Question

Q1. Although there is a popular belief that herbal remedies such as Ginkgo biloba and Ginseng may improve learning and memory in healthy adults, these effects are usually not supported by well-controlled research. In a typical study, a researcher obtains a sample of n 36 participants and has each person take the herbal supplements every day for 90 days. At the end of the 90 days, each person takes a standardized memory test. This researcher tries to demonstrate the herbal supplements can increase the memory test score. For the general population, scores from the memory test are normally distributed as X ~ N( -80, ơ-18). For the 36 participants in this study (ie, those who took the herbal supplements), they had an average of X-84, we set the Type I error rate α-0.05. (A) (B) (C) State the null hypothesis and alternative hypothesis. Calculate the p-value. Show your steps. What is your decision regarding the two hypotheses?
I need help on question #4

Answer 1

1.
Given that,
population mean(u)=80
standard deviation, sigma =18
sample mean, x =84
number (n)=36
null, Ho: μ=80
alternate, H1: μ>80
level of significance, alpha = 0.05
from standard normal table,right tailed z alpha/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 = 84-80/(18/sqrt(36)
zo = 1.333
| zo | = 1.333
critical value
the value of |z alpha| at los 5% is 1.645
we got |zo| =1.333 & | z alpha | = 1.645
make decision
hence value of |zo | < | z alpha | and here we do not reject Ho
p-value : right tail - ha : ( p > 1.333 ) = 0.091
hence value of p0.05 < 0.091, here we do not reject Ho
ANSWERS
---------------
a.
null, Ho: μ=80
alternate, H1: μ>80
test statistic: 1.333
critical value: 1.645
decision: do not reject Ho
b.
p-value: 0.091
c.
we do not have enough evidence to support the claim that herbal supplements can increase the memory test score.
conclusion:
here we are not finding the type 1 error because fail to reject the null hypothesis.

4.
Given that,
population mean(u)=40
standard deviation, sigma =12
sample mean, x =43.36
number (n)=100
null, Ho: μ=40
alternate, H1: μ>40
level of significance, alpha = 0.01
from standard normal table,right tailed z alpha/2 =2.326
since our test is right-tailed
reject Ho, if zo > 2.326
we use test statistic (z) = x-u/(s.d/sqrt(n))
zo = 43.36-40/(12/sqrt(100)
zo = 2.8
| zo | = 2.8
critical value
the value of |z alpha| at los 1% is 2.326
we got |zo| =2.8 & | z alpha | = 2.326
make decision
hence value of | zo | > | z alpha| and here we reject Ho
p-value : right tail - ha : ( p > 2.8 ) = 0.003
hence value of p0.01 > 0.003, here we reject Ho
ANSWERS
---------------
null, Ho: μ=40
alternate, H1: μ>40
test statistic: 2.8
critical value: 2.326
decision: reject Ho
p-value: 0.003
we have enough evidence to support the claim that self esteem scores for group participation adolescents are
higher than those of population.