When conducting a hypothesis test that compares two independent population proportions, the following characteristics should be present:
The two independent samples are simple random samples that are independent.
The number of successes is at least five, and the number of failures is at least five, for each of the samples.
Growing literature states that the population must be at least ten or 20 times the size of the sample. This keeps each population from being over-sampled and causing incorrect results.
Comparing two proportions, like comparing two means, is common. If two estimated proportions are different, it may be due to a difference in the populations or it may be due to chance. A hypothesis test can help determine if a difference in the estimated proportions reflects a difference in the population proportions.
Like the case of differences in sample means, we construct a sampling distribution for differences in sample proportions:
p'A=XAnAp'A=XAnA and p'B=XBnBp'B=XBnB are the sample proportions for the two sets of data in question XAXA and XBXB.
The difference of two proportions follows an approximate normal distribution. Generally, the null hypothesis states that the two proportions are the same. That is, H0: pA = pB. To conduct the test, we use a pooled proportion, pc.
Example 10.8
Problem
Two types of medication for hives are being tested to determine if there is a difference in the proportions of adult patient reactions. Twenty out of a random sample of 200 adults given medication A still had hives 30 minutes after taking the medication. Twelve out of another random sample of 200 adults given medication B still had hives 30 minutes after taking the medication. Test at a 1% level of significance.
Solution
The problem asks for a difference in proportions, making it a test of two proportions.
Let A and B be the subscripts for medication A and medication B, respectively. Then pA and pB are the desired population proportions.
Random Variable: P′A – P′B = difference in the proportions of adult patients who did not react after 30 minutes to medication A and to medication B.
H0: pA = pB
pA – pB = 0
Ha: pA ≠ pB
pA – pB ≠ 0
The words "is a difference" tell you the test is two-tailed.
Distribution for the test: Since this is a test of two binomial population proportions, the distribution is normal:
p c = x A + x B n A + n B = 20+12 200+200 =0.08 1– p c =0.92 p c = x A + x B n A + n B = 20+12 200+200 =0.08 1– p c =0.92
P ′ A – P ′ B ~N[ 0, (0.08)(0.92)( 1 200 + 1 200 ) ] P ′ A – P ′ B ~N[ 0, (0.08)(0.92)( 1 200 + 1 200 ) ]
P′A – P′B follows an approximate normal distribution.
Calculate the p-value using the normal distribution: p-value = 0.1404.
Estimated proportion for group A: p ′ A = x A n A = 20 200 =0.1 p ′ A = x A n A = 20 200 =0.1
Estimated proportion for group B: p ′ B = x B n B = 12 200 =0.06 p ′ B = x B n B = 12 200 =0.06
Graph:
P′A – P′B = 0.1 – 0.06 = 0.04.
Half the p-value is below –0.04, and half is above 0.04.
Compare α and the p-value: α = 0.01 and the p-value = 0.1404. α < p-value.
Make a decision: Since α < p-value, do not reject H0.
Conclusion: At a 1% level of significance, from the sample data, there is not sufficient evidence to conclude that there is a difference in the proportions of adult patients who did not react after 30 minutes to medication A and medication B.
Using the TI-83, 83+, 84, 84+ Calculator
Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 20 for x1, 200 for n1, 12 for x2, and 200 for n2. Arrow down to p1: and arrow to not equal p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is p = 0.1404 and the test statistic is 1.47. Do the procedure again, but instead of Calculate do Draw.
Try It 10.8
Two types of valves are being tested to determine if there is a difference in pressure tolerances. Fifteen out of a random sample of 100 of Valve A cracked under 4,500 psi. Six out of a random sample of 100 of Valve B cracked under 4,500 psi. Test at a 5% level of significance.
Example 10.9
Problem
A research study was conducted about gender differences regarding the use of seat belts in motor vehicles. The researcher believed that the proportion of women not wearing seat belts is less than the proportion of men not wearing seat belts. The data collected represents a random sample of U.S. adults and is summarized in Table 10.12. Is the proportion of women not wearing seat belts less than the proportion of men not wearing seat belts? Test at a 1% level of significance.
row: Men | Women
row: Does not wear seat belts | 183 | 156
row: Total number surveyed | 2231 | 2169
Solution
This is a test of two population proportions. Let M and F be the subscripts for men and women. Then pM and pF are the desired population proportions.
Random Variable: p′F − p′M = difference in the proportions of men and women who do not wear seat belts.
H0: pF = pM H0: pF – pM = 0
Ha: pF < pM Ha: pF – pM < 0
The words "less than" tell you the test is left-tailed.
Distribution for the test: Since this is a test of two population proportions, the distribution is normal:
p c = x F + x M n F + n M = 156+183 2169+2231 =0.077 p c = x F + x M n F + n M = 156+183 2169+2231 =0.077 1− p c =0.923 1− p c =0.923 Therefore, p ′ F – p ′ M ∼N( 0, (0.077)(0.923)( 1 2169 + 1 2231 ) ) p ′ F – p ′ M ∼N( 0, (0.077)(0.923)( 1 2169 + 1 2231 ) ) p′F – p′M follows an approximate normal distribution.
Calculate the p-value using the normal distribution: p-value = 0.1045 Estimated proportion for women: 0.0719 Estimated proportion for men: 0.082
Graph:
Decision: Since α < p-value, Do not reject H0
Conclusion: At the 1% level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of women not wearing seat belts is less than the proportion of men not wearing seat belts.
Try It 10.9
A survey was conducted about the favorable beverage as tea. The data collected is summarized in the table. Is the proportion of men favoring tea more than women favoring tea? Test at a 1% level of significance.
row: Men | Women
row: Favor tea | 16 | 18
row: Total surveyed | 230 | 218
Using the TI-83, 83+, 84, 84+ Calculator
Press STAT. Arrow over to TESTS and press 6:2-PropZTest. Arrow down and enter 156 for x1, 2169 for n1, 183 for x2, and 2231 for n2. Arrow down to p1: and arrow to less than p2. Press ENTER. Arrow down to Calculate and press ENTER. The p-value is P = 0.1045 and the test statistic is z = -1.256.
Example 10.10
Problem
A marketing firm claims that the proportion of younger adults who own electric vehicles is greater than the proportion of older adults who own electric vehicles. A random sample of U.S. adults was taken, and the results of the survey indicate the following:
Out of a sample of 232 older adults (aged 35 or older), 5% own electric vehicles.
Out of a sample of 1,343 young adults (aged 34 or younger), 10% own electric vehicles.
Test at the 5% level of significance. Is the proportion of younger adults greater than the proportion of older adults with respect to owning electric vehicles?
Solution
This is a test of two population proportions. Let Y and O be the subscripts for younger adults and older adults, respectively. Then pY and pO are the desired population proportions.
Random Variable: p’Y and p’O = difference in the proportions of younger and older adults who own electric vehicles.
The words "greater than" indicate that the test is right-tailed.
1. ਵੰਡ ਲਗਭਗ ਆਮ ਹੈ:
2. ਪੀ ਸੀ = ਐਕਸ ਵਾਈ + ਐਕਸ ਓ ਐਨ ਵਾਈ + ਐਨ ਓ = 134 + 12 1343 + 232 = 0 . 0927 ਪੀ ਸੀ = ਐਕਸ ਵਾਈ + ਐਕਸ ਓ ਐਨ ਵਾਈ + ਐਨ ਓ = 134 + 12 1343 + 232 = 0 . 0927
3. 1 - ਪੀ ਸੀ = 0 . 9073 1 - ਪੀ ਸੀ = 0 . 9073
4. ਇਸ ਲਈ,
5. ਪੀ 'ਵਾਈ - ਪੀ 'ਓ ~ ਐਨ 0 , 0 . 0927 ) ( 0 . 9073 ) ( 1 1343 + 1 232 ) ਪੀ 'ਵਾਈ - ਪੀ 'ਓ ~ ਐਨ 0 , 0 . 0927 ) ( 0 . 9073 ) ( 1 1343 + 1 232 )
6. p'y-p'Op'y-p'O ਲਗਭਗ ਆਮ ਵੰਡ ਦੀ ਪਾਲਣਾ ਕਰਦਾ ਹੈ।
7. ਆਮ ਵੰਡ ਦੀ ਵਰਤੋਂ ਕਰਕੇ ਪੀ-ਮੁੱਲ ਦੀ ਗਣਨਾ ਕਰੋ: ਪੀ-ਮੁੱਲ = 0.0077 ਸਮੂਹ Y ਲਈ ਅਨੁਮਾਨਿਤ ਅਨੁਪਾਤ: 0.10 ਸਮੂਹ O ਲਈ ਅਨੁਮਾਨਿਤ ਅਨੁਪਾਤ: 0.05
8. ਗ੍ਰਾਫ:
9. ਫੈਸਲਾ: ਕਿਉਂਕਿ > ਪੀ-ਮੁੱਲ, H0H0 ਨੂੰ ਰੱਦ ਕਰੋ।
10. ਸਿੱਟਾ: ਮਹੱਤਤਾ ਦੇ 5% ਪੱਧਰ 'ਤੇ, ਨਮੂਨਾ ਡਾਟਾ ਤੋਂ, ਇਹ ਸਿੱਟਾ ਕੱਢਣ ਲਈ ਕਾਫ਼ੀ ਸਬੂਤ ਹਨ ਕਿ ਪੁਰਾਣੇ ਬਾਲਗਾਂ ਦੀ ਤੁਲਨਾ ਵਿੱਚ ਨੌਜਵਾਨ ਬਾਲਗਾਂ ਦਾ ਇੱਕ ਵੱਡਾ ਅਨੁਪਾਤ ਇਲੈਕਟ੍ਰਿਕ ਵਾਹਨਾਂ ਦਾ ਮਾਲਕ ਹੈ।
11. TI-83, 83+, 84, 84+ ਕੈਲਕੂਲੇਟਰ ਦੀ ਵਰਤੋਂ ਕਰਦੇ ਹੋਏ
12. TI-83+ ਅਤੇ TI-84: STAT ਦਬਾਓ। TESTS ਵੱਲ ਐਰੋ ਕਰੋ ਅਤੇ 6:2-PropZTest ਦਬਾਓ। ਹੇਠਾਂ ਐਰੋ ਕਰੋ ਅਤੇ x1 ਲਈ 135, n1 ਲਈ 1343, x2 ਲਈ 12, ਅਤੇ n2 ਲਈ 232 ਦਰਜ ਕਰੋ। p1: ਤੱਕ ਹੇਠਾਂ ਐਰੋ ਕਰੋ ਅਤੇ p2 ਤੋਂ ਵੱਡਾ ਵੱਲ ਐਰੋ ਕਰੋ। ENTER ਦਬਾਓ। Calculate ਵੱਲ ਹੇਠਾਂ ਐਰੋ ਕਰੋ ਅਤੇ ENTER ਦਬਾਓ। P-ਮੁੱਲ P = 0.0092 ਹੈ ਅਤੇ ਟੈਸਟ ਸਟੈਟਿਸਟਿਕ Z = 2.33 ਹੈ।
13. ਇਸਨੂੰ ਅਜ਼ਮਾਓ 10.10
14. ਇੱਕ ਸਰਕਾਰੀ ਖੋਜਕਰਤਾ ਇਹ ਜਾਂਚ ਕਰ ਰਿਹਾ ਹੈ ਕਿ ਕੀ ਵੱਖ-ਵੱਖ ਭੂਗੋਲਿਕ ਖੇਤਰਾਂ ਵਿੱਚ ਮੋਟਰਸਾਈਕਲ ਸਵਾਰਾਂ ਦੁਆਰਾ ਹੈਲਮੇਟ ਦੀ ਵਰਤੋਂ ਵਿੱਚ ਕੋਈ ਅੰਤਰ ਹੈ, ਉਨ੍ਹਾਂ ਰਾਜਾਂ ਲਈ ਜਿੱਥੇ ਕਾਨੂੰਨ ਦੁਆਰਾ ਹੈਲਮੇਟ ਦੀ ਵਰਤੋਂ ਜ਼ਰੂਰੀ ਹੈ। ਖੋਜ ਦਰਸਾਉਂਦੀ ਹੈ ਕਿ ਉੱਤਰ-ਪੂਰਬੀ ਅਮਰੀਕਾ ਦੇ ਮੋਟਰਸਾਈਕਲ ਸਵਾਰਾਂ ਲਈ, 113,231 ਮੋਟਰਸਾਈਕਲ ਸਵਾਰਾਂ ਵਿੱਚੋਂ 7622 ਨੇ ਹੈਲਮੇਟ ਨਹੀਂ ਪਹਿਨੇ ਸਨ। ਦੱਖਣ-ਪੂਰਬੀ ਅਮਰੀਕਾ ਵਿੱਚ, 104,873 ਮੋਟਰਸਾਈਕਲ ਸਵਾਰਾਂ ਵਿੱਚੋਂ 7439 ਨੇ ਹੈਲਮੇਟ ਨਹੀਂ ਪਹਿਨੇ ਸਨ। 5% ਮਹੱਤਤਾ ਪੱਧਰ 'ਤੇ ਟੈਸਟ ਕਰੋ। ਹੇਠਾਂ ਦਿੱਤੇ ਸਵਾਲਾਂ ਦੇ ਜਵਾਬ ਦਿਓ:
15. ਏ. ਕੀ ਇਹ ਦੋ ਮਾਧਿਅਮਾਂ ਜਾਂ ਦੋ ਅਨੁਪਾਤਾਂ ਦਾ ਟੈਸਟ ਹੈ?
16. ਬੀ. ਟੈਸਟ ਕਰਨ ਲਈ ਤੁਸੀਂ ਕਿਹੜੀ ਵੰਡ ਦੀ ਵਰਤੋਂ ਕਰਦੇ ਹੋ?
17. ਸੀ. ਰੈਂਡਮ ਵੇਰੀਏਬਲ ਕੀ ਹੈ?
18. ਡੀ. ਸਿਫਰ ਅਤੇ ਵਿਕਲਪਕ ਪਰਿਕਲਪਨਾ ਕੀ ਹਨ? ਸਿਫਰ ਅਤੇ ਵਿਕਲਪਪਕ ਪਰਿਕਲਪਨਾ ਨੂੰ ਚਿੰਨ੍ਹਾਂ ਵਿੱਚ ਲਿਖੋ।
19. ਈ. ਕੀ ਇਹ ਟੈਸਟ ਸੱਜੇ-ਪਾਸੇ, ਖੱਬੇ-ਪਾਸੇ, ਜਾਂ ਦੋ-ਪਾਸੇ ਵਾਲਾ ਹੈ?
20. ਐਫ. ਪੀ-ਮੁੱਲ ਕੀ ਹੈ?
21. ਜੀ. ਕੀ ਤੁਸੀਂ ਸਿਫਰ ਪਰਿਕਲਪਨਾ ਨੂੰ ਰੱਦ ਕਰਦੇ ਹੋ ਜਾਂ ਨਹੀਂ?
22. ਐਚ. ___ ਮਹੱਤਤਾ ਦੇ ਪੱਧਰ 'ਤੇ, ਨਮੂਨਾ ਡਾਟਾ ਤੋਂ, ______ (ਹੈ/ਨਹੀਂ ਹੈ) ਇਹ ਸਿੱਟਾ ਕੱਢਣ ਲਈ ਕਾਫ਼ੀ ਸਬੂਤ ਹਨ ਕਿ ____________.