When we carry out a chi-square test of independence, the chi-square statistic is based on (rxc)-1...
When we carry out a chi-square test of independence, the chi-square statistic is based on (rxc)-1 degrees of freedom, where r and c denote, respectively, the number of rows and columns in the contingency table. True or false
Homework Answers
Solution :
False
When we carry out a chi-square test of independence, the chi-square statistic
is based on (r - 1) * (c - 1) degrees of freedom, where r and c denote, respectively,
the number of rows and columns in the contingency table .
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When we carry out a chi-square test of independence, the chi-square statistic is based on (rxc)-1...
10.00 points When we carry out a chi-square test of independence, the alternate hypothesis states that...
10.00 points When we carry out a chi-square test of independence, the alternate hypothesis states that the two relevant classifications O are mutually exclusive. O form a contingency table with rrows and c columns O have (1 - 1)(C-1) degrees of freedom are statistically dependent O O are normally distributed
When Chi-square distribution is used as a test of independence, the number of degrees of freedom...
When Chi-square distribution is used as a test of independence, the number of degrees of freedom is related to both the number of rows and the number of columns in the contingency table. Select one: True False Question 2 Answer saved Points out of 1.000 Flag question Question text A goodness of fit test can be used to determine if membership in categories of one variable is different as a function of membership in the categories of a second variable...
The chi-square goodness of fit test can be used when: Select one: a. We conduct a...
The chi-square goodness of fit test can be used when: Select one: a. We conduct a multinomial experiment. b. We perform a hypothesis test to determine if a population has a normal distribution. c. We perform a hypothesis test to determine if two population variances significantly differ from each other. d. We conduct a binomial experiment. The x statistic from a contingency table with 6 rows and five columns will have Select one: a. 24 degrees of freedom b. 50...
When we carry out a chi- square goodness-of-fit test for a normal distribution, the null hypothesis...
When we carry out a chi- square goodness-of-fit test for a normal distribution, the null hypothesis states that the population Does not have a normal distribution Has a normal distribution Has a chi-square distribution Does not have a chi-square distribution Has k-3 degrees of freedom
When we carry out a chi-square goodness-of-fit test for a normal distribution, the null hypothesis states...
When we carry out a chi-square goodness-of-fit test for a normal distribution, the null hypothesis states that the population: a) does not have a normal distribution. b) has a normal distribution. c) has a chi-square distribution. d) does not have a chi-square distribution. e) has k − 3 degrees of freedom.
For a chi-square test of independence for this contingency table, what is the number of degrees of freedom?
You observe 100 randomly selected college students to find out whether they arrive on time or late for their classes. The table below gives a two-way classification for these students.GenderOn TimeLateFemale359Male4313For a chi-square test of independence for this contingency table, what is the number of degrees of freedom?
When performing a chi squaredχ2 test for independence in a contingency table with r rows and...
When performing a chi squaredχ2 test for independence 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. b. alpha = 0.01, r = 5, c = 3 c. alpha = 0.01, r = 5, c = 4 d. alpha = 0.01, r = 3, c = 4 e. alpha = 0.01, r = 4, c = 5 QUESTION: what are the critical values for...
Does the data meet the conditions for the chi-square test? StatCrunch Instructions: Test of Independence Using...
Does the data meet the conditions for the chi-square test?
StatCrunch Instructions: Test of Independence Using
Technology
Next we will use StatCrunch to calculate the expected
counts:
Enter Yes and No in column var1.
Enter the observed counts as they appear in the table above
(not including the totals) into columns var2 and var3.
Rename: var1 as "911", var2 as "No Risk" and var3 as "M to S
Risk"
Choose Stat -> Tables -> Contingency -> with
summary
Select the...
For a chi-square test of independence, we calculate the degrees of freedom using which formula? A....
For a chi-square test of independence, we calculate the degrees of freedom using which formula? A. ??=???? × ??????? B. ??=???? + ??????? C. ??=(????−1)×(???????−1) D. ??=(????−1)+(???????−1)
a …… table has data arranged for the chi square independence test a) chi square b)...
a …… table has data arranged for the chi square independence test a) chi square b) contingency c) t d) z
10.00 points When we carry out a chi-square test of independence, the alternate hypothesis states that the two relevant classifications O are mutually exclusive. O form a contingency table with rrows and c columns O have (1 - 1)(C-1) degrees of freedom are statistically dependent O O are normally distributed
The chi-square goodness of fit test can be used when: Select one: a. We conduct a multinomial experiment. b. We perform a hypothesis test to determine if a population has a normal distribution. c. We perform a hypothesis test to determine if two population variances significantly differ from each other. d. We conduct a binomial experiment. The x statistic from a contingency table with 6 rows and five columns will have Select one: a. 24 degrees of freedom b. 50...
Does the data meet the conditions for the chi-square test?
StatCrunch Instructions: Test of Independence Using
Technology
Next we will use StatCrunch to calculate the expected
counts:
Enter Yes and No in column var1.
Enter the observed counts as they appear in the table above
(not including the totals) into columns var2 and var3.
Rename: var1 as "911", var2 as "No Risk" and var3 as "M to S
Risk"
Choose Stat -> Tables -> Contingency -> with
summary
Select the...
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