Question

Use the given categorical data to construct the relative frequency distribution.

Answer 1

Answer:

$P(M) = \frac{52}{390}= 0.133$

$P(T) = \frac{64}{390}= 0.164$

$P(W) = \frac{70}{390}= 0.179$

$P(Th) = \frac{59}{390}= 0.151$

$P(F) = \frac{60}{390}= 0.154$

$P(S) = \frac{43}{390}= 0.110$

$P(Su) = \frac{42}{390}= 0.109$

As we can see the relative frequencies are not too different the average for all the relative frequencies is $\frac{1}{7}= 0.143$ and the maximum absolute difference |0.143-0.109|= 0.034 is not enough higher to conclude that we have significant difference from the average or expected value.

Step-by-step explanation:

For this case we have the frequency of the number of times that the outcome happens per each day of the week:

Day                  Frequency

Monday               52

Tuesday              64

Wednesday        70  

Thursday             59

Friday                  60

Saturday             43

Sunday                42

We can begin findind the total for the frequency and we got:

52+64+70+59+60+43+42=390

And now we can calculate the relative frequency for each case like this:

$P(M) = \frac{52}{390}= 0.133$

$P(T) = \frac{64}{390}= 0.164$

$P(W) = \frac{70}{390}= 0.179$

$P(Th) = \frac{59}{390}= 0.151$

$P(F) = \frac{60}{390}= 0.154$

$P(S) = \frac{43}{390}= 0.110$

$P(Su) = \frac{42}{390}= 0.109$

As we can see the relative frequencies are not too different the average for all the relative frequencies is $\frac{1}{7}= 0.143$ and the maximum absolute difference |0.143-0.109|= 0.034 is not enough higher to conclude that we have significant difference from the average or expected value.

In order to test this more formally we can use a chi square test but the problem not ask for this.