# Solved (Free): Births Records of randomly selected births were obtained and categorized according to the day of the week that they occurred

#### ByDr. Raju Chaudhari

Apr 3, 2021

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 $\frac{1}{7}$.