50 points! Please help fast. I will mark brainliest. Random answers will be reported

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

3 years ago

Please help will give brainliest

Ivanshal [37]

The answer will be A

5 0

3 years ago

Translate the verbal phrase into a algebraic expression pls

erica [24]

2 × (x + 3) < 5
...................................................................

4 0

3 years ago

Can somebody help me with this question please and fast!! I’ll mark brainliest

Gnoma [55]

Answer:

The midrange of the data set shown is 85

8 0

3 years ago

Sue either travels by bus or walks when she visits the shops. The probability that she catches the bus from the shops is 0.7. Sh

Alina [70]

Corrected Question

Sue travels by bus or walks when she visits the shops. The probability that she catches the bus to the shops is 0.4. The probability she catches the bus from the shops is 0.7. Show the probability that Sue walks at one way is 0.72

Answer:

$P(A \cup B) =0.72$ (Proved)

Step-by-step explanation:

Sue travels by bus or walks when she visits the shops.

Let the event that she catches the bus to the shop=A

Let the event that she catches the bus from the shop=B

P(A)=0.4

P(B)=0.7

Both A and B are independent events.

Therefore,Probability that she catches the bus to and from the shop:

P(A∩B) = 0.4 X 0.7= 0.28

Probability Sue walks at least one way $P(A \cup B) = 1 - P(A \cap B)$

$= 1 - 0.28\\= 0.72$

Hence, the probability that Sue walks at least one way is 0.72.

8 0

3 years ago