The table below shows the distribution of births across the days of the week in a random sample of 141 births from local records in a large city:
Days Sun Mon Tue Wed Thu Fri Sat
Births 24 24 18 19 22 18 16
Are births evenly distributed across the days of the week? 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}}
Sun 24 3.8571 14.8772 0.7386
Mon 24 20.1429 14.8772 0.7386
Tue 18 20.1429 -2.1429 0.228
Wed 19 20.1429 -1.1429 1.3062
Thu 22 20.1429 1.8571 3.4488 0.1712
Fri 18 20.1429 -2.1429 4.592 0.228
Sat 16 20.1429 -4.1429 17.1636 0.8521
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.

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