BNAD 277 Exam 1 Conceptual

If there are five treatments under study, the number of pairwise comparisons is _______.

10

In a two-way ANOVA test, how many null hypotheses are tested?

2 or 3

Fisher's LSD method is applied when the _________________

ANOVA test has rejected the null hypothesis of equal population means.

An important application of the chi-square distribution is a. making inferences about a single population variance b. testing for goodness of fit c. testing for the independence of two variables d. All of these alternatives are correct.

All correct

Which of the following is the correct interpretation of the Fisher's 100(1 - α)% confidence interval for μi - μj? A. If the interval includes the value zero, the null hypothesis, that H0: μi - μj = 0, is rejected for at α level of significance B.If the interval does not include the value zero, the null hypothesis, that H0: μi - μj = 0, is rejected at 100(1 - α)% level of significance C. If the interval does not include the value zero, the null hypothesis, that H0: μi - μj = 0, is rejected at α level of significance D. If the interval includes the value zero, the null hypothesis, that H0: μi - μj = 0, is rejected at 100(1 - α)% level of significance

C. If the interval does not include the value zero, the null hypothesis, that H0: μi - μj = 0, is rejected at α level of significance

A goodness-of-fit test analyzes two qualitative variables whereas a chi-square test of a contingency table is for a single qualitative variable. T/F

False

ANOVA is a statistical technique used to determine if differences exist between the means of two populations. T/F

False

For a multinomial experiment with k categories, the goodness-of-fit test statistic is assumed to follow a chi-square distribution with k degrees of freedom. T/F

False

If the underlying populations cannot be assumed to be normal, then by the central limit theorem, the sampling distribution of 𝑿 ̅𝟏 − 𝑿 ̅𝟐 is approximately normal only if the sum of the sample observations is sufficiently large—that is, when 𝒏𝟏 + 𝒏𝟐 ≥ 𝟑𝟎. T/F

False

The between-treatments variability is the estimate of σ2 which is based on the variability due to chance. T/F

False

The same formulas are used to compute the test statistic whether the hypothesized difference between population proportions is zero or not.T/F

False

The t statistic is used to estimate the difference between two population proportions. T/F

False

Two random samples are considered independent if the observations in the first sample are different from the observations of the second sample. T/F

False

For the goodness-of- fit test for normality, the null and alternative hypotheses are _____________________. A. H 0 : Data does not follow a normal distribution, H A : Data follows a normal distribution B. H 0 : Data follows a normal distribution, H A : Data does not follow a normal distribution C. H 0 : Data follows a normal distribution, H A : Data are skewed right D. H 0 : Data follows a normal distribution, H A : Data are skewed left

H 0 : Data follows a normal distribution, H A : Data does not follow a normal distribution

A one-tailed hypothesis test of the population mean has . A. Only one critical value B. Two critical values, both positive C. Two critical values, both negative D. Two critical values, one positive and one negative

Only one critical value

Which of the following factors is used to conduct hypothesis tests regarding the population variance? A. Sample mean B. Population mean C. Sample proportion D. Sample variance

Sample variance

For a chi-square goodness-of-fit test, the expected category frequencies are calculated using the sample category proportions. T/F

True

For a chi-square test of a contingency table, the degrees of freedom are calculated as (r−1)(c−1) where r and c are the number of rows and columns in the contingency table. T/F

True

For the Jarque-Bera test for normality, the test statistic is assumed to have a chi-square distribution with two degrees of freedom.T/F

True

In the case when 𝝈𝟏 𝟐 and 𝝈𝟐 𝟐 are unknown and can be assumed equal, we can calculate a pooled estimate of the variance. T/F

True

One-way ANOVA analyzes the effect of one factor on the population mean and it is based on a completely randomized design. T/F

True

The between-treatments variability is the estimate of σ 2 based on variability between the sample means. The within-treatments variability is the estimate of σ 2 which is based on the variability of the data within each sample, or the variability due to chance. T/F

True

The chi-square test statistic measures the difference between the observed frequencies and the expected frequencies assuming the null hypothesis is true. T/F

True

The difference between the two sample means 𝑿 ̅𝟏 − 𝑿 ̅𝟐 is an interval estimator of the difference between two population means 𝝁𝟏 − 𝝁𝟐 . T/F

True

The interaction test is performed before making any conclusions based on the tests for the main effects. T/F

True

Tukey's honestly significant differences method is superior to Fisher's least significant difference method to determine which population means differ because Tukey's method accounts for the number of comparisons. T/F

True

We use ANOVA to test for differences between population means by examining the amount of variability between the samples relative to the amount of variability within the samples.T/F

True

When using Fisher's least difference (LSD) method at some stated significance level α, the probability of committing a Type I error increases as the number of pairwise comparisons increases. T/F

True

One of the disadvantages of Fishers least difference (LSD) method is that the probability of committing a:

Type I error increases as the number of pairwise comparisons increases.

For the chi-square test for normality, the expected frequencies for each interval must be __________.

at least 5

For the goodness-of- fit test, the chi-square test statistic will __________________.

be at least zero

One-Way ANOVA analyzes the effect of one factor on the population mean. It is based on a:

completely randomized design

In order to determine whether or not the means of two populations are equal, a. a t test must be performed b. an analysis of variance must be performed c. either a t test or an analysis of variance can be performed d. a chi-square test must be performed

either a t test or an analysis of variance can be performed

For the goodness-of- fit test, the expected category frequencies found are the _________________________.

hypothesized proportions

In testing for the equality of k population means, the number of treatments is

k

Tukeys honestly significant differences (HSD) method ensures that the probability of a Type I error remains fixed irrespective of the number of:

pairwise comparisons

When using Fisher's least significant difference (LSD) method at some stated significance level, the probability of committing a Type I error increases as the number of:

pairwise comparisons increases

If the amount of variability between treatments is significantly greater than the amount of variability within treatments, then:

reject the null hypothesis of equal population means

Between-treatments variability is based on a weighted sum of squared differences between the:

sample means and the overall mean of the data set

The required condition for using an ANOVA procedure on data from several populations is that the

sampled populations have equal variances

Tukey's honestly significant differences (HSD) method uses instead of when compared to Fishers least differences (LSD) method for pairwise comparisons.

studentized range values; t values

If we are testing for the equality of 3 population means, we should use the a. test statistic F b. test statistic t c. test statistic z d. test statistic x^2

test statistic F

In the analysis of variance procedure (ANOVA), factor refers to a. the dependent variable b. the independent variable c. different levels of a treatment d. the critical value of F

the independent variable

One-way ANOVA analyzes the effect of one factor on the population mean and it is based on a completely randomized design. T/F

true

The chi-square test of a contingency table is a test of independence for: A. A single qualitative variable B. Two qualitative variables C. Two quantitative variables D. Three or more quantitative variables

two qualitative variables

A goodness of fit test is always conducted as a

upper-tail test

The variability due to chance, also known as within-treatments variability, is the estimate of σ2 which is based on the:

variability of the data within each sample