2. There is some evidence that REM sleep, associated with dreaming, may also pla

Question

2. There is some evidence that REM sleep, associated with dreaming, may also play a role in learning and memory processing. For example, Smith and Lapp (1991) found increased REM activity for college students during exam periods. Suppose that REM activity for a sample of 16 students during the final exam period produces an average score of 143. Regular REM activity for the college population averages 110 with a standard deviation of 50. Do the data provide evidence for increase in REM activity during exams? a. Use the five steps of hypothesis testing. Assume a p level of .05. (Be sure to include everything that we did in class). Label each step. Don’t forget the decision (e.g., reject or fail to reject the null) and conclusion (4 points) b. Now calculate the 95% and 99% confidence intervals for question 2. Then, hypothesis test with the 99% confidence intervals. Does this change your conclusion? Why or why not?

Explanation / Answer

Q1.
Given that,
population mean(u)=110
standard deviation, =50
sample mean, x =143
number (n)=16
null, Ho: =110
alternate, H1: >110
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 = 143-110/(50/sqrt(16)
zo = 2.64
| zo | = 2.64
critical value
the value of |z | at los 5% is 1.645
we got |zo| =2.64 & | z | = 1.645
make decision
hence value of | zo | > | z | and here we reject Ho
p-value : right tail - ha : ( p > 2.64 ) = 0.00415
hence value of p0.05 > 0.00415, here we reject Ho

ANSWERS
---------------
null, Ho: <110
alternate, H1: >110
test statistic: 2.64
critical value: 1.645
decision: reject Ho
p-value: 0.00415
b.
AT 95% CI.
Confidence Interval
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=143
Standard deviation( sd )=50
Sample Size(n)=16
Confidence Interval = [ 143 ± Z a/2 ( 50/ Sqrt ( 16) ) ]
= [ 143 - 1.96 * (12.5) , 143 + 1.96 * (12.5) ]
= [ 118.5,167.5 ]

AT 99% CI
CI = x ± Z a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=143
Standard deviation( sd )=50
Sample Size(n)=16
Confidence Interval = [ 143 ± Z a/2 ( 50/ Sqrt ( 16) ) ]
= [ 143 - 2.58 * (12.5) , 143 + 2.58 * (12.5) ]
= [ 110.75,175.25 ]

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