Question

10. In ANOVA, if the null hypothesis is true, the F_obt value will be _____.

               a.            0

               b.            1

               c.            a negative number

               d.            a high number

11.       The null hypothesis for the one-way ² is _____.

a.            expected frequencies equal observed frequencies

b.            expected frequencies are equal to each other

c.            observed frequencies are equal to each other

d.            f_os ? f_es

Imagine that a general estimate of elementary-school students who have lice is 8%. The principal of Clean Kids Elementary suspects that there is a lice epidemic at her school. Indeed, the nurse examined a sample of 150 children and found that 15 of them had lice.

12.       How would you analyze these data?

13.       What are your hypotheses for this study?

14.       Analyze the data, and include results, reject or fail to reject the null hypothesis, and a statement of significance.

15.       Prepare a brief report for the principal (100 words or less).

16 Suppose you ran a chi-square test of independence on a 2??2 contingency table.

If the null hypothesis is true for this chi-square analysis, then you would expect

A. that the observed and expected frequencies in a cell would be about equal.

B. that the observed and expected frequencies in a cell would be very

different from each other.

C. the sample means for each of the populations sampled to be about equal to

each other.

D. the cell means should be about equal to each other.

17. Suppose you ran a chi-square test of independence on a 2??2 contingency table.

If the null hypothesis is true for this chi-square analysis, then you would expect

A. reject the null hypothesis and accept the alternative.

B. fail to reject the null hypothesis and not accept the alternative.

C. fail to reject H0: 1 2 3 ? ? ? A A A ????.

D. conclude the observed and expected cell frequencies differ significantly

from each other.

18. Suppose the results of a chi-square test of independence are summarized as ?2 (2,

N = 180) = 11.46, p <.05. From this report you would know that the total number

of participants involved was ____ and the null hypothesis was

A. 181; rejected.

B. 180; not rejected.

C. 179; rejected.

D. 180; rejected.

19. The Mann-Whitney U test is a nonparametric test that may be used to analyze

data from a ____ design with ____ levels of the independent variable.

A. within-subjects; three or more

B. between-subjects; two

C. between-subjects; three or more

D. within-subjects; two

20. To calculate the U statistic, you would find the

A. number of times that the rank of a score in one group precedes the rank of

a score in the other group.

B. ratio of MSA to MSError.

C. expected and observed frequencies in a contingency table.

D. smaller of the two sums of ranks of the two treatment condi

Answer 1

Solution 10:

Option B is correct.

Explanation - If the null hypothesis is true, we expect there is no variability due to treatment, hence we divide it by error variance and get 1.

Solution 11:

Option A is correct.

Explanation - Under the null hypothesis, one would assume that the observed frequencies (f_o) in a given situation will be equal to the frequencies expected by chance (f_e). It assumed that there is no difference between the observed frequency and expected frequency.

Solution 12:

This data is not able to be analyzed without a proportion equation

Solution 13:

Null Hypothesis (Ho): p = 0.08

Alternative Hypothesis (Ha): p 0.08

Solution 14:

Sample proportion, p' = x/n

Sample proportion, p' = 15/150 = 0.10

Test Statistics

Z = (0.10 - 0.08)/0.08 (1 - 0.08)/150

Z = 0.02/0.02215

Z = 0.90

Using Z-tables, the p-value is

P [Z < -0.90] + P [Z > 0.90] = 0.1841 + 0.1841 = 0.3682 (since it is two-tailed)

Since p-value is greater than 0.05 significance level, we fail to reject Ho.

Solution 15:

Since we fail to reject the null hypothesis. Therefore, we can conclude that the general estimate of elementary-school students who have lice is 8%.