A cheerleading squad received a mean rating (out of 100 possible points) of 69 ±

Question

A cheerleading squad received a mean rating (out of 100 possible points) of 69 ± 7 ( ± ) in competitions over the previous three seasons. The same cheerleading squad performed in 16 local competitions this season with a mean rating equal to 71 in competitions. Suppose we conduct a one-independent sample z-test to determine whether mean ratings increased this season (compared to the previous three seasons) at a 0.05 level of significance.

State the value of the test statistic. (Round your answer to two decimal places.)

Compute effect size using Cohen's d. (Round your answer to two decimal places.)

Explanation / Answer

alpha=0.05 Xbar=69 sigma=7 (given)

H0: u<=69
H1:u>69
Right tailed test

Sample size=16

So, the test statistic is Z=(Xbar-u)/(sigma/sqrt(n))

= (71-69)/(7/4) =1.14

Value of test statistic=1.14

z0.05=1.645

Therefore the test statistic lies in the acceptance zone. So we accept the null hypothesis

Cohen's d is the difference between the two means divided by the standard deviation for the data.

So the effect size= (X1bar- X2 bar)/sigma

= 71-69/7

=0.285 ~ 0.29

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