## MATH1015 Quizzes

Quiz 12: Chi-square tests
Question 1 Questions
If $P\left({\chi }_{5}^{2}>a\right)=0.1$, find $a$. Exactly one option must be correct)
 a) $a=9.236$ b) $a=1.610$ c) $a=7.779$ d) $a=15.086$

Choice (a) is correct!
Use the chi-square tables with $\nu =5$, and $p=.1$.
Choice (b) is incorrect
$P\left({\chi }_{5}^{2}>1.610\right)=0.9$.
Choice (c) is incorrect
You have used the wrong value for $\nu$.
Choice (d) is incorrect
$P\left({\chi }_{5}^{2}>15.086\right)=0.01$.
Provide bounds for $p=P\left({\chi }_{25}^{2}>38.5\right)$. Exactly one option must be correct)
 a) $0\le p\le 0.01$ b) $0.05\le p\le 0.025$ c) $0.025\le p\le 0.05$ d) $0.05\le p\le 0.1$

Choice (a) is incorrect
Try again.
Choice (b) is incorrect
You have reversed the upper and lower limits. Obviously $P\left({\chi }_{25}^{2}>38.5\right)$ must exceed 0.025.
Choice (c) is correct!
Draw a chi-square diagram and mark in the value 38.5 on the $x$-axis. From the tables, $P\left({\chi }_{25}^{2}>40.646\right)=0.025$ and $P\left({\chi }_{25}^{2}>37.652\right)=0.05.$ Mark both 37.652 and 40.646 on the $x$-axis as well, and it will be obvious that $P\left({\chi }_{25}^{2}>38.5\right)$ has bounds (0.025, 0.05).
Choice (d) is incorrect
Draw a chi-square diagram and mark in the value 38.5 on the $x$-axis. From the tables, $P\left({\chi }_{25}^{2}>37.652\right)=0.05.$ Now mark 37.652 on your diagram, and it will be obvious that $P\left({\chi }_{25}^{2}>38.5\right)$ must be less than 0.05.
Questions 3 to 6 use the same information.
The observed frequencies of genotypes A, B and C of 100 progeny from a genetic cross are ${O}_{i}$: 18, 55 and 27 respectively. A model states that A, B and C occur in the ratio 1:2:1.
Under the model, the expected frequencies (${E}_{i}$) are: Exactly one option must be correct)
 a) ${E}_{i}$: 10, 20, 10 b) ${E}_{i}$: 25, 50, 25 c) ${E}_{i}$: 25, 25, 50 d) ${E}_{i}$: 50, 25, 25

Choice (a) is incorrect
The frequencies must satisfy ${\sum }_{i}{E}_{i}={\sum }_{i}{O}_{i}$.
Choice (b) is correct!
Under the model, the proportions for A, B and C are respectively $\frac{1}{4},\phantom{\rule{1em}{0ex}}\frac{1}{2},\phantom{\rule{1em}{0ex}}\frac{1}{4}$. Multiplying by 100 gives the expected frequencies 25, 50, 25.
Choice (c) is incorrect
These are not in the correct order.
Choice (d) is incorrect
These are not in the correct order.
Questions 3 to 6 use the same information.
The observed frequencies of genotypes A, B and C of 100 progeny from a genetic cross are ${O}_{i}$: 18, 55 and 27 respectively. A model states that A, B and C occur in the ratio 1:2:1.
Writing ${E}_{i}$ for the corresponding expected frequencies under the model, calculate ${\tau }_{obs}={\sum }_{i}\frac{{\left({O}_{i}-{E}_{i}\right)}^{2}}{{E}_{i}}$, the observed value of the chi-square test statistic. Exactly one option must be correct)
 a) 102.62 b) 2.62 c) 0.262 d) 0.78

Choice (a) is incorrect
You have found ${\sum }_{i}\frac{{O}_{i}^{2}}{{E}_{i}}$.
Choice (b) is correct!
${\tau }_{obs}=\frac{{\left(18-25\right)}^{2}}{25}+\frac{{\left(55-50\right)}^{2}}{50}+\frac{{\left(27-25\right)}^{2}}{25}=2.62$.
Choice (c) is incorrect
Choice (d) is incorrect
${\tau }_{obs}={\sum }_{i}\frac{{\left({O}_{i}-{E}_{i}\right)}^{2}}{{E}_{i}}$. You have found instead, $\frac{{\sum }_{i}{\left({O}_{i}-{E}_{i}\right)}^{2}}{{\sum }_{i}{E}_{i}}$.
Questions 3 to 6 use the same information.
The observed frequencies of genotypes A, B and C of 100 progeny from a genetic cross are ${O}_{i}$: 18, 55 and 27 respectively. A model states that A, B and C occur in the ratio 1:2:1.
State the appropriate chi-square variable, ${\chi }_{\nu }^{2}$ , for testing the goodness of fit of this model. Exactly one option must be correct)
 a) ${\chi }_{2}^{2}$ b) ${\chi }_{1}^{2}$ c) ${\chi }_{3}^{2}$ d) ${\chi }_{99}^{2}$

Choice (a) is correct!
There are three categories, so $\nu =2$.
Choice (b) is incorrect
Try again. There are three categories
Choice (c) is incorrect
$\nu =k-1$ where $k$ is the number of categories.
Choice (d) is incorrect
There are 3 categories, not 100.
Questions 3 to 6 use the same information.
The observed frequencies of genotypes A, B and C of 100 progeny from a genetic cross are ${O}_{i}$: 18, 55 and 27 respectively. A model states that A, B and C occur in the ratio 1:2:1.
Test the model for goodness of fit, providing an expression for the P-value. Exactly one option must be correct)
 a) P-value = $P\left({\chi }_{99}^{2}\ge 2.62\right)>0.99.$ The model is a good fit. b) P-value = $P\left({\chi }_{1}^{2}\ge 2.62\right)<0.1.$ The data provide evidence against the model. c) P-value = $P\left({\chi }_{2}^{2}\le 2.62\right)$ is very large. The data are consistent with the model. d) P-value = $P\left({\chi }_{2}^{2}\ge 2.62\right)>0.1.$ The data are consistent with the model.

Choice (a) is incorrect
You have used the wrong chi-square distribution.
Choice (b) is incorrect
You have used the wrong chi-square distribution.
Choice (c) is incorrect
The P-value is not the lower tail of ${\chi }_{2}^{2}$.
Choice (d) is correct!
Since there are 3 categories and since ${\tau }_{obs}=2.62$, the appropriate chi-square distribution is ${\chi }_{2}^{2}$ and P-value = $P\left({\chi }_{2}^{2}\ge 2.62\right)>0.1$. This means that the data are fairly typical of what is expected under the model.
Questions 7 to 10 use the same information.
A market researcher wishes to assess consumers’ preference among four different colours available on a name-brand dishwasher. In a sample of 198 recent sales, 61 were for stainless steel, 55 for white, 41 for black and 41 for cream.
To test the hypothesis that all four colours are equally preferred, a chi-square test is used. Writing ${E}_{i}$ for the corresponding expected frequencies under the model, calculate ${\tau }_{obs}={\sum }_{i}\frac{{\left({O}_{i}-{E}_{i}\right)}^{2}}{{E}_{i}}$, the observed value of the chi-square test statistic. Exactly one option must be correct)
 a) ${\tau }_{obs}=6.2$ b) ${\tau }_{obs}=10.3$ c) ${\tau }_{obs}=206.7$ d) ${\tau }_{obs}=204.2$

Choice (a) is correct!
${\tau }_{obs}=\frac{{\left(61-49.5\right)}^{2}}{49.5}+\frac{{\left(55-49.5\right)}^{2}}{49.5}+\frac{{\left(41-49.5\right)}^{2}}{49.5}+\frac{{\left(41-49.5\right)}^{2}}{49.5}=6.2$.
Choice (b) is incorrect
Choice (c) is incorrect
Choice (d) is incorrect
You have found ${\sum }_{i}\frac{{O}_{i}^{2}}{{E}_{i}}$ instead of ${\sum }_{i}\frac{{\left({O}_{i}-{E}_{i}\right)}^{2}}{{E}_{i}}$
Questions 7 to 10 use the same information.
A market researcher wishes to assess consumers’ preference among four different colours available on a name-brand dishwasher. In a sample of 198 recent sales, 61 were for stainless steel, 55 for white, 41 for black and 41 for cream.
To test the hypothesis that all four colours are equally preferred, a chi-square test is used. Write an expression for the P-value. Exactly one option must be correct)
 a) P-value = $P\left({\chi }_{2}^{2}\ge 6.2\right)>0.05$ b) P-value = $P\left({\chi }_{3}^{2}\ge 6.2\right)>0.1$ c) P-value = $P\left({\chi }_{3}^{2}\ge 10.3\right)<0.025.$ d) P-value = $P\left({\chi }_{197}^{2}\ge 6.2\right)>0.99$

Choice (a) is incorrect
You have used the wrong chi-square distribution.
Choice (b) is correct!
Under this model, the ${E}_{i}$ are all 49.5, and ${\tau }_{obs}=6.2$. Since there are 4 categories, the appropriate chi-square is ${\chi }_{3}^{2}$, and P-value = $P\left({\chi }_{3}^{2}\ge 6.2\right)>0.1$
Choice (c) is incorrect
Check your calculation of ${\tau }_{obs}$.
Choice (d) is incorrect
Try again. You have used the wrong chi-square distribution.
Questions 7 to 10 use the same information.
A market researcher wishes to assess consumers’ preference among four different colours available on a name-brand dishwasher. In a sample of 198 recent sales, 61 were for stainless steel, 55 for white, 41 for black and 41 for cream.
Writing ${E}_{i}$ for the expected frequencies under a model which states that the preferences are in the ratio 6:5:4:3, calculate ${\tau }_{obs}={\sum }_{i}\frac{{\left({O}_{i}-{E}_{i}\right)}^{2}}{{E}_{i}}$, the observed value of the chi-square test statistic. Exactly one option must be correct)
 a) ${\tau }_{obs}=200.52$ b) ${\tau }_{obs}=8$ c) ${\tau }_{obs}=0.495$ d) ${\tau }_{obs}=2.52$

Choice (a) is incorrect
You have found ${\sum }_{i}\frac{{O}_{i}^{2}}{{E}_{i}}$ instead of ${\sum }_{i}\frac{{\left({O}_{i}-{E}_{i}\right)}^{2}}{{E}_{i}}$
Choice (b) is incorrect
The expected frequencies under this model are 66, 55, 44, 33. Using these, recalculate ${\tau }_{obs}$.
Choice (c) is incorrect
${\tau }_{obs}={\sum }_{i}\frac{{\left({O}_{i}-{E}_{i}\right)}^{2}}{{E}_{i}}$. You have found instead, $\frac{{\sum }_{i}{\left({O}_{i}-{E}_{i}\right)}^{2}}{{\sum }_{i}{E}_{i}}$.
Choice (d) is correct!
The expected frequencies under this model are 66, 55, 44, 33 (for example the model indicates that the proportion preferring stainless steel is $\frac{6}{6+5+4+3}=\frac{1}{3}$ Multiplying by 198 gives 66. similarly for the other three colours.
Thus, ${\tau }_{obs}=\frac{{\left(61-66\right)}^{2}}{66}+\frac{{\left(55-55\right)}^{2}}{55}+\frac{{\left(41-44\right)}^{2}}{44}+\frac{{\left(41-33\right)}^{2}}{33}=2.52$.
Questions 7 to 10 use the same information.
A market researcher wishes to assess consumers’ preference among four different colours available on a name-brand dishwasher. In a sample of 198 recent sales, 61 were for stainless steel, 55 for white, 41 for black and 41 for cream.
To test the hypothesis that the preferences are in the ratio 6:5:4:3, a chi-square test is used. Write an expression for the P-value. Exactly one option must be correct)
 a) P-value = $P\left({\chi }_{3}^{2}\ge 2.52\right)>0.1$ b) P-value = $P\left({\chi }_{3}^{2}\ge 8\right)<0.05$ c) P-value = $P\left({\chi }_{3}^{2}\ge 0.495\right)>0.9$ d) P-value = $P\left({\chi }_{197}^{2}\ge 2.52\right)>.99$

Choice (a) is correct!
Under this model, the ${E}_{i}$ are respectively 66, 55, 44, 33 and ${\tau }_{obs}=2.52$. Since there are 4 categories, the appropriate chi-square is ${\chi }_{3}^{2}$ and the P-value is $P\left({\chi }_{3}^{2}\ge 2.52\right)>0.1$.
Choice (b) is incorrect
Check your calculation of ${\tau }_{obs}$.
Choice (c) is incorrect
Check your calculation of ${\tau }_{obs}$.
Choice (d) is incorrect
You have used the wrong chi-square distribution.