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/

To test the assumption of equal average frequencies we use the test statistic

𝑍=𝑅1−𝑅2

√ 𝑅1

𝑡 1+

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 = = = 2.40

(IV) CRITICAL REGION

Reject Ho if 2.40

(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:

604,

= 39.61

( I ) STATE THE NULL AND ALTERNATIVE HYPOTHESIS

Ho : µ1 = µ2

Ha : µ1 µ2

(II) LEVEL OF SIGNIFICANCE

(III) TEST STATISTIC

Z-test for comparing two counts (Poisson distribution)

Z = = = 0.31

(IV) CRITICAL REGION

Reject Ho if 0.31<

(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!

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