Question

A local bank wants to evaluate the usage of its ATM. Currently the bank manager assumes that the ATM is used consistently throughout the week, including weekends. She decides to use statistical inference with a 0.025 level of significance to test a customer’s claim that the ATM is much busier on some days of the week than it is on other days. During a randomly selected week, the bank recorded the number of times the ATM was used on each day.

Monday 38

Tuesday 33

Wednesday 41

Thursday 25

Friday 22

Saturday 38

Sunday 45

Answer 1

Chi-Square Goodness of Fit test

(1) Null and Alternative Hypotheses The following null and alternative hypotheses need to be tested: H0: p1 =0.1429, p2 =0.1429, p3​=0.1429, p4​=0.1429, p5 =0.1429, p6=0.1429, p7 =0.1429 Ha​: Some of the population proportions differ from the values stated in the null hypothesis This corresponds to a Chi-Square test for Goodness of Fit. (2) Degrees of Freedom The number of degrees of freedom is df=n-1=7-1=6 (3) Test Statistics The Chi-Squared statistic is computed as follows: (4)Critical Value and Rejection Region Based on the information provided, the significance level is α=0.025, the number of degrees of freedom is df=n-1=7-1=6, so the critical value becomes 14.4494. Then the rejection region for this test is R={χ2:χ2>14.4494}. (5)P-value The P-value is the probability that a chi-square statistic having 6 degrees of freedom is more extreme than 14.4494. The p-value is p=Pr(χ2>12.3141)=0.0553 (6) The decision about the null hypothesis Since it is observed that χ2=12.3141≤χc2​=14.4494, it is then concluded that the null hypothesis is not rejected. (7) Conclusion It is concluded that the null hypothesis Ho is not rejected. Therefore, there is NOT enough evidence to claim that some of the population proportions differ from those stated in the null hypothesis, at the α=0.025 significance level.

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