Mathematics, 12.08.2020 06:01 bear342

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

Monday 17

Tuesday 13

Wednesday 14

Thursday 16

He expected to have 15 customers each day. To answer whether the number of customers follows a uniform distribution, a chi-square test for goodness of fit should be performed. (alpha = 0.10)

What is the chi-squared test statistic? Answers are rounded to the nearest hundredth.

Answers: 2

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Answer from: austind9027

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.

Answer from: Quest

The correct answer is b

Answer from: Quest

32.5 is the answer i got thats the answer

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

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