Question

Many customers like to do their banking in person. to decide staffing needs (tellers), one bank...

Many customers like to do their banking in person. to decide staffing needs (tellers), one bank branch tracks the number of customrs whoe bank face-to-face over the course of its normal banking hours, from 9am to 6pm, a period of 9 hours total. the customer visits by times of the working day are given by the following frequesncy counts:

Time 910am 10-11am 11-1pm 1-3pm 3-6pm

customers 30 12 50 20 50

test the hypothesis that the bank customers are distributed equi-preportionally by hour for time periods of the day. Use a significance level of 0.05.

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Answer #1

Here' the answer to the question. Please let me know in case you've doubts.

We will do the ChiSquare test of homogeneity to solve this problem.

Lets write down the hypothesis first:

Hypothesis:

Ho ( Null hypothesis) : The bank customers are distributed equi-preportionally by hour for time periods of the day.

Ha: (Alternate hypo) : bank customers are NOT distributed equi-preportionally by hour for time periods of the day.

Calculating Chi Square test:

Below is how I have calculated the Chi-Square coefficient , from the data provided in the question.

The classes, are to the left, and the calculation has been calculated in the extreme bottom right corner, Chi-Square coefficient is = 36.889

Observed Expected Difference Difference Sq. Diff. Sq. / Exp Fr.
9-10 30 32.4   -2.40   5.76   0.18
10-11 12 32.4   -20.40   416.16   12.84
11-1 50 32.4   17.60   309.76   9.56
1-3 20 32.4   -12.40   153.76   4.75
3-6 50 32.4   17.60   309.76   9.56
Chi-Square Coefficient:   36.889

Lets calculate the p-value with respect to this ChiSquare coefficient of 36.889

Calculate df

df = Number of categories -1 = 5-1 = 4

Calculate p-value

It is = 1 - CHISQ.DIST(ChiSquare stat , df ,TRUE) [this is the Excel formula to convert Chi Square statistic to a p-value]

= 1- CHISQ.DIST(36.9,4,TRUE)

= 1.89 * 10^-7

Drawing a Conclusion

This is << .05 (significance level asked to be assumed in the question )  

Hence, we reject null hypothesis and conclude that "bank customers are distributed equi-preportionally by hour for time periods of the day. " and conclude that "bank customers are NOT distributed equi-preportionally by hour for time periods of the day"

Answer: "bank customers are NOT distributed equi-preportionally by hour for time periods of the day"

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