Question

Individuals experience various mood states or emotions at varying times every day, both subtle changes and obvious changes, that can result in attitude alternations with both negative and positive effects and impair or facilitate performance. The connections in the brain between emotion and memory have been particularly interesting to researchers given that neural networks that are activated by mood states are linked with neural networks that process basic perceptual functions such as word processing. Dr. Siebert has conducted a number of research experiments inducing various mood states through the use of music and then studying the effects of mood on lexical processing. In this study, higher scores on the Mood State measure reflect higher levels of lexical processing. The average Mood State score is μ = 3. (Dr. Pennie S. Siebert)

5.20, 1.20, 6.10, 6.50, 3.20, 2.10, 7.10, 4.10, 5.80, 4.90   

1. Write null and alternative hypotheses (in words and notation) appropriate for this research scenario.

2. Compute a single sample t-test to test the hypotheses, use an α = .05, two-tailed.

3. Calculate Cohen's d measure of effect size and 95% confidence interval.

4. Write a results paragraph that interprets and explains the results of your tests.

Answer 1

Given that,
population mean(u)=3
sample mean, x =4.62
standard deviation, s =1.944
number (n)=10
null, Ho: μ=3
alternate, H1: μ!=3
level of significance, alpha = 0.05
from standard normal table, two tailed t alpha/2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use test statistic (t) = x-u/(s.d/sqrt(n))
to =4.62-3/(1.944/sqrt(10))
to =2.635
| to | =2.635
critical value
the value of |t alpha| with n-1 = 9 d.f is 2.262
we got |to| =2.635 & | t alpha | =2.262
make decision
hence value of | to | > | t alpha| and here we reject Ho
p-value :two tailed ( double the one tail ) - Ha : ( p != 2.6352 ) = 0.0271
hence value of p0.05 > 0.0271,here we reject Ho
ANSWERS
---------------
1.
null, Ho: μ=3
alternate, H1: μ!=3
2.
test statistic: 2.635
critical value: -2.262 , 2.262
decision: reject Ho
p-value: 0.0271
we have enough evidence to support the claim that population mean is 3.
3.
cohen's d size is d
d = sample mean - population mean /standard deviation
d = (4.62-3)/1.944
d =0.833
larger size because it is more than 0.80
4.
TRADITIONAL METHOD
given that,
sample mean, x =4.62
standard deviation, s =1.944
sample size, n =10
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.944/ sqrt ( 10) )
= 0.615
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 9 d.f is 2.262
margin of error = 2.262 * 0.615
= 1.391
III.
CI = x ± margin of error
confidence interval = [ 4.62 ± 1.391 ]
= [ 3.229 , 6.011 ]
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DIRECT METHOD
given that,
sample mean, x =4.62
standard deviation, s =1.944
sample size, n =10
level of significance, α = 0.05
from standard normal table, two tailed value of |t α/2| with n-1 = 9 d.f is 2.262
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 4.62 ± t a/2 ( 1.944/ Sqrt ( 10) ]
= [ 4.62-(2.262 * 0.615) , 4.62+(2.262 * 0.615) ]
= [ 3.229 , 6.011 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 3.229 , 6.011 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
4.
Answers:
cohen d size =0.833
95% sure that the interval [ 3.229 , 6.011 ]