Respuesta :
Answer:
Day           Relative Frequency
Monday       52/400=0.13 = 13%
Tuesday       64/400=0.16=16%
Wednesday    71/400=0.1775=17.75%
Thursday       57/400=0.1425=14.25%
Friday         57/400=0.1425=14.25%
Saturday       46/400=0.115=11.5%
Sunday        53/400=0.1325=13.25%
Total             1.00=100%
After we see the results obtained we can conclude that the relative frequencies are very similar for each day analyzed. But in order to have a precise answer is possible to use a chi square test.
[tex]\chi^2 = \frac{(13-14.286)^2}{14.286}+\frac{(16-14.286)^2}{14.286}+\frac{(17.75-14.286)^2}{14.286}+\frac{(14.25-14.286)^2}{14.286}+\frac{(14.25-14.286)^2}{14.286}+\frac{(11.5-14.286)^2}{14.286}+\frac{(13.25-14.286)^2}{14.286} =1.78[/tex]
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=(Categories-1)=7-1=6[/tex]
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{6} >1.78)=0.939[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(1.78,6,TRUE)"
Since the p value is higher than any significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that such births occur on the days of the week with equal​ frequency.
Step-by-step explanation:
Data given
Day        Frequency
Monday       52
Tuesday       64
Wednesday    71
Thursday       57
Friday         57
Saturday       46
Sunday        53
The first step on this case is find the total sum of frequencies given by:
52+64+71+57+57+46+53=400.
A relative frequency is "the fraction of times an answer occurs. To find the relative frequencies, we just need to divide each frequency by the total number", and we can express it as % or as fraction.
If we apply this definition and we find the relative frequencies we have:
Day           Relative Frequency
Monday       52/400=0.13 = 13%
Tuesday       64/400=0.16=16%
Wednesday    71/400=0.1775=17.75%
Thursday       57/400=0.1425=14.25%
Friday         57/400=0.1425=14.25%
Saturday       46/400=0.115=11.5%
Sunday        53/400=0.1325=13.25%
Total             1.00=100%
We can check the operations with the total, if the total is not equal to 1 we need to review the calculations since the total sum needs to be 1 or 100%.
After we see the results obtained we can conclude that the relative frequencies are very similar for each day analyzed. But in order to have a precise answer is possible to use a chi square test.
The statistic to check the hypothesis is given by:
[tex]\sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}[/tex]
The expected frequencies for each day would be 100%/7 =14.286%
[tex]\chi^2 = \frac{(13-14.286)^2}{14.286}+\frac{(16-14.286)^2}{14.286}+\frac{(17.75-14.286)^2}{14.286}+\frac{(14.25-14.286)^2}{14.286}+\frac{(14.25-14.286)^2}{14.286}+\frac{(11.5-14.286)^2}{14.286}+\frac{(13.25-14.286)^2}{14.286} =1.78[/tex]
Now we can calculate the degrees of freedom for the statistic given by:
[tex]df=(Categories-1)=7-1=6[/tex]
And we can calculate the p value given by:
[tex]p_v = P(\chi^2_{6} >1.78)=0.939[/tex]
And we can find the p value using the following excel code:
"=1-CHISQ.DIST(1.78,6,TRUE)"
Since the p value is higher than any significance level we FAIL to reject the null hypothesis at 5% of significance, and we can conclude that such births occur on the days of the week with equal​ frequency.
