A cheerleading squad received a mean rating (out of 100 possible points) of 68 ±

Question

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

68 ± 9 ( ± )

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.

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

State whether to retain or reject the null hypothesis.

Retain the null hypothesis.Reject the null hypothesis.    

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

Explanation / Answer

PART A.
Given that,
population mean(u)=68
standard deviation, =9
sample mean, x =71
number (n)=16
null, Ho: =68
alternate, H1: >68
level of significance, = 0.05
from standard normal table,right tailed z /2 =1.645
since our test is right-tailed
reject Ho, if zo > 1.645
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 71-68/(9/sqrt(16)
zo = 1.33333
| zo | = 1.33333
critical value
the value of |z | at los 5% is 1.645
we got |zo| =1.33333 & | z | = 1.645
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : right tail - ha : ( p > 1.33333 ) = 0.09121
hence value of p0.05 < 0.09121, here we do not reject Ho
ANSWERS
---------------
null, Ho: =68
alternate, H1: >68
test statistic: 1.33333 ~ 1.33
critical value: 1.645
decision: do not reject Ho
p-value: 0.09121

PART B.
d = | x-u/sigma |
x= sample mean
u = null hypothesis population mean
sigma = null hypothesis population standard deviation
Note that Cohen’s d is derived from the population standard deviation of the null
hypothesis: it does not use the standard error of the mean (i.e., the standard deviatio
the sampling distribution of the mean).
coheren's d=|68-71/9|
=|-3/9|
=|-0.333|
=0.333

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