Question

Carl recorded the number of customers who visited his new store during the week:

Answer 1

Answer:

The chi - square test can be $\approx$ 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:

$Chi -square = \dfrac{(observed \ value - expected \ value )^2}{expected \ value}$

For Monday:

$Chi -square = \dfrac{(17 - 15 )^2}{15}$

$Chi -square = \dfrac{(2)^2}{15}$

$Chi - square = \dfrac{4}{15}$

chi - square = 0.2666666667

For Tuesday :

$Chi -square = \dfrac{(13- 15 )^2}{15}$

$Chi -square = \dfrac{(-2)^2}{15}$

$Chi - square = \dfrac{4}{15}$

chi - square = 0.2666666667

For Wednesday :

$Chi -square = \dfrac{(14- 15 )^2}{15}$

$Chi -square = \dfrac{(-1 )^2}{15}$

$Chi -square = \dfrac{(1 )}{15}$

chi - square = 0.06666666667

For Thursday:

$Chi -square = \dfrac{(16- 15 )^2}{15}$

$Chi -square = \dfrac{(1 )^2}{15}$

$Chi -square = \dfrac{(1 )}{15}$

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 $\approx$ 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.