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 10 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.
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.
The pooled proportion is calculated as follows:
The distribution for the differences is
The test statistic (z-score) is
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 percent 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:
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 percent 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 percent level of significance.
Example 10.9
Problem
A research study was conducted about gender differences in texting. The researcher believed that the proportion of girls involved in texting is less than the proportion of boys involved. The data collected in spring 2010 among a random sample of middle and high school students in a large school district in the southern United States is summarized in Table 10.10. Is the proportion of girls sending texts less than the proportion of boys texting? Test at a 1 percent level of significance.
row: Males | Females
row: Sent texts | 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 males and females. Then, pM and pF are the desired population proportions.
Random variable:
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 females: 0.0719 Estimated proportion for males: 0.082
Graph:
Decision: Since α < p-value, do not reject H0.
Conclusion: At the 1 percent level of significance, from the sample data, there is not sufficient evidence to conclude that the proportion of girls sending texts is less than the proportion of boys sending texts.
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
Researchers conducted a study of smartphone use (Phone A versus Phone B) among adults. A cell phone company claimed that Phone B smartphones are more popular with whites (non-Hispanic) than with African Americans. The results of the survey indicate that of the 232 African American cell phone owners randomly sampled, 5 percent own Phone B. Of the 1,343 white cell phone owners randomly sampled, 10 percent own Phone B. Test at the 5 percent level of significance. Is the proportion of white Phone B owners greater than the proportion of African American Phone B owners?
Solution
This is a test of two population proportions. Let W and A be the subscripts for the whites and African Americans. Then, pW and pA are the desired population proportions.
Random variable:
H0: pW = pA H0: pW – pA = 0
Ha: pW > pA Ha: pW – pA > 0
The words more popular indicate that the test is right-tailed.
Distribution for the test: The distribution is approximately normal.
p c = x W + x A n W + n A = 134+12 1343+232 = 0.0927 p c = x W + x A n W + n A = 134+12 1343+232 = 0.0927
1− p c =0.9073 1− p c =0.9073
Therefore,
p ′ W – p ′ A ∽N( 0, ( 0.0927 )( 0.9073 )( 1 1343 + 1 232 ) ) p ′ W – p ′ A ∽N( 0, ( 0.0927 )( 0.9073 )( 1 1343 + 1 232 ) )
1. p ′ W – p ′ A p ′ W – p ′ A ਲਗਭਗ ਇੱਕ ਆਮ ਵੰਡ ਦਾ ਪਾਲਣ ਕਰਦਾ ਹੈ।
2. ਆਮ ਵੰਡ ਦੀ ਵਰਤੋਂ ਕਰਕੇ p-ਮੁੱਲ ਦੀ ਗਣਨਾ ਕਰੋ:
3. ਗ੍ਰਾਫ:
4. ਫੈਸਲਾ: ਕਿਉਂਕਿ α > p-ਮੁੱਲ, H0 ਨੂੰ ਰੱਦ ਕਰੋ।
5. ਸਿੱਟਾ: ਮਹੱਤਤਾ ਦੇ 5 ਪ੍ਰਤੀਸ਼ਤ ਪੱਧਰ 'ਤੇ, ਨਮੂਨਾ ਡਾਟਾ ਤੋਂ, ਇਹ ਸਿੱਟਾ ਕੱਢਣ ਲਈ ਕਾਫ਼ੀ ਸਬੂਤ ਹਨ ਕਿ ਅਫਰੀਕੀ ਅਮਰੀਕਨਾਂ ਨਾਲੋਂ ਚਿੱਟੇ ਸੈੱਲ ਫੋਨ ਮਾਲਕਾਂ ਦਾ ਇੱਕ ਵੱਡਾ ਅਨੁਪਾਤ ਫੋਨ ਬੀ ਦੀ ਵਰਤੋਂ ਕਰਦਾ ਹੈ।
6. TI-83, 83+, 84, 84+ ਕੈਲਕੁਲੇਟਰ ਦੀ ਵਰਤੋਂ ਕਰਨਾ
7. 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: ਵੱਲ 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।
8. ਇਸਨੂੰ ਅਜ਼ਮਾਓ 10.10
9. ਨਾਗਰਿਕਾਂ ਦੇ ਇੱਕ ਸਮੂਹ ਨੇ ਇਹ ਜਾਣਨਾ ਚਾਹਿਆ ਕਿ ਕੀ ਉਨ੍ਹਾਂ ਦੇ ਛੋਟੇ ਸ਼ਹਿਰ ਵਿੱਚ ਘਰਾਂ ਦੇ ਮਾਲਕਾਂ ਦਾ ਅਨੁਪਾਤ 2011 ਵਿੱਚ 2010 ਨਾਲੋਂ ਵੱਖਰਾ ਸੀ। ਉਨ੍ਹਾਂ ਦੀ ਖੋਜ ਨੇ ਦਿਖਾਇਆ ਕਿ 2010 ਵਿੱਚ ਉਨ੍ਹਾਂ ਦੇ ਸ਼ਹਿਰ ਵਿੱਚ ਉਪਲਬਧ 113,231 ਘਰਾਂ ਵਿੱਚੋਂ, 7,622 ਘਰ ਉਨ੍ਹਾਂ ਪਰਿਵਾਰਾਂ ਦੇ ਸਨ ਜੋ ਉੱਥੇ ਰਹਿੰਦੇ ਸਨ। 2011 ਵਿੱਚ, 104,873 ਉਪਲਬਧ ਘਰਾਂ ਵਿੱਚੋਂ 7,439 ਘਰ ਸ਼ਹਿਰ ਦੇ ਵਸਨੀਕਾਂ ਦੇ ਸਨ। 5 ਪ੍ਰਤੀਸ਼ਤ ਮਹੱਤਤਾ ਪੱਧਰ 'ਤੇ ਟੈਸਟ ਕਰੋ। ਹੇਠਾਂ ਦਿੱਤੇ ਸਵਾਲਾਂ ਦੇ ਜਵਾਬ ਦਿਓ:
10. ਏ. ਕੀ ਇਹ ਦੋ ਮਾਧਾਂ ਜਾਂ ਦੋ ਅਨੁਪਾਤਾਂ ਦਾ ਟੈਸਟ ਹੈ?
11. ਬੀ. ਟੈਸਟ ਕਰਨ ਲਈ ਤੁਸੀਂ ਕਿਹੜੀ ਵੰਡ ਦੀ ਵਰਤੋਂ ਕਰਦੇ ਹੋ?
12. ਸੀ. ਰੈਂਡਮ ਵੇਰੀਏਬਲ ਕੀ ਹੈ?
13. ਡੀ. ਸਿਫਰ ਅਤੇ ਬਦਲਵੇਂ ਪਰਿਕਲਪਨਾ ਕੀ ਹਨ? ਸਿਫਰ ਅਤੇ ਬਦਲਵੇਂ ਪਰਿਕਲਪਨਾ ਨੂੰ ਚਿੰਨ੍ਹਾਂ ਵਿੱਚ ਲਿਖੋ।
14. ਈ. ਕੀ ਇਹ ਟੈਸਟ ਸੱਜੇ, ਖੱਬੇ, ਜਾਂ ਦੋ-ਪੂਛ ਵਾਲਾ ਹੈ?
15. ਐਫ. p-ਮੁੱਲ ਕੀ ਹੈ?
16. ਜੀ. ਕੀ ਤੁਸੀਂ ਸਿਫਰ ਪਰਿਕਲਪਨਾ ਨੂੰ ਰੱਦ ਕਰਦੇ ਹੋ ਜਾਂ ਨਹੀਂ?
17. ਐਚ. ਮਹੱਤਤਾ ਦੇ ______ ਪੱਧਰ 'ਤੇ, ਨਮੂਨਾ ਡਾਟਾ ਤੋਂ, ______ (ਹੈ/ਨਹੀਂ ਹੈ) ਇਹ ਸਿੱਟਾ ਕੱਢਣ ਲਈ ਕਾਫ਼ੀ ਸਬੂਤ ਕਿ ______.