given below is the smoking status by level of...
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:
$13.964$