had to take 2 pictures.. 20. Chapter 8, Problem 18 Aa Aa Researchers have noted
ID: 2929604 • Letter: H
Question
had to take 2 pictures..
Explanation / Answer
Given that,
population mean(u)=45
standard deviation, =9
sample mean, x =50.2
number (n)=16
null, Ho: =45
alternate, H1: !=45
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
reject Ho, if zo < -1.96 OR if zo > 1.96
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 50.2-45/(9/sqrt(16)
zo = 2.31111
| zo | = 2.31111
critical value
the value of |z | at los 5% is 1.96
we got |zo| =2.31111 & | z | = 1.96
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : two tailed ( double the one tail ) - ha : ( p != 2.31111 ) = 0.02083
hence value of p0.05 > 0.02083, here we reject Ho
ANSWERS
---------------
null, Ho: =45
alternate, H1: !=45
test statistic: 2.31111
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.02083
Cohen's D = |X-U/| = |(50.2-45)/9| = 0.5777
the effective size is medium
the supplement has a significant effect on cognitive skill
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