When we carry out a chi-square test for independence, the null hypothesis states that the two relevant classifications
|A)||are mutually exclusive.|
|B)||form a contingency table with r rows and c columns.|
|C)||have (r –1) and (c –1) degrees of freedom where r and c are the number of rows and columns, respectively.|
|D)||are statistically independent.|
|E)||are normally distributed.|
The test statistic for the goodness of fit test for multinomial probabilities has the following degrees for freedom
|A)||k – 1|
|B)||k – 3|
|C)||(r –1)(c – 1)|
|D)||k – 1 – m|
A chi-square goodness of fit test is considered to be valid if each of the expected cell frequencies is
|A)||greater than 0.|
|B)||less than 5.|
|C)||between 0 and 5.|
|D)||at most 1.|
|E)||at least 5.|
Which of the following is not a characteristic of a multinomial experiment with k possible outcomes on each trial?
|A)||There is a fixed number of trials, n.|
|B)||The trials in the experiment are dependent.|
|C)||The probabilities of the k outcome – p1, p2, …, and pk – stay constant from trial to trial.|
|D)||The results of the experiment are observed frequencies (counts) of the number of trials that result in each of the k possible outcomes.|
You wish to test against You take a sample of size 500 and compute your test statistic to be = 8.52. At , what conclusion would you reach?
|A)||Reject H0 and accept Ha. Conclude that at least one of p1, p2, p3, and p4 exceeds 0.25.|
|B)||Reject H0 and accept Ha. Conclude that p1, p2, p3, and p4 all exceed 0.25.|
|C)||Reject H0 and accept Ha. Conclude that p1, p2, p3, and p4 are all less than 0.25.|
|D)||Do not reject H0. Conclude that p1 = p2 = p3 = p4 = 0.25.|
In a test for independence, the statistic based on a contingency table with 6 rows and 5 columns will have ____ degrees of freedom.
For a chi-square goodness of fit test, the rejection point is
|A)||in the left tail of a chi-square curve with an area of to the right of it.|
|B)||in both the left and right tails of a chi-square curve with an area of to the left and right, respectively.|
|C)||in the left tail of a chi-square curve with an area of to the left of it.|
|D)||in the right tail of a chi-square curve with an area of to the right of it.|
|E)||in the right tail of a chi-square curve with an area of to the left of it.|
The test statistic is being used to test whether the assumption of normality is reasonable for a given population distribution. The sample consists of 5000 observations and is divided into 6 categories (intervals). What are the degrees of freedom associated with the test statistic?
The actual counts in the cells of a contingency table are referred to as the expected cell frequencies.
Homogeneity is a test of the null hypothesis that all multinomial probabilities are equal.
In a contingency table, when all the expected frequencies equal the observed frequencies, the calculated statistic equals zero.
When using the chi-square goodness of fit test with multinomial probabilities, the rejection of the null hypothesis indicates that at least one of the multinomial probabilities is not equal to the value stated in the null hypothesis.
A fastener manufacturing company uses a chi-square goodness of fit test to determine if the population of all lengths of 1/4 inch bolts it manufactures is distributed according to a normal distribution. If we reject the null hypothesis, it is reasonable to assume that the population distribution is at least approximately normally distributed.
When using the chi-square goodness of fit test, the smaller the value of the chi-square test statistic, the more likely we are to reject the null hypothesis.