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# When performing a χ^2 test of independent in a contingency table with r rows and c columns, determine the upper-tail critical value of the test statistic in each of the following circumstances: a. α=0.05, r=4 rows, c=5 columns b. α=0.01, r=4 rows, c=5 columns c. α=0.01, r=4 rows, c=6 columns d. α=0.01, r=3 rows, c=6 columns e. α=0.01, r=6 rows, c=3 columns # When performing a χ^2 test of independent in a contingency table with r rows and c columns, determine the upper-tail critical value of the test statistic in each of the following circumstances: a. α=0.05, r=4 rows, c=5 columns b. α=0.01, r=4 rows, c=5 columns c. α=0.01, r=4 rows, c=6 columns d. α=0.01, r=3 rows, c=6 columns e. α=0.01, r=6 rows, c=3 columns

Question
Two-way tables asked 2021-03-02
When performing a $$\displaystyleχ^{{2}}$$ test of independent in a contingency table with r rows and c columns, determine the upper-tail critical value of the test statistic in each of the following circumstances:
a. α=0.05, r=4 rows, c=5 columns
b. α=0.01, r=4 rows, c=5 columns
c. α=0.01, r=4 rows, c=6 columns
d. α=0.01, r=3 rows, c=6 columns
e. α=0.01, r=6 rows, c=3 columns

## Answers (1) 2021-03-03
(a) Given:
r = Number of rows in table = 4
$$\displaystyle¢$$ = Number of columns in table = 5
a = Significance level = 0.05
The degrees of freedom is the product of the number ofrow and the number of columns, both decreased by 1.
df = (r—1)(c— 1) = (4-1)(5-1) = 3(4) = 12
Determine the critical value in the row with df = 12 and in the column with a = 0.05 in the chi-square distribution table in the appendix.
$$\displaystyle{x}^{{2}}={21.026}$$
(b) Given:
r = Number of rows in table = 4
$$\displaystyle¢$$ = Number of columns in table = 5
a = Significance level = 0.01
The degrees of freedom is the product of the number ofrow and the number of columns, both decreased by 1.
df = (r—1)(c— 1) = (4-1)(5-1) = 3(4) = 12
Determine the critical value in the row with df = 12 and in the column with a = 0.01 in the chi-square distribution table in the appendix.
$$\displaystyle{x}^{{2}}={26.217}$$
(c) Given:
r = Number of rows in table = 4
$$\displaystyle¢$$ = Number of columns in table = 6
a = Significance level = 0.01
The degrees of freedom is the product of the number ofrow and the number of columns, both decreased by 1.
df = (r—1)(c— 1) = (4-1)(6-1) = 3(5) = 15
Determine the critical value in the row with df = 15 and in the column with a = 0.01 in the chi-square distribution table in the appendix.
$$\displaystyle{x}^{{2}}={30.578}$$
(d) Given:
r = Number of rows in table = 3
$$\displaystyle¢$$ = Number of columns in table = 6
a = Significance level = 0.01
The degrees of freedom is the product of the number ofrow and the number of columns, both decreased by 1.
df = (r—1)(c— 1) = (3-1)(6-1) = 2(5) = 10
Determine the critical value in the row with df = 10 and in the column with a = 0.01 in the chi-square distribution table in the appendix.
$$\displaystyle{x}^{{2}}={23.209}$$
(e) Given:
r = Number of rows in table = 6
$$\displaystyle¢$$ = Number of columns in table = 3
a = Significance level = 0.01
The degrees of freedom is the product of the number ofrow and the number of columns, both decreased by 1.
df = (r—1)(c— 1) = (6-1)(3-1) = 5(2) = 10
Determine the critical value in the row with df = 10 and in the column with a = 0.01 in the chi-square distribution table in the appendix.
$$\displaystyle{x}^{{2}}={23.209}$$
a. 21.026. b. 26.217. c. 30.578. d. 23.209. e. 23.209.

### Relevant Questions asked 2021-05-05

A random sample of $$n_1 = 14$$ winter days in Denver gave a sample mean pollution index $$x_1 = 43$$.
Previous studies show that $$\sigma_1 = 19$$.
For Englewood (a suburb of Denver), a random sample of $$n_2 = 12$$ winter days gave a sample mean pollution index of $$x_2 = 37$$.
Previous studies show that $$\sigma_2 = 13$$.
Assume the pollution index is normally distributed in both Englewood and Denver.
(a) State the null and alternate hypotheses.
$$H_0:\mu_1=\mu_2.\mu_1>\mu_2$$
$$H_0:\mu_1<\mu_2.\mu_1=\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1<\mu_2$$
$$H_0:\mu_1=\mu_2.\mu_1\neq\mu_2$$
(b) What sampling distribution will you use? What assumptions are you making? NKS The Student's t. We assume that both population distributions are approximately normal with known standard deviations.
The standard normal. We assume that both population distributions are approximately normal with unknown standard deviations.
The standard normal. We assume that both population distributions are approximately normal with known standard deviations.
The Student's t. We assume that both population distributions are approximately normal with unknown standard deviations.
(c) What is the value of the sample test statistic? Compute the corresponding z or t value as appropriate.
(Test the difference $$\mu_1 - \mu_2$$. Round your answer to two decimal places.) NKS (d) Find (or estimate) the P-value. (Round your answer to four decimal places.)
(e) Based on your answers in parts (i)−(iii), will you reject or fail to reject the null hypothesis? Are the data statistically significant at level \alpha?
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are not statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we fail to reject the null hypothesis and conclude the data are statistically significant.
At the $$\alpha = 0.01$$ level, we reject the null hypothesis and conclude the data are not statistically significant.
(f) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is insufficient evidence that there is a difference in mean pollution index for Englewood and Denver.
Fail to reject the null hypothesis, there is sufficient evidence that there is a difference in mean pollution index for Englewood and Denver. (g) Find a 99% confidence interval for
$$\mu_1 - \mu_2$$.
(Round your answers to two decimal places.)
lower limit
upper limit
(h) Explain the meaning of the confidence interval in the context of the problem.
Because the interval contains only positive numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, we can not say that the mean population pollution index for Englewood is different than that of Denver.
Because the interval contains both positive and negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is greater than that of Denver.
Because the interval contains only negative numbers, this indicates that at the 99% confidence level, the mean population pollution index for Englewood is less than that of Denver. asked 2020-12-14

A chi-square value of 6.15 is calculated from data in a $$4x5$$ contingency table. Assuming $$\alpha=0.05$$, identify the critical value.
26.217
31.410
21.026
16.919 asked 2021-01-17
A new thermostat has been engineered for the frozen food cases in large supermarkets. Both the old and new thermostats hold temperatures at an average of $$25^{\circ}F$$. However, it is hoped that the new thermostat might be more dependable in the sense that it will hold temperatures closer to $$25^{\circ}F$$. One frozen food case was equipped with the new thermostat, and a random sample of 21 temperature readings gave a sample variance of 5.1. Another similar frozen food case was equipped with the old thermostat, and a random sample of 19 temperature readings gave a sample variance of 12.8. Test the claim that the population variance of the old thermostat temperature readings is larger than that for the new thermostat. Use a $$5\%$$ level of significance. How could your test conclusion relate to the question regarding the dependability of the temperature readings? (Let population 1 refer to data from the old thermostat.)
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What are the degrees of freedom?
$$df_{N} = ?$$
$$df_{D} = ?$$
What assumptions are you making about the original distribution?
The populations follow independent normal distributions. We have random samples from each population.The populations follow dependent normal distributions. We have random samples from each population.The populations follow independent normal distributions.The populations follow independent chi-square distributions. We have random samples from each population.
(c) Find or estimate the P-value of the sample test statistic. (Round your answer to four decimal places.)
(d) Based on your answers in parts (a) to (c), will you reject or fail to reject the null hypothesis?
At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we fail to reject the null hypothesis and conclude the data are statistically significant. At the ? = 0.05 level, we reject the null hypothesis and conclude the data are not statistically significant.At the ? = 0.05 level, we reject the null hypothesis and conclude the data are statistically significant.
(e) Interpret your conclusion in the context of the application.
Reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings.Fail to reject the null hypothesis, there is sufficient evidence that the population variance is larger in the old thermostat temperature readings. Fail to reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings.Reject the null hypothesis, there is insufficient evidence that the population variance is larger in the old thermostat temperature readings. asked 2021-02-25
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You select a person at random. What is the probability the person does not take vitamins regularly?

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