Mathematics

Mathematics, 21.02.2020 19:32, savage5447

A walk-in medical clinic believes that arrivals are uniformly distributed over weekdays (Monday through Friday). It has collected the following data based on a random sample of 100 days. Assuming that a goodness-of-fit test is to be conducted using a 0.10 level of significance, what is the critical value?

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answered: stevencardona12mejia

$\chi^2 = \frac{(25-20)^2}{20}+\frac{(22-20)^2}{20}+\frac{(19-20)^2}{20}+\frac{(18-20)^2}{20}+ \frac{(16-20)^2}{20} =2.5$

Now we can calculate the degrees of freedom for the statistic given by:

df=(categories-1)=(5-1)=4

$p_v = P(\chi^2_{4} 2.5)=0.64$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(2.5,4,TRUE)"

Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 10% of significance, and we can conclude that we NO have significant differences in the arrivals from Monday to Friday.

Step-by-step explanation:

For this case we assume the following data for the observed frequencies:

Monday 25, Tuesday 22, Wednesday 19 , Thursday 18, Friday 16

Previous concepts

A chi-square goodness of fit test "determines if a sample data matches a population".

A chi-square test for independence "compares two variables in a contingency table to see if they are related. In a more general sense, it tests to see whether distributions of categorical variables differ from each another".

The observed values are given by:

Monday 25, Tuesday 22, Wednesday 19 , Thursday 18, Friday 16

We need to conduct a chi square test in order to check the following hypothesis:

H0: The arrivals are uniformly distributed over weekdays (Monday through Friday

H1:The arrivals are NOT uniformly distributed over weekdays (Monday through Friday

The level of significance assumed for this case is $\alpha=0.1$

The statistic to check the hypothesis is given by:

$\chi^2 = \sum_{i=1}^n \frac{(O_i -E_i)^2}{E_i}$

Now we just need to calculate the expected values with the following formula $E_i = \frac{total}{n}= \frac{100}{5}= 20$

And now we can calculate the statistic:

$\chi^2 = \frac{(25-20)^2}{20}+\frac{(22-20)^2}{20}+\frac{(19-20)^2}{20}+\frac{(18-20)^2}{20}+ \frac{(16-20)^2}{20} =2.5$

Now we can calculate the degrees of freedom for the statistic given by:

df=(categories-1)=(5-1)=4

And we can calculate the p value given by:

$p_v = P(\chi^2_{4} 2.5)=0.64$

And we can find the p value using the following excel code:

"=1-CHISQ.DIST(2.5,4,TRUE)"

Since the p value is higher than the significance level we FAIL to reject the null hypothesis at 10% of significance, and we can conclude that we NO have significant differences in the arrivals from Monday to Friday.

Answer

answered: Guest

yes

step-by-step explanation:

Answer

answered: Guest

Idon’t know the answer to this

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