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.
