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Taya2010 [7]
3 years ago
15

Explain how you can tell, without computing, whether the quotient of 1/2 divided by 6 is greater than 1 or less than 1.

Mathematics
2 answers:
Oliga [24]3 years ago
7 0
You can tell the quotient is less than one without computing because if you start with 1/2 which is smaller than one when you split it into 6 groups they are going to be smaller groups because when you divide something it gets smaller which has to be less than one because 1/2 is less than one
Gnesinka [82]3 years ago
4 0
You already know that the quetioent of 1/2 divided by 6 would be more or less than one because, 1/2 is equal to 50% and 50% of 6 is 3. So, you would know how the answer would turn out.
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Answer:

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Step-by-step explanation:

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A personnel manager is concerned about absenteeism. She decides to sample employee records to determine if absenteeism is distri
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Answer:

df=categor-1=6-1=5

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got \chi^2_{critc}= 11.0705

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And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

Step-by-step explanation:

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".

Solution to the problem

For this case we want to test:

H0: Absenteeism is distributed evenly throughout the week

H1: Absenteeism is NOT distributed evenly throughout the week

We have the following data:

Monday  Tuesday  Wednesday Thursday Friday Saturday    Total

 12             9                 11                 10           9            9              60

The level of significance assumed for this case is \alpha=0.05

The statistic to check the hypothesis is given by:

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

The table given represent the observed values, we just need to calculate the expected values with the following formula E_i = \frac{60}{6}= 10 and the expected value is the same for all the days since that's what we want to test.

now we can calculate the statistic:

\chi^2 = \frac{(12-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(11-10)^2}{10}+\frac{(10-10)^2}{10}+\frac{(9-10)^2}{10}+\frac{(9-10)^2}{10}=0.8

Now we can calculate the degrees of freedom (We know that we have 6 categories since we have information for 6 different days) for the statistic given by:

df=categor-1=6-1=5

The critical value can be founded with the following Excel formula:

=CHISQ.INV(1-0.05,5)

And we got \chi^2_{critc}= 11.0705

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And since our calculated value is lower than the critical we FAIL to reject the null hypothesis at 5% of significance

4 0
3 years ago
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aleksklad [387]
The answer is A the first choice
7 0
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Read 2 more answers
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Zielflug [23.3K]

Answer:

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I don’t wanna fail if you know the answer pls help me :(
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Step-by-step explanation: hope it helps

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