Births Records of randomly selected births were obtained and categorized according to the day of the week that they occurred (based on data from the National Center for Health Statistics). Because babies are unfamiliar with our schedule of weekdays, a reasonable claim is that births occur on the different days with equal frequency. Use a 0.01 significance level to test that claim.

Day | Sun | Mon | Tues | Wed | Thurs | Fri | Sat |
---|---|---|---|---|---|---|---|

No. of Births | 77 | 110 | 124 | 122 | 120 | 123 | 97 |

#### Solution

The observed data is

Day | Obs. Freq.$(O)$ | Prop. |
---|---|---|

Sun | 77 | 0.1429 |

Mon | 110 | 0.1429 |

Tues | 124 | 0.1429 |

Wed | 122 | 0.1429 |

Thurs | 120 | 0.1429 |

Fri | 123 | 0.1429 |

Sat | 97 | 0.1429 |

##### Step 1 Setup the hypothesis

The null and alternative hypothesis are as follows:

`$H_0:p_{Sun}=p_{Mon} =\cdots = p_{Sat} =\frac{1}{7}$`

$H_1:$ At least one of the proportion is different from `$\frac{1}{7}$`

.

##### Step 2 Test statistic

The test statistic for testing above hypothesis is

` $$ \begin{equation*} \chi^2= \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ \end{equation*} $$ `

##### Step 3 Level of Significance

The level of significance is $\alpha =0.05$.

##### Step 4 Critical value of $\chi^2$

The level of significance is $\alpha =0.05$. Degrees of freedom $df=k-1=7-1 =6$.

The critical value of $\chi^2$ for $df=6$ and $\alpha=0.05$ level of significance is $\chi^2 =12.5916$.

##### Step 5 Test Statistic

The expected frequencies can be calculated as

` $$ \begin{equation*} E_{i} =N*p_i \end{equation*} $$ `

For example, $E_{1}$ is given by

` $$ \begin{eqnarray*} E_{1} & = &N*p_1\\ &=& 773*0.1429\\ &=&110.4617. \end{eqnarray*} $$ `

Day | Obs. Freq.$(O)$ | Prop. $p_i$ | Expe.Freq.$(E)$ | $(O-E)^2/E$ |
---|---|---|---|---|

Sun | 77 | 0.1429 | 110.4617 | 10.136 |

Mon | 110 | 0.1429 | 110.4617 | 0.002 |

Tues | 124 | 0.1429 | 110.4617 | 1.659 |

Wed | 122 | 0.1429 | 110.4617 | 1.205 |

Thurs | 120 | 0.1429 | 110.4617 | 0.824 |

Fri | 123 | 0.1429 | 110.4617 | 1.423 |

Sat | 97 | 0.1429 | 110.4617 | 1.641 |

The test statistic is

` $$ \begin{eqnarray*} \chi^2&=& \sum \frac{(O_{i} -E_{i})^2}{E_{i}} \sim \chi^2_{(k-1)}\\ &=&\frac{(77-110.4617)^2}{110.4617}+\cdots + \frac{(97-110.46)^2}{110.46}\\ &=& 10.136 +\cdots + 1.641\\ &=& 16.89. \end{eqnarray*} $$ `

##### Step 6 Decision (Traditional approach)

The test statistic is $\chi^2 =16.89$ which falls $inside$ the critical region bounded by the critical value $12.5916$, we $\textit{reject}$ the null hypothesis.

**OR**

##### Step 6 Decision ($p$-value approach)

The p-value is $P(\chi^2_{6}>16.89) =0.0097$.

As the p-value $0.0097$ is $\textit{less than}$ the significance level of $\alpha = 0.05$, we $\textit{reject}$ the null hypothesis.

We conclude that at least one of the proportion is different from $\frac{1}{7}$.

#### Further Reading

- Statistics
- Descriptive Statistics
- Probability Theory
- Probability Distribution
- Hypothesis Testing
- Confidence interval
- Sample size determination
- Non-parametric Tests
- Correlation Regression
- Statistics Calculators