What value of z solves the equation

4. You want to know if there's an association between college students' spring break destinations and what year they're in. You

Pachacha [2.7K]

Answer:

$\chi^2 =27.356$

$p_v = P(\chi^2_{9} >27.356)=0.00122$

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

"=1-CHISQ.DIST(27.356,9,TRUE)"

Since the p value is lower than the significance level assumed 0.05 we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

Step-by-step explanation:

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".

Assume the following dataset:

Amusement Parks Mexico Home Other Total

Freshman 23 21 43 21 108

Sophomore 34 23 14 26 97

Junior 25 30 23 26 104

Senior 27 33 17 19 96

Total 109 107 97 92 405

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

H0: There is independence between the two random variables

H1: There is dependence between the two variables

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

The statistic to check the hypothesis is given by:

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

The table given represent the observed values, we just need to calculate the expected values with the following formula $E_i = \frac{total col * total row}{grand total}$

And the calculations are given by:

$E_{1} =\frac{109*108}{405}=29.07$

$E_{2} =\frac{107*108}{405}=28.53$

$E_{3} =\frac{97*108}{405}=25.87$

$E_{4} =\frac{92*108}{405}=24.53$

$E_{5} =\frac{109*97}{405}=26.11$

$E_{6} =\frac{107*97}{405}=25.63$

$E_{7} =\frac{97*97}{405}=23.23$

$E_{8} =\frac{92*97}{405}=22.03$

$E_{9} =\frac{109*104}{405}=27.99$

$E_{10} =\frac{107*104}{405}=27.48$

$E_{11} =\frac{97*104}{405}=24.91$

$E_{12} =\frac{92*104}{405}=23.62$

$E_{13} =\frac{109*96}{405}=25.84$

$E_{14} =\frac{107*96}{405}=25.36$

$E_{15} =\frac{97*96}{405}=22.99$

$E_{16} =\frac{92*96}{405}=21.81$

And the expected values are given by:

Amusement Parks Mexico Home Other Total

Freshman 29.07 28.53 25.87 24.53 108

Sophomore 26.11 25.63 23.23 22.03 97

Junior 27.99 27.48 24.91 23.62 104

Senior 25.84 25.36 22.99 21.81 96

Total 109 107 97 92 405

And now we can calculate the statistic:

$\chi^2 =27.356$

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

$df=(rows-1)(cols-1)=(4-1)(4-1)=9$

And we can calculate the p value given by:

$p_v = P(\chi^2_{9} >27.356)=0.00122$

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

"=1-CHISQ.DIST(27.356,9,TRUE)"

Since the p value is lower than the significance level assumed 0.05 we can reject the null hypothesis at 5% of significance, and we can conclude that we have association between the two variables analyzed.

7 0

2 years ago

ℜ 0, -180 is the same rotation as _______.

andreyandreev [35.5K]

270 degrees or -90 degrees.

7 0

3 years ago

Read 2 more answers

Factor of x^3-x^2-24x-36

Basile [38]

The trial and error method is used to find an initial factor:
If we let f(x) = x³ - x² - 24x - 36 and all we have to do is sub' in values of x until
f(x) = 0, we can use this to find an initial factor by the factor theorem:
f(1) = (1)³ - (1)² - 24(1) - 36 = -60
f(2) = (2)³ - (2)² - 24(2) - 36 = -80
f(5) = (5)³ - (5)² - 24(5) - 36 = -56
*** f(6) = (6)³ - (6)² - 24(6) - 36 = 0 ***

f(6) = 0 so (x - 6) is a factor of f(x).
This means that: f(x) = x³ - x² - 24x - 36 = (x - 6)(ax² + bx + c).
To find a,b and c, use long division (or inspection) to divide x³ - x² - 24x - 36 by x - 6.
The other 2 factors of f(x) can then be found by factorizing the
ax² + bx + c quadratic the way you would with any other quadratic (i.e. by quadratic formula, CTS or inspection).

5 0

3 years ago

What was the problem at kenya airways described in this case? what management, organization, and technology factors contributed

MatroZZZ [7]

The management, organisation, and technology factors contributed to this problem are listed below.

<h3>What was the problem at Kenya Airways ?</h3>

The problem in the airways was that corporation didn't know its customers, the airline hasn't been able to take use of its market opportunity in recent years.

Airways was unable to evaluate and keep track of its marketing efforts.

The technology factor that contributed were:

No reliable systems for tracking and accounting.
The technology used was neither accurate nor consistent.

The Organisation factors that contributed were

No communication between the organisation and the customers

No track record of the online campaigns and advertisement output

Customer Relations Needed to be improved.

The Management factors that contributed were

The management never gave reviews to the organisation about the failing system
The management even didn't take reviews from the customers and from the people working.

To know more about problem at Kenya Airways

brainly.com/question/27974672

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8 0

1 year ago

How many numbers are from k to n? (Including k and n.)

shepuryov [24]

Answer:

n - k + 1

Step-by-step explanation:

This is assuming (because you did not say) that n and k are integers and n is greater than k.

Example: from 2 to 5 {2, 3, 4, 5} includes 5 - 2 + 1 = 4 numbers.

Example: from -6 to 4 includes 4 - (-6) + 1 = 11 numbers, namely {-6, -5, -4, -3, -2, -1, 0, 1, 2, 3, 4}

7 0

2 years ago