. Bibliotherapy is a common tool to use with children who have a hard time bondi

Question

. Bibliotherapy is a common tool to use with children who have a hard time bonding with a therapist and opening up during therapy sessions. Dr. Contos wants to examine if children who participate in bibliotherapy (M=16.48, SD=2.67) score lower on levels of anxiety (0-25) than children who participate in play therapy (M=18.11, SD=2.58). She interviews 25 children in each group for a total of 50 children. (Total 29 pts)

A. What are the independent and dependent variables and how are each scaled (1 pt each; Total 4 pts)?

Independent:________________________ Scaling:_____________________________

Dependent:__________________________ Scaling:_____________________________

B. What test should Dr. Contos use (4 pts)?__________________________________________________

C. Why that test (2 pts)?

D. What are the null and alternative hypotheses for that test (4 pts)?

E. What are that tests assumptions (3 pts)?

F. What are the df (2 pts)?________________________

G. What is the critical value for p=.05 (2 pts)? _________________________

H. If Audi calculates a test statistic of 4.76, should they (2 pts)? Retain H0 Reject H0

I. Interpret the findings (6 pts):

Explanation / Answer

Solution:-

A)

Independent- Participitation

Dependent - score on levels of anxiety

B) Test should Dr. Contos use  two-sample t-test.

C) We are using  two-sample t-test becuase each sample is less than 30 and population standard deviation is unknown.

D)

State the hypotheses. The first step is to state the null hypothesis and an alternative hypothesis.

Null hypothesis: Biblio> Play
Alternative hypothesis: Biblio < Play

Note that these hypotheses constitute a one-tailed test. The null hypothesis will be rejected if the mean difference between sample means is too small.

Formulate an analysis plan. For this analysis, the significance level is 0.05. Using sample data, we will conduct a two-sample t-test of the null hypothesis.

Analyze sample data. Using sample data, we compute the standard error (SE), degrees offreedom (DF), and the t statistic test statistic (t).

SE = sqrt[(s12/n1) + (s22/n2)]
SE = 0.743
DF = 48

t = [ (x1 - x2) - d ] / SE

t = - 2.19

tcritical = - 1.68

where s1 is the standard deviation of sample 1, s2 is the standard deviation of sample 2, n1 is thesize of sample 1, n2 is the size of sample 2, x1 is the mean of sample 1, x2 is the mean of sample 2, d is the hypothesized difference between population means, and SE is the standard error.

The observed difference in sample means produced a t statistic of - 2.19. We use the t Distribution Calculator to find P(t < - 2.19).

Therefore, the P-value in this analysis is 0.017.

Interpret results. Since the P-value (0.017) is less than the significance level (0.05), we have to reject the null hypothesis.

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