Question

question 49 of 54 the monterey bay aquarium, founded in 1984, is situated on the beautiful coast of monterey bay in the historic cannery row district. in 1985, the aquarium began a survey program that involved randomly sampling visitors as they exit for the day. the survey included visitor demographic information, use of social media, and opinions on their aquarium visit. for each visitor sampled during 2013–2015, the distribution of the number of children in their group is given. number of children 0 1 2 3 or more 2013 1855 585 599 515 2014 1751 636 601 506 2015 1998 591 483 289 to access the complete data set, click the link for your preferred software format: csv excel jmp mac-text minitab14-18 minitab18+ pc-text r spss ti crunchit! in this exercise, we will answer the question of whether there a significant difference in the distribution of the number of children in the group over this three-year period, and if so, how the distribution has changed. solve: calculate the test statistic for the chi-square test of homogeneity. give your answer to three decimal places. \\(chi^2 = \\)

Explanation:

Step1: Find Row and Column Totals

First, we calculate the row totals (for each year) and column totals (for each number of children).

For the number of children = 0:
2013: 1855, 2014: 1751, 2015: 1998
Column total ($C_0$) = 1855 + 1751 + 1998 = 5604

For the number of children = 1:
2013: 585, 2014: 636, 2015: 591
Column total ($C_1$) = 585 + 636 + 591 = 1812

For the number of children = 2:
2013: 599, 2014: 601, 2015: 483
Column total ($C_2$) = 599 + 601 + 483 = 1683

For the number of children = 3 or more:
2013: 515, 2014: 506, 2015: 289
Column total ($C_3$) = 515 + 506 + 289 = 1310

Now, row totals:
2013 total ($R_{2013}$) = 1855 + 585 + 599 + 515 = 3554
2014 total ($R_{2014}$) = 1751 + 636 + 601 + 506 = 3494
2015 total ($R_{2015}$) = 1998 + 591 + 483 + 289 = 3361

Grand total ($N$) = 3554 + 3494 + 3361 = 10409

Step2: Calculate Expected Frequencies

The formula for expected frequency $E_{ij}$ for row $i$ and column $j$ is $E_{ij} = \frac{R_i \times C_j}{N}$

For (2013, 0): $E_{2013,0} = \frac{3554 \times 5604}{10409} \approx \frac{3554\times5604}{10409} \approx \frac{19916616}{10409} \approx 1913.35$

For (2013, 1): $E_{2013,1} = \frac{3554 \times 1812}{10409} \approx \frac{3554\times1812}{10409} \approx \frac{6440848}{10409} \approx 618.79$

For (2013, 2): $E_{2013,2} = \frac{3554 \times 1683}{10409} \approx \frac{3554\times1683}{10409} \approx \frac{5981382}{10409} \approx 574.63$

For (2013, 3): $E_{2013,3} = \frac{3554 \times 1310}{10409} \approx \frac{3554\times1310}{10409} \approx \frac{4655740}{10409} \approx 447.23$

For (2014, 0): $E_{2014,0} = \frac{3494 \times 5604}{10409} \approx \frac{3494\times5604}{10409} \approx \frac{19580376}{10409} \approx 1881.07$

For (2014, 1): $E_{2014,1} = \frac{3494 \times 1812}{10409} \approx \frac{3494\times1812}{10409} \approx \frac{6331128}{10409} \approx 608.23$

For (2014, 2): $E_{2014,2} = \frac{3494 \times 1683}{10409} \approx \frac{3494\times1683}{10409} \approx \frac{5880402}{10409} \approx 564.90$

For (2014, 3): $E_{2014,3} = \frac{3494 \times 1310}{10409} \approx \frac{3494\times1310}{10409} \approx \frac{4577140}{10409} \approx 439.89$

For (2015, 0): $E_{2015,0} = \frac{3361 \times 5604}{10409} \approx \frac{3361\times5604}{10409} \approx \frac{18835044}{10409} \approx 1810.58$

For (2015, 1): $E_{2015,1} = \frac{3361 \times 1812}{10409} \approx \frac{3361\times1812}{10409} \approx \frac{6090132}{10409} \approx 585.08$

For (2015, 2): $E_{2015,2} = \frac{3361 \times 1683}{10409} \approx \frac{3361\times1683}{10409} \approx \frac{5659563}{10409} \approx 543.67$

For (2015, 3): $E_{2015,3} = \frac{3361 \times 1310}{10409} \approx \frac{3361\times1310}{10409} \approx \frac{4402910}{10409} \approx 422.88$

Step3: Calculate Chi-Square Statistic

The formula for chi-square is $\chi^2 = \sum \frac{(O_{ij} - E_{ij})^2}{E_{ij}}$

For each cell:

- (2013, 0): $\frac{(1855 - 1913.35)^2}{1913.35} \approx \frac{(-58.35)^2}{1913.35} \approx \frac{3404.72}{1913.35} \approx 1.78$
- (2013, 1): $\frac{(585 - 618.79)^2}{618.79} \approx \frac{(-33.79)^2}{618.79} \approx \frac{1130.06}{618.79} \approx 1.83$
- (2013, 2): $\frac{(599 - 574.63)^2}{574.63} \approx \frac{(24.37)^2}{574.63} \approx \frac{593.80}{574.63} \approx 1.03$
- (2013, 3): $\frac{(515 - 447.23)^2}{447.23} \approx \frac{(67.77)^2}{447.23} \approx \frac{4592.77}{447.23} \approx 10.27$
- (2014, 0): $\frac{(1751 - 1881.07)^2}{1881.07} \approx \frac{(-130.07)^2}{1881.07} \approx \frac{16918.20}{1881.07} \approx 9.00$
- (2014, 1): $\frac{(636 - 608.23)^2}{608.23} \approx \frac{(27.77)^2}{608.23} \approx \frac{770.17}{608.23} \approx 1.27$
- (2014, 2): $\frac{(601 - 564.90)^2}{564.90} \approx \frac{(36.10)^2}{564.90} \approx \frac{1303.21}{564.90} \approx 2.31$
- (2014, 3): $\frac{(506 - 439.89)^2}{439.89} \approx \frac{(66.11)^2}{439.89} \approx \frac{4370.53}{439.89} \approx 9.93$
- (2015, 0): $\frac{(1998 - 1810.58)^2}{1810.58} \approx \frac{(187.42)^2}{1810.58} \approx \frac{35126.26}{1810.58} \approx 19.40$
- (2015, 1): $\frac{(591 - 585.08)^2}{585.08} \approx \frac{(5.92)^2}{585.08} \approx \frac{35.05}{585.08} \approx 0.06$
- (2015, 2): $\frac{(483 - 543.67)^2}{543.67} \approx \frac{(-60.67)^2}{543.67} \approx \frac{3680.85}{543.67} \approx 6.77$
- (2015, 3): $\frac{(289 - 422.88)^2}{422.88} \approx \frac{(-133.88)^2}{422.88} \approx \frac{17924.85}{422.88} \approx 42.39$

Now, sum all these values:

1.78 + 1.83 + 1.03 + 10.27 + 9.00 + 1.27 + 2.31 + 9.93 + 19.40 + 0.06 + 6.77 + 42.39

Let's add step by step:

1.78 + 1.83 = 3.61

3.61 + 1.03 = 4.64

4.64 + 10.27 = 14.91

14.91 + 9.00 = 23.91

23.91 + 1.27 = 25.18

25.18 + 2.31 = 27.49

27.49 + 9.93 = 37.42

37.42 + 19.40 = 56.82

56.82 + 0.06 = 56.88

56.88 + 6.77 = 63.65

63.65 + 42.39 = 106.04

Answer:

106.04 (rounded to two decimal places, but the problem asks for three, so 106.040 or as calculated)