is more common n?
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Estim n (sample size) 25468 Number that meet the outcome studied 13173 -hat ? 0.517 Does it meet conditions? Value n -5>/0 *? 13168 n +5</n *? 13178 Sample Calculations Point estimate hat) (? 0.517 Standard Error= sqrt hat(1 hat)/n] [? -? 0.00 Confidence interval Confidence interval (enter proportion) 0.95 Z multiple/Critical Z 1.96 95%Confidence Interval Lower Limit 0.5111 Upper Limit 0.5234 Confidence Interval Length 0.0123 Hypothesis Testing 0.5 Z statistic (Zsample) 5.502 Expected proportion (p o )
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Test (One Sample) Ha p≠0.5 p<0.5 p>0.5 Test (Choose from drop down list) Right-tailed test Critical Z value 1.645 Z statistic (from your sampling) 5.502 p-value 0.0000 Decision reject Ho Two sided test( H 0 : p=p o , H a : p≠p o ) Left tailed test (H o : p≥p o , H a : p<p o ) Right tailed test (H o : p≤p o , H a : p>p o )
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mating Proportions (One Sample) Enter Enter Meets Condition? yes yes Enter l Enter Test Statistic Z= ((𝑝 ̂−𝑝 0 ))/√((𝑝 0 (1−𝑝 0 ))/𝑛)
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95% Conf Interval Z Critical Value Lower Bound Upper Bound p-value +/- 1.960 0.5111 0.5234 0.0000 -1.645 - ∞ 0.5224 1.0000 1.645 0.5121 + ∞ 0.0000
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Newborn babies are more likely to be boys A random sample found 13,173 boys were b newborn children. The sample proportion o Is this sample evidence that the birth of boy than the birth of girls in the entire populatio po=0.5 if boys=girls, 𝑝 ̂ =51.7%, Z test statisti 2-sided test: Ha: proportion≠0.50, Critical Z= because Z score more extreme than critical and 95% CI (0.5111,0.5234) does not contai supports that the proportion of boys is not supports that it is >0.50. Notice that the enti 0.50. 1 tailed Right sided test: Ha: p>0.50, Critical Ho because Z score more extreme than criti <<0.05 and 95% CI (Lower bound 0.512) doe Our data supports that the proportion of bo
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Decision reject Ho cannot reject Ho reject Ho
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than girls. born among 25,468 of boys was 0.5172. ys is more common on? tic=5.502 =1.96. Reject Ho Z, p-value <<0.05 in 0.5. Our data 0.50. In fact, it tire CI is above l Z=1.645. Reject tical Z, p-value es not contain 0.5. oys is >0.50.
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Estim n (sample size) Number that meet the outcome studied -hat ? #DIV/0! Does it meet conditions? Value n -5>/0 *? #DIV/0! n +5</n *? #DIV/0! Sample Calculations Point estimate hat) (? #DIV/0! Standard Error= sqrt hat(1 hat)/n] [? -? #DIV/0! Confidence interval Confidence interval (enter proportion) Z multiple/Critical Z 0.00 0%Confidence Interval Lower Limit #DIV/0! Upper Limit #DIV/0! Confidence Interval Length #DIV/0! Hypothesis Testing Z statistic (Zsample) #DIV/0! Expected proportion (p o )
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Test (One Sample) Ha p≠ p< p> Test (Choose from drop down list) 2-tailed test Critical Z value +/- 0.000 Z statistic (from your sampling) #DIV/0! p-value #DIV/0! Decision #DIV/0! Two sided test( H 0 : p=p o , H a : p≠p o ) Left tailed test (H o : p≥p o , H a : p<p o ) Right tailed test (H o : p≤p o , H a : p>p o )
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mating Proportions (One Sample) Enter Enter Meets Condition? #DIV/0! #DIV/0! Enter l Enter Test Statistic Z= ((𝑝 ̂−𝑝 0 ))/√((𝑝 0 (1−𝑝 0 ))/𝑛) In 1995, 40% of adults ag schools. On June 1, 2005, 1004 adults aged 18 year Does the evidence sugges having “a great deal” of c
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0% Conf Interval Z Critical Value Lower Bound Upper Bound p-value +/- 0.000 #DIV/0! #DIV/0! #DIV/0! #VALUE! - ∞ #VALUE! #DIV/0! #VALUE! #VALUE! + ∞ #DIV/0! Friendly reminder that critical values for 2 sided test are +/- Critical Z
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ged 18 years or older reported that they had “a great deal” of co , the Gallup Organization ( www.gallup.com ) released results of a rs or older stated that they had “a great deal” of confidence in pu est at the α =0.05 significance level that the proportion of adults confidence in the public schools is significantly lower in 2005 tha
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Decision #DIV/0! #DIV/0! #DIV/0!
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onfidence in the public a poll in which 372 of ublic schools. aged 18 years or older an the 1995 proportion?
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Estim n (sample size) 1004 Number that meet the outcome studied 372 -hat ? 0.371 Does it meet conditions? Value n -5>/0 *? 367 n +5</n *? 377 Sample Calculations Point estimate hat) (? 0.371 Standard Error= sqrt hat(1 hat)/n] [? -? 0.02 Confidence interval Confidence interval (enter proportion) 0.95 Z multiple/Critical Z 1.96 95%Confidence Interval Lower Limit 0.3406 Upper Limit 0.4004 Confidence Interval Length 0.0597 Hypothesis Testing 0.4 Z statistic (Zsample) -1.907 Expected proportion (p o )
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Test (One Sample) Ha p≠0.4 p<0.4 p>0.4 Test (Choose from drop down list) Left-tailed test Critical Z value -1.645 Z statistic (from your sampling) -1.907 p-value 0.0283 Decision reject Ho Two sided test( H 0 : p=p o , H a : p≠p o ) Left tailed test (H o : p≥p o , H a : p<p o ) Right tailed test (H o : p≤p o , H a : p>p o )
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mating Proportions (One Sample) Enter Enter Meets Condition? yes yes Enter l Enter Test Statistic Z= ((𝑝 ̂−𝑝 0 ))/√((𝑝 0 (1−𝑝 0 ))/𝑛)
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95% Conf Interval Z Critical Value Lower Bound Upper Bound p-value +/- 1.960 0.3406 0.4004 0.0565 -1.645 - ∞ 0.3956 0.0283 1.645 0.3454 + ∞ 0.9717
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In 1995, 40% of adults aged 18 y public schools. On June 1, 2005, which 372 of 1004 adults aged 1 public schools. Does the evidence suggest at the older having “a great deal” of con proportion? This can be run either as a 2-taile 2-sided test (more conservative) Ha: proportion ≠ 0.4 95% CI: (0.3406, 0.4004); Cannot p-value: 0.0565; Cannot reject H test statistic -1.907; Cannot rejec Decision: cannot reject Ho thus w proportion of adults aged 18 yea "different" from 1995. 1-sided test, left (more liberal) Ha: proportion < 0.4 95% CI: (-∞, 0.3956); Can reject H p-value: 0.0283; Can reject Ho in test statistic -1.907; Can reject H Decision: our 2005 sample suppo adults aged 18 years or older hav lower" from 1995
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Decision cannot reject Ho reject Ho cannot reject Ho
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years or older reported that they had “a great deal” of confidenc the Gallup Organization ( www.gallup.com ) released results of a 18 years or older stated that they had “a great deal” of confidenc e α =0.05 significance level that the proportion of adults aged 18 nfidence in the public schools is significantly lower in 2005 than ed or left-tailed test and the decisions would be different. t reject Ho because 95% CI contains 0.40 Ho because p-value greater than ⍺ ct Ho because -1.907 is not more extreme than -1.96 would conclude that the 2005 sample data does not support tha ars or older having a great deal of confidence in the public schoo Ho in favor of Ha because 95% CI does NOT contains 0.40 n favor of Ha because p-value is less than ⍺ Ho in favor of Ha because -1.907 is more extreme than -1.645 ort supports that we can reject Ho in favor of Ha that the propo ving a great deal of confidence in the public schools is "significan
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ce in the poll in ce in 8 years or n the 1995 at the ols is ortion of ntly
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