four-year colleges in the United States, To whether this average has changed in
ID: 3332669 • Letter: F
Question
four-year colleges in the United States, To whether this average has changed in re d explain cha64 of today's researcher selected a sample of nyears college students and obtained an average hours.|fthe standard deviation for thedistribut=12.5 = 4.8 hours per week, does this sample ind significant change in the number of hours sn ing? Use a two-tailed test with = .05. , spent study. Childhood participation in sports, cultural groups youth groups appears to be related to im esteem for adolescents (McGee, Williams, Howden. Chapman, Martin, & Kawachi, 2006). In a repres tive study, a sample of n100 adolescents with a history of group participation is given a standardized self-esteem questionnaire. For the general popula- tion of adolescents, scores on this questionnaire form a normal distribution with a mean of -50 and a standard deviation of 15. The sample of group- participation adolescents had an average of M = 53.8. a. Does this sample provide enough evidence to co- .01. of M 12.5 ributionS itical I error? the le mean ole enta students mput- dents ages ample on clude that self-esteem scores for these adolescents are significantly different from those of the population? Use a two-tailed test with aD difference. hly o use b. Compute Cohen's d to measure the size or t ng are c. Write a sentence describing the hypothesis test and the measure of effect size outcome of the ould appear in a research report psychology department is gradually changing is 9, curriculum hyExplanation / Answer
PART A.
Given that,
population mean(u)=50
standard deviation, =15
sample mean, x =53.8
number (n)=100
null, Ho: =50
alternate, H1: !=50
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 = 53.8-50/(15/sqrt(100)
zo = 2.53333
| zo | = 2.53333
critical value
the value of |z | at los 5% is 1.96
we got |zo| =2.53333 & | 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 != 2.53333 ) = 0.0113
hence value of p0.05 > 0.0113, here we reject Ho
ANSWERS
---------------
null, Ho: =50
alternate, H1: !=50
test statistic: 2.53333
critical value: -1.96 , 1.96
decision: reject Ho
p-value: 0.0113
PART B.
Cohens d = | x - u |/ = 53.8 - 50 / 15 = 0.253
PART C.
we have enough evidence to conclude that
self esteem scores for these adolescents are significantly diffrent from general
population
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.