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

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)"

Answer from: Quest

step-by-step explanation:

because first you have to add the 4x, 9x, and the 7x together and you get 20x and then you add the last numbers up and you get 5and then you put the subtraction sign in there and that's how you get the answer.

Need , 5. find the perimeter. a. 40x – 10 b. 12x – 5 c. 20x – 5 d. 20x – 3

Answer from: Quest

step-by-step explanation:

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