given below is the smoking status by level of education for residents 18 years old or older living in a…

given below is the smoking status by level of education for residents 18 years old or older living in a certain country from a random sample of 1016 residents. complete parts (a) and (b). click the icon to view the chi - square table of critical values. (a) test whether smoking status and level of education are independent at the α = 0.01 level of significance. what are the hypotheses? a. h₀: the three smoking statuses are independent. h₁: the three smoking statuses are dependent. b. h₀: smoking status and level of education are independent. h₁: smoking status and level of education are dependent. c. h₀: smoking status and level of education are dependent. h₁: smoking status and level of education are independent. find the test statistic. χ₀² = □ (round to three decimal places as needed.)

given below is the smoking status by level of education for residents 18 years old or older living in a certain country from a random sample of 1016 residents. complete parts (a) and (b). click the icon to view the chi - square table of critical values. (a) test whether smoking status and level of education are independent at the α = 0.01 level of significance. what are the hypotheses? a. h₀: the three smoking statuses are independent. h₁: the three smoking statuses are dependent. b. h₀: smoking status and level of education are independent. h₁: smoking status and level of education are dependent. c. h₀: smoking status and level of education are dependent. h₁: smoking status and level of education are independent. find the test statistic. χ₀² = □ (round to three decimal places as needed.)

Answer

Explanation:

Step1: Calculate row - totals

Row 1 total: $168 + 94+194 = 456$ Row 2 total: $104 + 76+144 = 324$ Row 3 total: $45 + 22+52 = 119$ Row 4 total: $35 + 35+47 = 117$

Step2: Calculate column - totals

Column 1 total: $168+104 + 45+35=352$ Column 2 total: $94 + 76+22+35 = 227$ Column 3 total: $194+144 + 52+47 = 437$

Step3: Calculate the grand - total

Grand - total $=1016$

Step4: Calculate expected frequencies

For cell in row 1, column 1: $E_{11}=\frac{456\times352}{1016}\approx158.278$ For cell in row 1, column 2: $E_{12}=\frac{456\times227}{1016}\approx102.530$ For cell in row 1, column 3: $E_{13}=\frac{456\times437}{1016}\approx195.192$ For cell in row 2, column 1: $E_{21}=\frac{324\times352}{1016}\approx112.658$ For cell in row 2, column 2: $E_{22}=\frac{324\times227}{1016}\approx72.638$ For cell in row 2, column 3: $E_{23}=\frac{324\times437}{1016}\approx138.704$ For cell in row 3, column 1: $E_{31}=\frac{119\times352}{1016}\approx41.398$ For cell in row 3, column 2: $E_{32}=\frac{119\times227}{1016}\approx26.574$ For cell in row 3, column 3: $E_{33}=\frac{119\times437}{1016}\approx51.028$ For cell in row 4, column 1: $E_{41}=\frac{117\times352}{1016}\approx40.666$ For cell in row 4, column 2: $E_{42}=\frac{117\times227}{1016}\approx26.258$ For cell in row 4, column 3: $E_{43}=\frac{117\times437}{1016}\approx50.076$

Step5: Calculate the chi - square test statistic

[ \begin{align*} \chi_{0}^{2}&=\sum\frac{(O - E)^{2}}{E}\ &=\frac{(168 - 158.278)^{2}}{158.278}+\frac{(94 - 102.530)^{2}}{102.530}+\frac{(194 - 195.192)^{2}}{195.192}+\frac{(104 - 112.658)^{2}}{112.658}+\frac{(76 - 72.638)^{2}}{72.638}+\frac{(144 - 138.704)^{2}}{138.704}+\frac{(45 - 41.398)^{2}}{41.398}+\frac{(22 - 26.574)^{2}}{26.574}+\frac{(52 - 51.028)^{2}}{51.028}+\frac{(35 - 40.666)^{2}}{40.666}+\frac{(35 - 26.258)^{2}}{26.258}+\frac{(47 - 50.076)^{2}}{50.076}\ &\approx13.964 \end{align*} ]

Answer:

$13.964$