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TEST 6
Z-TEST FOR COMPARING TWO COUNTS
OBJECTIVE
To investigate the
significance of the difference
between two counts.
LIMITATIONS
The test is approximate and assumes
that the counts are large enough for
the normal approximation to the
Poisson to apply.
METHOD
Let n1 and n2 be the two counts taken over times
t1 and t2, respectively. Then the two average
frequencies are R1 = n1/t1 and R2 = n2 t2 . To
test the assumption of equal average frequencies
we use the test statistic
𝑍 =
𝑅1 − 𝑅2
𝑅1
𝑡1
+
𝑅2
𝑡2
This may be compared with a
standard normal distribution using
either a one-tailed or two-tailed test.
EXAMPLES
1.) Two traffic roundabouts are compared for intensity
of traffic at non-peak times with steady conditions.
Roundabout one has 952 arrivals over 22 minutes and
roundabout two has 1168 arrivals over 20 minutes.
The arrival rates per minute are : 952/22 (43.27) and
1168/30 (38. 93) respectively. What do these results
say about the two arrival rates or frequency taken over
the two time intervals?
GIVEN:
n1 = 952 , n2 = 1168 ,
t1 = 22 mins., t2 = 30 mins.,
R1 = n1/t1 = 43.27 , R2 = n1/t2 = 38.93
(I) STATE THE NULL AND
ALTERNATIVE HYPOTHESIS
Ho : µ1 = µ2
Ha : µ1 ≠ µ2
(II) LEVEL OF SIGNIFICANCE
𝛼 =
0.05
2
=
0.025
(III) TEST STATISTIC
 Z-test for comparing two counts
(Poisson distribution)
Z =
𝑅1 − 𝑅2
𝑅1 + 𝑅2
𝑡1 + 𝑡2
=
4.34
3.26
=
4.344
1.81
=
2.40
(IV) CRITICAL REGION
 Reject Ho if 𝑍 𝑐 > 𝑍𝑡
2.40 > 1.96
(V) DECISION
 Reject the Ho o f no difference
between two rates at 5% level of
significance.
(VI) CONCLUSION
 Therefore, the roundabout one has
an intensity of arrival significantly
higher than the roundabout two.
2.) Mila and Christine are law students; they
are being compared of their capacity to
memorized laws and articles over times. Mila
was able to memorized 604 words over 15
minutes, and Christine memorized 911 words
over 23 minutes. The average frequency per
minute are: 604/15 (40.27) and 911/23(39.01),
respectively. At 5% level of significance, are
the two counts differ significantly?
GIVEN:
n1 = 604, n2 = 911,
t1 = 15 mins., t2 = 23 mins. ,
R1 = n1 t1 = 604 15 = 40.27 ,
R2 = n2 t2 = 911 23 = 39.61
(I) STATE THE NULL AND
ALTERNATIVE HYPOTHESIS
Ho : µ1 = µ2
Ha : µ1 ≠ µ2
𝛼 =
0.05
2
= 0.025
(II) LEVEL OF SIGNIFICANCE
(III) TEST STATISTIC
 Z-test for comparing two counts
(Poisson distribution)
Z =
𝑅1 − 𝑅2
𝑅1 + 𝑅2
𝑡1 + 𝑡2
=
𝑜.66
4.41
=
𝑜.66
2.10
= 0.31
(IV) CRITICAL REGION
 Reject Ho if 𝑍 𝑐 > 𝑍𝑡
0.31 < 1.96
(V) DECISION
Fail to reject the Ho at
5% level of significance.
(VI) CONCLUSION
Therefore, there is no
significant difference
between the two counts.
THANK YOU!
Nothing follows……

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100 Statistical Tests - Test 6

  • 1. TEST 6 Z-TEST FOR COMPARING TWO COUNTS
  • 2. OBJECTIVE To investigate the significance of the difference between two counts.
  • 3. LIMITATIONS The test is approximate and assumes that the counts are large enough for the normal approximation to the Poisson to apply.
  • 4. METHOD Let n1 and n2 be the two counts taken over times t1 and t2, respectively. Then the two average frequencies are R1 = n1/t1 and R2 = n2 t2 . To test the assumption of equal average frequencies we use the test statistic
  • 5. 𝑍 = 𝑅1 − 𝑅2 𝑅1 𝑡1 + 𝑅2 𝑡2 This may be compared with a standard normal distribution using either a one-tailed or two-tailed test.
  • 7. 1.) Two traffic roundabouts are compared for intensity of traffic at non-peak times with steady conditions. Roundabout one has 952 arrivals over 22 minutes and roundabout two has 1168 arrivals over 20 minutes. The arrival rates per minute are : 952/22 (43.27) and 1168/30 (38. 93) respectively. What do these results say about the two arrival rates or frequency taken over the two time intervals?
  • 8. GIVEN: n1 = 952 , n2 = 1168 , t1 = 22 mins., t2 = 30 mins., R1 = n1/t1 = 43.27 , R2 = n1/t2 = 38.93
  • 9. (I) STATE THE NULL AND ALTERNATIVE HYPOTHESIS Ho : µ1 = µ2 Ha : µ1 ≠ µ2
  • 10. (II) LEVEL OF SIGNIFICANCE 𝛼 = 0.05 2 = 0.025
  • 11. (III) TEST STATISTIC  Z-test for comparing two counts (Poisson distribution) Z = 𝑅1 − 𝑅2 𝑅1 + 𝑅2 𝑡1 + 𝑡2 = 4.34 3.26 = 4.344 1.81 = 2.40
  • 12. (IV) CRITICAL REGION  Reject Ho if 𝑍 𝑐 > 𝑍𝑡 2.40 > 1.96
  • 13. (V) DECISION  Reject the Ho o f no difference between two rates at 5% level of significance.
  • 14. (VI) CONCLUSION  Therefore, the roundabout one has an intensity of arrival significantly higher than the roundabout two.
  • 15. 2.) Mila and Christine are law students; they are being compared of their capacity to memorized laws and articles over times. Mila was able to memorized 604 words over 15 minutes, and Christine memorized 911 words over 23 minutes. The average frequency per minute are: 604/15 (40.27) and 911/23(39.01), respectively. At 5% level of significance, are the two counts differ significantly?
  • 16. GIVEN: n1 = 604, n2 = 911, t1 = 15 mins., t2 = 23 mins. , R1 = n1 t1 = 604 15 = 40.27 , R2 = n2 t2 = 911 23 = 39.61
  • 17. (I) STATE THE NULL AND ALTERNATIVE HYPOTHESIS Ho : µ1 = µ2 Ha : µ1 ≠ µ2
  • 18. 𝛼 = 0.05 2 = 0.025 (II) LEVEL OF SIGNIFICANCE
  • 19. (III) TEST STATISTIC  Z-test for comparing two counts (Poisson distribution) Z = 𝑅1 − 𝑅2 𝑅1 + 𝑅2 𝑡1 + 𝑡2 = 𝑜.66 4.41 = 𝑜.66 2.10 = 0.31
  • 20. (IV) CRITICAL REGION  Reject Ho if 𝑍 𝑐 > 𝑍𝑡 0.31 < 1.96
  • 21. (V) DECISION Fail to reject the Ho at 5% level of significance.
  • 22. (VI) CONCLUSION Therefore, there is no significant difference between the two counts.