Researcher recorded the amount of time each bird spent

Reference no: EM13489395

The researcher recorded the amount of time each bird spent in the plain chamber during a 60-minute session. Suppose the study produced a mean of M = 37 minutes in the plain chamber with SS = 228 for a sample of n = 9 birds. (Note: If the eye-spots have a no affect, then the birds should spend an average of U = 30 minutes in each chambers.)

Step 1. State the hypothesis

HO: U = 30 there will be a change

HO: U does not equal 30

Step 2. Locate the critical area

a. Is this sample sufficient influence on the birds’ behavior? Use a two-tail method

a = .05

df = n-1

df = 9-1 = 8

t= plus or minus 2.306

Step 3. Compute the test statistic

S (squared) = SS/n-1 = 228/8 = 28.5

Sm = inside the square root bracket = s2/n = 28.5/9 = 3.16

T = m-u/ sm = 37-30/3.16 = 7/3.16 = 2.21

Step 4. make a decision about the H0 and state a conclusion.

T= 2.21 is in the critical region.

T(8) = 2.21, p <.05

c. Construct the 95 % confidence interval to estimate the mean among of time spent on the plain side for the population of birds.

If I subtract estimated standard error and the t-statistic: 3.16-2.21= 0.95 which relates and says that 95 % of the birds prefer the plain side vs. the other .05 prefer the moth eye room? Can you tell me if this is correct. Or do I need to do my Estimating of Cohen’s d and compute for r2?

Reference no: EM13489395

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