Question 1

To compute a pooled sample proportion, each of the sample proportions is weighted by the size of the population from which the sample was selected.

Accepted Answer

To compute a pooled sample proportion, each of the sample proportions is weighted by the size of the sample from which it was drawn, not the size of the population.

Question 2

The degrees of freedom for in a test of independence involving a contingency table with 10 rows and 11 columns are 90.

Accepted Answer

The degrees of freedom for a contingency table is calculated as (rows - 1) * (columns - 1). For a table with 10 rows and 11 columns, it is (10-1)*(11-1) = 9*10 = 90.

Question 3

One hundred people were sampled from each of three populations and asked a question.The responses are shown in the table:
  If the three populations represented here contain the same proportion of yes responses, we would have expected to see 23.33 yes responses in each sample.

Accepted Answer

True

Question 4

The degrees of freedom for a test of independence involving a contingency table with 10 rows and 8 columns are 63.

Accepted Answer

The answer of The degrees of freedom for a test...

Question 5

In a chi-square calculation involving 6 independent terms (that is, with df = 6), there is a 5% probability that the result will be less than 1.635.

Accepted Answer

The chi-square distribution with 6 degrees of freedom has a critical value of 1.635 for a 5% significance level. This means that there is a 5% probability of obtaining a chi-square value less than 1.635 if the null hypothesis is true.

Question 6

Two hundred items were sampled from each of two recent shipments.The results are shown in the table:
  If the two shipments contain the same proportion of defective items, we would have expected to see 14 defective items in each sample.

Accepted Answer

The answer of Two hundred items were sampled from each...

Question 7

The table-based approach to testing for differences in population proportions yields more accurate results than the squared standardized normal random variable approach.

Accepted Answer

The table-based approach and the squared standardized normal random variable approach (which essentially refers to using the chi-square test for independence or goodness of fit) both have their own contexts and applications. Accuracy is not inherently better in one over the other; rather, it depends on the specific conditions and assumptions of the test being applied.

Question 8

Samples of equal size have been selected from each of three populations, producing sample proportions of 0.4, 0.3, and 0.5.To compute a pooled sample proportion, we can compute the simple average of the three sample proportions.

Accepted Answer

The pooled sample proportion is the average of the sample proportions from the different populations. In this case, the pooled sample proportion would be (0.4 + 0.3 + 0.5) / 3 = 0.4.

Question 9

Samples of equal size have been selected from each of three populations, producing sample proportions of 0.4, 0.3, and 0.5.To compute a pooled sample proportion, each of the sample proportions is weighted by size of the population from which the sample was selected.

Accepted Answer

A pooled sample proportion is computed by combining the successes and total observations from all samples, not by weighting by the population sizes. The size of the populations is irrelevant when calculating a pooled sample proportion; what matters is the total number of successes and the total sample size across all samples.

Question 10

The degrees of freedom for test of independence involving a contingency table with 12 rows and 12 columns are 144.

Accepted Answer

The degrees of freedom for a test of independence in a contingency table are calculated as (number of rows - 1) * (number of columns - 1), so for a 12x12 table, it would be (12-1)*(12-1) = 11*11 = 121.

Question 11

In a chi-square distribution, $\chi$² is the sum of squared standardized normal random variables with degrees of freedom equal to the number of independent terms included in the sum.

Accepted Answer

A chi-square distribution is indeed the distribution of a sum of the squares of k independent standard normal random variables, where k represents the degrees of freedom.

Question 12

In a chi-square test of proportion differences, we will reject the "all proportions are equal" null hypothesis if the p-value for the chi-square statistic is less than the significance level for the test.

Accepted Answer

If the p-value for the chi-square statistic is less than the chosen significance level (often 0.05), it indicates that the observed differences in proportions are statistically significant, leading to the rejection of the null hypothesis that all proportions are equal.

Question 13

Samples of equal size have been selected from each of three populations, producing sample proportions of 0.5, 0.7, and 0.6.The pooled sample proportion would be 0.65.

Accepted Answer

The pooled sample proportion is calculated by adding the number of successes across all samples and dividing by the total number of observations across all samples. Given the proportions alone without the sample sizes, we cannot determine the pooled proportion directly, but assuming equal sample sizes, the correct pooled proportion would be the average of the three proportions: (0.5 + 0.7 + 0.6) / 3 = 0.6, not 0.65.

Question 14

The degrees of freedom in a goodness of fit test for a multinomial distribution with 5 categories are 4.

Accepted Answer

In a goodness of fit test for a multinomial distribution, the degrees of freedom are calculated as the number of categories minus one, so for 5 categories, it would be 5 - 1 = 4.

Question 15

In a chi-square calculation involving 5 independent terms (that is, with df = 5),
there is a 5% probability that the result will be greater than 16.750.

Accepted Answer

For a chi-square distribution with 5 degrees of freedom, the critical value at a 5% significance level (p = 0.05) is approximately 11.070. A result greater than 16.750 would have a lower probability than 5%, indicating it falls further into the tail of the distribution.

Question 16

The degrees of freedom in a goodness of fit test for a multinomial distribution with 5 categories are (5 − 2) = 3.

Accepted Answer

The degrees of freedom in a goodness of fit test for a multinomial distribution with $k$ categories are $k - 1$. So, for 5 categories, it would be $5 - 1 = 4$.

Question 17

In a chi-square calculation involving 3 independent terms (that is, with df = 3),
there is a 5% probability that the result will be greater than 7.815.

Accepted Answer

In chi-square distributions, the critical value for a 5% probability level (p = 0.05) with 3 degrees of freedom (df = 3) is indeed 7.815. This means there is a 5% chance that the chi-square statistic will exceed this value if the null hypothesis is true.

Question 18

Below is a contingency table showing the results of a survey in which 200 voters were asked to comment on current economic conditions.
  If "view of the economy" is independent of "gender," we would have expected 30% of the male responses to be "good."

Accepted Answer

The answer of Below is a contingency table showing the...

Question 19

Four hundred people were sampled from each of two populations and asked a question.The responses are shown in the table:
  If the two populations contain the same proportion of yes responses, we would have expected to see 55 yes responses in each sample.

Accepted Answer

The answer of Four hundred people were sampled from each...

Question 20

In a chi-square test regarding the equality of 3 population proportions, we should form the null and alternative hypotheses as follows: H0: $\pi$₁ = $\pi$₂ = $\pi$₃ (All three proportions are equal.) Ha: $\pi$₁ ≠ $\pi$₂ ≠ $\pi$₃ (All three proportions are different.)

Accepted Answer

In a chi-square test for equality of proportions, the alternative hypothesis (Ha) should state that at least two of the proportions are different, not necessarily that all three must be different.

Quiz 16: Chi-Square Tests

To compute a pooled sample proportion, each of the sample proportions is weighted by the size of the population from which the sample was selected.

The degrees of freedom for in a test of independence involving a contingency table with 10 rows and 11 columns are 90.

One hundred people were sampled from each of three populations and asked a question.The responses are shown in the table: If the three populations represented here contain the same proportion of yes responses, we would have expected to see 23.33 yes responses in each sample.

The degrees of freedom for a test of independence involving a contingency table with 10 rows and 8 columns are 63.

In a chi-square calculation involving 6 independent terms (that is, with df = 6), there is a 5% probability that the result will be less than 1.635.

Two hundred items were sampled from each of two recent shipments.The results are shown in the table: If the two shipments contain the same proportion of defective items, we would have expected to see 14 defective items in each sample.

The table-based approach to testing for differences in population proportions yields more accurate results than the squared standardized normal random variable approach.

Samples of equal size have been selected from each of three populations, producing sample proportions of 0.4, 0.3, and 0.5.To compute a pooled sample proportion, we can compute the simple average of the three sample proportions.

Samples of equal size have been selected from each of three populations, producing sample proportions of 0.4, 0.3, and 0.5.To compute a pooled sample proportion, each of the sample proportions is weighted by size of the population from which the sample was selected.

The degrees of freedom for test of independence involving a contingency table with 12 rows and 12 columns are 144.

In a chi-square distribution, $\chi$ ² is the sum of squared standardized normal random variables with degrees of freedom equal to the number of independent terms included in the sum.

In a chi-square test of proportion differences, we will reject the "all proportions are equal" null hypothesis if the p-value for the chi-square statistic is less than the significance level for the test.

Samples of equal size have been selected from each of three populations, producing sample proportions of 0.5, 0.7, and 0.6.The pooled sample proportion would be 0.65.

The degrees of freedom in a goodness of fit test for a multinomial distribution with 5 categories are 4.

In a chi-square calculation involving 5 independent terms (that is, with df = 5), there is a 5% probability that the result will be greater than 16.750.

The degrees of freedom in a goodness of fit test for a multinomial distribution with 5 categories are (5 − 2) = 3.

In a chi-square calculation involving 3 independent terms (that is, with df = 3), there is a 5% probability that the result will be greater than 7.815.

Below is a contingency table showing the results of a survey in which 200 voters were asked to comment on current economic conditions. If "view of the economy" is independent of "gender," we would have expected 30% of the male responses to be "good."

Four hundred people were sampled from each of two populations and asked a question.The responses are shown in the table: If the two populations contain the same proportion of yes responses, we would have expected to see 55 yes responses in each sample.

In a chi-square test regarding the equality of 3 population proportions, we should form the null and alternative hypotheses as follows: H0: $\pi$ ₁ = $\pi$ ₂ = $\pi$ ₃ (All three proportions are equal.) Ha: $\pi$ ₁ ≠ $\pi$ ₂ ≠ $\pi$ ₃ (All three proportions are different.)

Quiz 16: Chi-Square Tests

To compute a pooled sample proportion, each of the sample proportions is weighted by the size of the population from which the sample was selected.

The degrees of freedom for in a test of independence involving a contingency table with 10 rows and 11 columns are 90.

One hundred people were sampled from each of three populations and asked a question.The responses are shown in the table: If the three populations represented here contain the same proportion of yes responses, we would have expected to see 23.33 yes responses in each sample.

The degrees of freedom for a test of independence involving a contingency table with 10 rows and 8 columns are 63.

In a chi-square calculation involving 6 independent terms (that is, with df = 6), there is a 5% probability that the result will be less than 1.635.

Two hundred items were sampled from each of two recent shipments.The results are shown in the table: If the two shipments contain the same proportion of defective items, we would have expected to see 14 defective items in each sample.

The table-based approach to testing for differences in population proportions yields more accurate results than the squared standardized normal random variable approach.

Samples of equal size have been selected from each of three populations, producing sample proportions of 0.4, 0.3, and 0.5.To compute a pooled sample proportion, we can compute the simple average of the three sample proportions.

Samples of equal size have been selected from each of three populations, producing sample proportions of 0.4, 0.3, and 0.5.To compute a pooled sample proportion, each of the sample proportions is weighted by size of the population from which the sample was selected.

The degrees of freedom for test of independence involving a contingency table with 12 rows and 12 columns are 144.

In a chi-square distribution, χ\chiχ 2 is the sum of squared standardized normal random variables with degrees of freedom equal to the number of independent terms included in the sum.

In a chi-square test of proportion differences, we will reject the "all proportions are equal" null hypothesis if the p-value for the chi-square statistic is less than the significance level for the test.

Samples of equal size have been selected from each of three populations, producing sample proportions of 0.5, 0.7, and 0.6.The pooled sample proportion would be 0.65.

The degrees of freedom in a goodness of fit test for a multinomial distribution with 5 categories are 4.

In a chi-square calculation involving 5 independent terms (that is, with df = 5), there is a 5% probability that the result will be greater than 16.750.

The degrees of freedom in a goodness of fit test for a multinomial distribution with 5 categories are (5 − 2) = 3.

In a chi-square calculation involving 3 independent terms (that is, with df = 3), there is a 5% probability that the result will be greater than 7.815.

Below is a contingency table showing the results of a survey in which 200 voters were asked to comment on current economic conditions. If "view of the economy" is independent of "gender," we would have expected 30% of the male responses to be "good."

Four hundred people were sampled from each of two populations and asked a question.The responses are shown in the table: If the two populations contain the same proportion of yes responses, we would have expected to see 55 yes responses in each sample.

In a chi-square test regarding the equality of 3 population proportions, we should form the null and alternative hypotheses as follows: H0: π\piπ 1 = π\piπ 2 = π\piπ 3 (All three proportions are equal.) Ha: π\piπ 1 ≠ π\piπ 2 ≠ π\piπ 3 (All three proportions are different.)

In a chi-square distribution, $\chi$ ² is the sum of squared standardized normal random variables with degrees of freedom equal to the number of independent terms included in the sum.

In a chi-square test regarding the equality of 3 population proportions, we should form the null and alternative hypotheses as follows: H0: $\pi$ ₁ = $\pi$ ₂ = $\pi$ ₃ (All three proportions are equal.) Ha: $\pi$ ₁ ≠ $\pi$ ₂ ≠ $\pi$ ₃ (All three proportions are different.)