Question 1

Following are observed frequencies. The null hypothesis is $H _ { 0 } : p _ { 1 } = 0.35 , p _ { 2 } = 0.3$, $p _ { 3 } = 0.1 , p _ { 4 } = 0.15 , p _ { 5 } = 0.1$.
$\begin{array}{l|lllll}
\text { Category } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Observed } & 51 & 32 & 11 & 13 & 16
\end{array}$

Test the hypothesis that the distribution of the observed frequencies is as given by the nul hypothesis. Use the $\alpha = 0.01$ level of significance.
A) Reject $H _ { 0 }$.
B) Do not reject $H _ { 0 }$.

Accepted Answer

To test the hypothesis, we first calculate the expected frequencies for each category based on the null hypothesis proportions and the total sample size. The total sample size is $51 + 32 + 11 + 13 + 16 = 123$. The expected frequencies are calculated as follows:- For category 1: $0.35 \times 123 = 43.05$- For category 2: $0.3 \times 123 = 36.9$- For category 3: $0.1 \times 123 = 12.3$- For category 4: $0.15 \times 123 = 18.45$- For category 5: $0.1 \times 123 = 12.3$Next, we use the chi-square test statistic formula:$$\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$$where $O_i$ is the observed frequency and $E_i$ is the expected frequency for category $i$.Calculating the chi-square statistic:$$\chi^2 = \frac{(51 - 43.05)^2}{43.05} + \frac{(32 - 36.9)^2}{36.9} + \frac{(11 - 12.3)^2}{12.3} + \frac{(13 - 18.45)^2}{18.45} + \frac{(16 - 12.3)^2}{12.3}$$$$\chi^2 = \frac{(7.95)^2}{43.05} + \frac{(-4.9)^2}{36.9} + \frac{(-1.3)^2}{12.3} + \frac{(-5.45)^2}{18.45} + \frac{(3.7)^2}{12.3}$$$$\chi^2 = 1.47 + 0.65 + 0.14 + 1.61 + 1.11 = 4.98$$With 4 degrees of freedom (5 categories - 1), we look up the critical value of $\chi^2$ for $\alpha = 0.01$, which is approximately 13.28. Since our calculated $\chi^2$ value of 4.98 is less than the critical value, we do not have enough evidence to reject the null hypothesis at the 0.01 level of significance.

Question 2

$$\text { For four degrees of freedom, the value of } \alpha \text { shown in the } \chi ^ { 2 } \text { distribution below is } 0.05 \text {. }$$

Accepted Answer

False

Question 3

Before computing the test value, one should state the hypothesis.

Accepted Answer

Stating the hypothesis before computing the test value is essential in hypothesis testing to avoid bias and to clearly define what is being tested.

Question 4

In a goodness of fit test between one observed frequency distribution and one expected
frequency distribution, the degrees of freedom are equal to the number of categories
minus two.

Accepted Answer

The degrees of freedom for a goodness-of-fit test are equal to the number of categories minus one, not minus two.

Question 5

A biology professor claims that, on the average, $15 \%$ of her students get a grade of $A$, $25 \%$ get a B, $40 \%$ get a C , $10 \%$ get a D, and $10 \%$ get an $\mathrm { F }$. The grades of a random sample of 131 students were recorded. The following table presents the results.
$\begin{array}{c|lllll}
\text { Grade } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Observed } & 13 & 21 & 73 & 11 & 13
\end{array}$

What is the value of $\chi ^ { 2 }$ ?
A) 14.902
B) 9.488
C) 11.143
D) 16.873

Accepted Answer

The chi-square statistic is calculated using the formula $\chi^2 = \sum \frac{(O_i - E_i)^2}{E_i}$, where $O_i$ is the observed frequency and $E_i$ is the expected frequency. The expected frequencies are calculated based on the professor's claims: $E_A = 19.65$, $E_B = 32.75$, $E_C = 52.4$, $E_D = 13.1$, $E_F = 13.1$. Calculating $\chi^2$ gives approximately 14.902.

Question 6

The chi-square distribution is continuous, whereas the goodness-of-fit test is
.

Accepted Answer

The answer of The chi-square distribution is continuous, whereas the...

Question 7

The formula for finding the chi-square test value is .

Accepted Answer

The answer of The formula for finding the chi-square test...

Question 8

$$\text { Find the } \alpha = 0.05 \text { critical value for the chi-square statistic with } 10 \text { degrees of freedom. }$$
A) 3.940
B) 16.919
C) 20.483
D) 18.307

Accepted Answer

The critical value for a chi-square distribution with 10 degrees of freedom at $\alpha = 0.05$ is approximately 18.307.

Question 9

An assumption for the goodness-of-fit test is that the frequency for
each category must be five or more.

Accepted Answer

The answer of An assumption for the goodness-of-fit test is...

Question 10

The chi-square goodness-of-fit test is always one-tailed left.

Accepted Answer

The chi-square goodness-of-fit test is always one-tailed left because it tests whether the observed frequencies of a categorical variable differ significantly from the expected frequencies. If the observed frequencies are significantly lower than the expected frequencies, then the test will be significant.

Question 11

Following are observed frequencies. The null hypothesis is $H _ { 0 } : p _ { 1 } = 0.35 , p _ { 2 } = 0.2$, $p _ { 3 } = 0.15 , p 4 = 0.15 , p _ { 5 } = 0.15$.
$\begin{array}{l|lllll}
\text { Category } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Observed } & 38 & 26 & 22 & 4 & 16
\end{array}$

Compute the value of $\chi ^ { 2 }$.
A) 14.332
B) 13.277
C) 15.086
D) 12.356

Accepted Answer

The answer of Following are observed frequencies. The null hypothesis...

Question 12

An anatomy teacher hypothesizes that the final grades in her class are distributed as $10 \%$ A's, $23 \%$ B's, $45 \%$ C's, 14\% D's, and $8 \%$ F's. At the end of the semester, she has the follo grades.
$$\begin{array} { l l l l l } 
\text { A } & \text { B } & \text { C } & \text { D } & \text { F } \
6 & 14 & 22 & 8 & 4
\end{array}$$
Find the critical value. Use $\alpha = 0.05$.
A) 9.488
B) 9.236
C) 7.779
D) 11.070

Accepted Answer

The critical value for a chi-square test with 4 degrees of freedom (categories - 1) at $\alpha = 0.05$ is 9.488.

Question 13

Frequencies obtained by calculation are called expected frequencies.

Accepted Answer

Expected frequencies are calculated based on a theoretical distribution or model, rather than observed or measured directly from data.

Question 14

In a chi-square goodness-of-fit test when there is close agreement between the observed
frequency and the expected frequency, the chi-square test value will be small.

Accepted Answer

A small chi-square test value indicates that the observed frequencies are very close to the expected frequencies, suggesting a good fit between the observed data and the model or theoretical expectation.

Question 15

A biology professor claims that, on the average, $15 \%$ of her students get a grade of A, $25 \%$ get a B, $35 \%$ get a C , $10 \%$ get a D, and $15 \%$ get an F. The grades of a random sample of 119 students were recorded. The following table presents the results.
$\begin{array}{c|lllll}
\text { Grade } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Observed } & 28 & 11 & 38 & 13 & 29
\end{array}$

Compute the expected frequencies.
A) 17.85, 29.75, 41.65, 11.9, 17.85
B) 28, 11, 38, 13, 29
C) 0.235, 0.092, 0.319, 0.109, 0.244
D) 23.8, 23.8, 23.8, 23.8, 23.8

Accepted Answer

The expected frequencies are calculated by multiplying the total number of students (119) by the percentage expected for each grade: A (15%), B (25%), C (35%), D (10%), and F (15%). This gives: A = 119 * 0.15 = 17.85, B = 119 * 0.25 = 29.75, C = 119 * 0.35 = 41.65, D = 119 * 0.10 = 11.9, F = 119 * 0.15 = 17.85.

Question 16

$$\text { The null hypothesis should not be rejected for the data given in the figure below. }$$

Accepted Answer

The answer of \[\text { The null hypothesis should not...

Question 17

Following are observed frequencies. The null hypothesis is $H _ { 0 } : p _ { 1 } = 0.35$, $p _ { 2 } = 0.25 , p _ { 3 } = 0.2 , p _ { 4 } = 0.05 , p _ { 5 } = 0.15$.
$\begin{array}{l|lllll}
\text { Category } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Observed } & 38 & 31 & 24 & 23 & 4
\end{array}$

i). Compute the expected frequencies.
ii). Compute the value of $\chi ^ { 2 }$.
iii). How many degrees of freedom are there?
iv). Test the hypothesis that the distribution of the observed frequencies is as giver null hypothesis. Use the $\alpha = 0.05$ level of significance.

Accepted Answer

The answer of Following are observed frequencies. The null hypothesis...

Question 18

The critical chi-square value for 15 degrees of freedom when $\alpha = 0.05$ and the test is right-tailed is $19.675$.

Accepted Answer

The critical chi-square value for 15 degrees of freedom at $\alpha = 0.05$ for a right-tailed test is approximately 24.996.

Question 19

Following are observed frequencies. The null hypothesis is $H _ { 0 } : p _ { 1 } = 0.35 , p _ { 2 } = 0.25$, $p _ { 3 } = 0.1 , p _ { 4 } = 0.1 , p _ { 5 } = 0.2$.
$\begin{array}{l|lllll}
\text { Category } & 1 & 2 & 3 & 4 & 5 \
\hline \text { Observed } & 43 & 20 & 10 & 4 & 31
\end{array}$

Compute the expected frequencies.
A) 37.8, 27, 10.8, 10.8, 21.6
B) 0.398, 0.185, 0.093, 0.037, 0.287
C) 43, 20, 10, 4, 31
D) 21.6, 21.6, 21.6, 21.6, 21.6

Accepted Answer

To calculate the expected frequencies, multiply the total number of observations by the probability of each category under the null hypothesis. The total number of observations is $43 + 20 + 10 + 4 + 31 = 108$. Therefore, the expected frequencies are:- For category 1: $108 \times 0.35 = 37.8$- For category 2: $108 \times 0.25 = 27$- For category 3: $108 \times 0.1 = 10.8$- For category 4: $108 \times 0.1 = 10.8$- For category 5: $108 \times 0.2 = 21.6$

Question 20

Find the area to the right of 5.578 under the chi-square distribution with 11 degrees of freedom.
A) 0.9
B) 0.1
C) 1.35
D) 0.45

Accepted Answer

The area to the right of 5.578 under the chi-square distribution with 11 degrees of freedom is approximately 0.9.

Quiz 11: Other Chi-Square Tests

$\text { For four degrees of freedom, the value of } \alpha \text { shown in the } \chi ^ { 2 } \text { distribution below is } 0.05 \text {. }$ $\text { For four degrees of freedom, the value of } \alpha \text { shown in the } \chi ^ { 2 } \text { distribution below is } 0.05 \text {. }$

Before computing the test value, one should state the hypothesis.

In a goodness of fit test between one observed frequency distribution and one expected frequency distribution, the degrees of freedom are equal to the number of categories minus two.

The chi-square distribution is continuous, whereas the goodness-of-fit test is .

The formula for finding the chi-square test value is .

$\text { Find the } \alpha = 0.05 \text { critical value for the chi-square statistic with } 10 \text { degrees of freedom. }$

An assumption for the goodness-of-fit test is that the frequency for each category must be five or more.

The chi-square goodness-of-fit test is always one-tailed left.

Frequencies obtained by calculation are called expected frequencies.

In a chi-square goodness-of-fit test when there is close agreement between the observed frequency and the expected frequency, the chi-square test value will be small.

$\text { The null hypothesis should not be rejected for the data given in the figure below. }$ $\text { The null hypothesis should not be rejected for the data given in the figure below. }$

The critical chi-square value for 15 degrees of freedom when $\alpha = 0.05$ and the test is right-tailed is $19.675$ .

Find the area to the right of 5.578 under the chi-square distribution with 11 degrees of freedom.

Quiz 11: Other Chi-Square Tests

Before computing the test value, one should state the hypothesis.

In a goodness of fit test between one observed frequency distribution and one expected frequency distribution, the degrees of freedom are equal to the number of categories minus two.

The chi-square distribution is continuous, whereas the goodness-of-fit test is .

The formula for finding the chi-square test value is .

An assumption for the goodness-of-fit test is that the frequency for each category must be five or more.

The chi-square goodness-of-fit test is always one-tailed left.

Frequencies obtained by calculation are called expected frequencies.

In a chi-square goodness-of-fit test when there is close agreement between the observed frequency and the expected frequency, the chi-square test value will be small.

The null hypothesis should not be rejected for the data given in the figure below. \text { The null hypothesis should not be rejected for the data given in the figure below. } The null hypothesis should not be rejected for the data given in the figure below.

The critical chi-square value for 15 degrees of freedom when α=0.05\alpha = 0.05α=0.05 and the test is right-tailed is 19.67519.67519.675 .

Find the area to the right of 5.578 under the chi-square distribution with 11 degrees of freedom.

$\text { The null hypothesis should not be rejected for the data given in the figure below. }$ $\text { The null hypothesis should not be rejected for the data given in the figure below. }$

The critical chi-square value for 15 degrees of freedom when $\alpha = 0.05$ and the test is right-tailed is $19.675$ .