Respuesta :
Answer:
The chi - square test can be [tex]\approx[/tex] 0.667
Step-by-step explanation:
From the given data :
The null hypothesis and the alternative hypothesis can be computed as:
Null hypothesis: The number of customers does follow a uniform distribution
Alternative hypothesis: The number of customers does not follow a uniform distribution
We learnt that: Carl recorded the number of customers who visited his new store during the week:
Day Customers
Monday 17
Tuesday 13
Wednesday 14
Thursday 16
The above given data was the observed value.
However, the question progress by stating that : He expected to have 15 customers each day.
Now; we can have an expected value for each customer as:
Observed Value Expected Value
Day Customers
Monday 17 15
Tuesday 13 15
Wednesday 14 15
Thursday 16 15
The Chi square corresponding to each data can be determined by using the formula:
[tex]Chi -square = \dfrac{(observed \ value - expected \ value )^2}{expected \ value}[/tex]
For Monday:
[tex]Chi -square = \dfrac{(17 - 15 )^2}{15}[/tex]
[tex]Chi -square = \dfrac{(2)^2}{15}[/tex]
[tex]Chi - square = \dfrac{4}{15}[/tex]
chi - square = 0.2666666667
For Tuesday :
[tex]Chi -square = \dfrac{(13- 15 )^2}{15}[/tex]
[tex]Chi -square = \dfrac{(-2)^2}{15}[/tex]
[tex]Chi - square = \dfrac{4}{15}[/tex]
chi - square = 0.2666666667
For Wednesday :
[tex]Chi -square = \dfrac{(14- 15 )^2}{15}[/tex]
[tex]Chi -square = \dfrac{(-1 )^2}{15}[/tex]
[tex]Chi -square = \dfrac{(1 )}{15}[/tex]
chi - square = 0.06666666667
For Thursday:
[tex]Chi -square = \dfrac{(16- 15 )^2}{15}[/tex]
[tex]Chi -square = \dfrac{(1 )^2}{15}[/tex]
[tex]Chi -square = \dfrac{(1 )}{15}[/tex]
chi - square = 0.06666666667
Observed Value Expected Value chi - square
Day Customers
Monday 17 15 0.2666666667
Tuesday 13 15 0.2666666667
Wednesday 14 15 0.06666666667
Thursday 16 15 0.06666666667
Total : 0.6666666668
The chi - square test can be [tex]\approx[/tex] 0.667
At level of significance ∝ = 0.10
degree of freedom = n - 1
degree of freedom = 4 - 1
degree of freedom = 3
At ∝ = 0.10 and df = 3
The p - value for the chi - square test statistics is 0.880937
Decision rule: If the p - value is greater than the level of significance , we fail to reject the null hypothesis
Conclusion: Since the p - value is greater than the level of significance , we fail to reject the null hypothesis and conclude that there is insufficient evidence to show that the number of customers does not follows a uniform distribution.
