The numbers of false fire alarms were counted each month at a number of sites. The results are given in the following table.
Month Number of Alarms
January 35
February 40
March 42
April 28
May 41
June 33
July 45
August 29
Septemeber 45
October 38
November 29
December 36
Test the hypothesis that false alarms are equally likely to occur in any month. Use 5% level of significance.

Procedure:

Assumptions: (select everything that applies)

Step 1. Hypotheses Set-Up:

 H0:\displaystyle {H}_{{0}}:
 Ha:\displaystyle {H}_{{a}}: , and the test is

Step 2. The significance level α=\displaystyle \alpha= %

Step 3. Compute the value of the test statistic using the table below: (Round the answers to 4 decimal places)

Category [O]bserved [E]xpected O−E\displaystyle {O}-{E}  (O−E)2\displaystyle {\left({O}-{E}\right)}^{{2}}  (O−E)2E\displaystyle \frac{{\left({O}-{E}\right)}^{{2}}}{{E}} 
January 35 -1.75 3.0625 0.0833
February 40 36.75 10.5625 0.2874
March 42 36.75 5.25 0.75
April 28 36.75 -8.75 76.5625
May 41 36.75 4.25 18.0625 0.4915
June 33 36.75 -3.75 14.0625 0.3827
July 45 36.75 8.25 68.0625 1.852
August 29 36.75 -7.75 60.0625 1.6344
Septemeber 45 36.75 8.25 68.0625 1.852
October 38 36.75 1.25 1.5625 0.0425
November 29 36.75 -7.75 60.0625 1.6344
December 36 36.75 -0.75 0.5625 0.0153
Total:  

 =∑(O−E)2E=\displaystyle =\sum\frac{{\left({O}-{E}\right)}^{{2}}}{{E}}= 
        df=

 Step 4. Testing Procedure: (Round the answers to 3 decimal places)

CVA PVA
Provide the critical value(s) for the Rejection Region: Compute the P-value of the test statistic:
left CV is and right CV is P-value is

Step 5. Decision:

CVA PVA
Is the test statistic in the rejection region? Is the P-value less than the significance level?

Conclusion:

Step 6. Interpretation:

At 5% significance level we have sufficient evidence to reject the null hypothesis in favor of the alternative hypothesis.

License