A researcher plans to use a standardized memory test to evaluate the claim that

Question

A researcher plans to use a standardized memory test to evaluate the claim that certain herbs can improve human memory. Scores on the standardized test form a normal-shaped distribution with a mean of 70 and a standard deviation of 15. The researcher obtains a sample of 25 people and has each person take an herbal supplement every day for 30 days. At the end of the 30-day period, each person takes the standardized memory test and the mean score for the sample is calculated to be 75. Using the .01 level of significance, what decision should the researcher conclude about the herbal supplement?

Explanation / Answer

Given that,
population mean(u)=70
standard deviation, =15
sample mean, x =75
number (n)=25
null, Ho: = 70
alternate, certain herbs can improve human memory H1: >70
level of significance, = 0.01
from standard normal table,right tailed z /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 = 75-70/(15/sqrt(25)
zo = 1.66667
| zo | = 1.66667
critical value
the value of |z | at los 1% is 2.326
we got |zo| =1.66667 & | z | = 2.326
make decision
hence value of |zo | < | z | and here we do not reject Ho
p-value : right tail - ha : ( p > 1.66667 ) = 0.04779
hence value of p0.01 < 0.04779, here we do not reject Ho
ANSWERS
---------------
null, Ho: <70
alternate, H1: >70
test statistic: 1.66667
critical value: 2.326
decision: do not reject Ho
p-value: 0.04779

we don't have evidence that certain herbs can improve human memory

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